#set document(title: "5.2 Solving Systems of Equations by Substitution", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 5.2#h(0.6em)Solving Systems of Equations by Substitution #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Simplify #math.equation(block: false, alt: "−5 open parenthesis 3 minus x close parenthesis")[$−5 ( 3 − x )$]. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "minus 15 plus 5 x")[$− 15 + 5 x$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Simplify #math.equation(block: false, alt: "4 minus 2 open parenthesis n plus 5 close parenthesis")[$4 − 2 ( n + 5 )$]. #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "minus 2 n minus 6")[$− 2 n − 6$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve for #math.equation(block: false, alt: "y")[$y$]: #math.equation(block: false, alt: "8 y minus 8 equals 32 minus 2 y")[$8 y − 8 = 32 − 2 y$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "y equals 4")[$y = 4$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve for #math.equation(block: false, alt: "x")[$x$]: #math.equation(block: false, alt: "3 x minus 9 y equals −3")[$3 x − 9 y = −3$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "x equals 3 y minus 1")[$x = 3 y − 1$] ] ] Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with #emph[x] and #emph[y] both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph. In this section, we will solve systems of linear equations by the substitution method. === Solve a System of Equations by Substitution We will use the same system we used first for graphing. #math.equation(block: true, alt: "{ 2 x plus y equals 7; x minus 2 y equals 6")[$\{ 2 x + y = 7 \ x − 2 y = 6$]We will first solve one of the equations for either #emph[x] or #emph[y]. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy. Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those! After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true. We’ll fill in all these steps now in . #examplebox("Example 1")[How to Solve a System of Equations by Substitution][ Solve the system by substitution. #math.equation(block: false, alt: "{ 2 x plus y equals 7; x minus 2 y equals 6")[$\{ 2 x + y = 7 \ x − 2 y = 6$] #solutionbox[ #figure(figph[This figure has three columns and six rows. The first row says, “Step 1. Solve one of the equations for either variable.” To the right of this, the middl row reads, “We’ll solve the first equation for y.” The third column shows the two equations: 2x + y = 7 and x – 2y = 6. It shows that 2x + y = 7 becomes y = 7 – 2x.], alt: "This figure has three columns and six rows. The first row says, “Step 1. Solve one of the equations for either variable.” To the right of this, the middl row reads, “We’ll solve the first equation for y.” The third column shows the two equations: 2x + y = 7 and x – 2y = 6. It shows that 2x + y = 7 becomes y = 7 – 2x.", caption: none) #figure(figph[The second row reads, “Step 2. Substitute the expression from Step 1 into the other equation.” Then, “We replace y in the second equation with the expression 7 – 2x.” It then shows the x – 2y = 6 becomes x – 2(7 – 2x) = 6.], alt: "The second row reads, “Step 2. Substitute the expression from Step 1 into the other equation.” Then, “We replace y in the second equation with the expression 7 – 2x.” It then shows the x – 2y = 6 becomes x – 2(7 – 2x) = 6.", caption: none) #figure(figph[The third row says, “Step 3: Solve the resulting equation.” Then “Now we have an equation with just 1 variable. WE know how to solve this!” It then shows that x – 2(7 – 2x) = 6 becomes x – 14 + 4x = 6 which becomes 5x = 20. Thus x = 4.], alt: "The third row says, “Step 3: Solve the resulting equation.” Then “Now we have an equation with just 1 variable. WE know how to solve this!” It then shows that x – 2(7 – 2x) = 6 becomes x – 14 + 4x = 6 which becomes 5x = 20. Thus x = 4.", caption: none) #figure(figph[The fourth row says, “Step 4. Substitute the solution in Step 3 into one of the original quaitons to find the other variable.” Then, “We’ll use the first equation and replace x with 4.” Then it shows that 2x + y = 7 becomes 2(4) + y = 7. This becomes 8 + y = 7, and thus y = −1.], alt: "The fourth row says, “Step 4. Substitute the solution in Step 3 into one of the original quaitons to find the other variable.” Then, “We’ll use the first equation and replace x with 4.” Then it shows that 2x + y = 7 becomes 2(4) + y = 7. This becomes 8 + y = 7, and thus y = −1.", caption: none) #figure(figph[The fifth row reads, “Step 5. Write the solution as an ordered pair.” Then “The ordered air is (x, y).” Then (4, −1).], alt: "The fifth row reads, “Step 5. Write the solution as an ordered pair.” Then “The ordered air is (x, y).” Then (4, −1).", caption: none) #figure(figph[The sixth row reads, “Step 6. Check that the order pair is a solution to both original equations.” Then, “Substitute (4, −1) into both equations and make sure they are both true.” It then shows that 2x + y = 7 becomxe 2(4) + −1 = 7, and thus 7 = 7. It also shows that x – 2y = 6 becomes 4 – 2(−1) = 6, and thus 6−6. It also states, “Both equations are ture. (4, −1) is the solution to the system.”], alt: "The sixth row reads, “Step 6. Check that the order pair is a solution to both original equations.” Then, “Substitute (4, −1) into both equations and make sure they are both true.” It then shows that 2x + y = 7 becomxe 2(4) + −1 = 7, and thus 7 = 7. It also shows that x – 2y = 6 becomes 4 – 2(−1) = 6, and thus 6−6. It also states, “Both equations are ture. (4, −1) is the solution to the system.”", caption: none) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ −2 x plus y equals −11; x plus 3 y equals 9")[$\{ −2 x + y = −11 \ x + 3 y = 9$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 6 , 1 close parenthesis")[$( 6 , 1 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ x plus 3 y equals 10; 4 x plus y equals 18")[$\{ x + 3 y = 10 \ 4 x + y = 18$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 4 , 2 close parenthesis")[$( 4 , 2 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solve a system of equations by substitution.] + Solve one of the equations for either variable. + Substitute the expression from Step 1 into the other equation. + Solve the resulting equation. + Substitute the solution in Step 3 into one of the original equations to find the other variable. + Write the solution as an ordered pair. + Check that the ordered pair is a solution to #strong[both] original equations. ] If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in . #examplebox("Example 2")[][ Solve the system by substitution. #math.equation(block: true, alt: "{ x plus y equals −1; y equals x plus 5")[$\{ x + y = −1 \ y = x + 5$] #solutionbox[ The second equation is already solved for #emph[y]. We will substitute the expression in place of #emph[y] in the first equation. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A system of two linear equations is displayed. The first equation is x + y = -1, and the second equation is y = x + 5. A light blue brace on the left groups the two equations together.], alt: "A system of two linear equations is displayed. The first equation is x + y = -1, and the second equation is y = x + 5. A light blue brace on the left groups the two equations together.", caption: none)]), [The second equation is already solved for #emph[y]. #linebreak() We will substitute into the first equation.], [], [Replace the #emph[y] with #emph[x] + 5.], [#figure(figph[A system of two linear equations is displayed. The first equation, y = x + 5, is circled in red, with an arrow pointing to the second equation, x + y = -1, suggesting substitution.], alt: "A system of two linear equations is displayed. The first equation, y = x + 5, is circled in red, with an arrow pointing to the second equation, x + y = -1, suggesting substitution.", caption: none)], [Solve the resulting equation for #emph[x].], [#figure(figph[A math equation showing 'X + X + 5 = -1' with the second 'X' and '5' in a reddish hue, displayed against a white background.], alt: "A math equation showing 'X + X + 5 = -1' with the second 'X' and '5' in a reddish hue, displayed against a white background.", caption: none)], [], [#figure(figph[A mathematical equation is displayed, showing '2x + 5 = -1' in a clear, digital font against a white background.], alt: "A mathematical equation is displayed, showing '2x + 5 = -1' in a clear, digital font against a white background.", caption: none)], [], [#figure(figph[A mathematical equation is displayed on a white background, reading '2x = -6' in gray text. The equation shows two times 'x' equals negative six.], alt: "A mathematical equation is displayed on a white background, reading '2x = -6' in gray text. The equation shows two times 'x' equals negative six.", caption: none)], [Substitute #emph[x] = −3 into #emph[y] = #emph[x] + 5 to find #emph[y].], [#figure(figph[An algebraic problem illustrating the substitution of x = -3 into the equation y = x + 5.], alt: "An algebraic problem illustrating the substitution of x = -3 into the equation y = x + 5.", caption: none)], [], [#figure(figph[A mathematical equation is displayed on a white background, which reads 'y = -3 + 5'. The number '3' is highlighted in red, indicating a negative value in the sum.], alt: "A mathematical equation is displayed on a white background, which reads 'y = -3 + 5'. The number '3' is highlighted in red, indicating a negative value in the sum.", caption: none)], [The ordered pair is (−3, 2).], [#figure(figph[The equation y=2 is displayed on a white background, representing a horizontal line in a Cartesian coordinate system where the y-coordinate is consistently 2.], alt: "The equation y=2 is displayed on a white background, representing a horizontal line in a Cartesian coordinate system where the y-coordinate is consistently 2.", caption: none)], [Check the ordered pair in both equations: #linebreak() #linebreak() #math.equation(block: false, alt: "x plus y, equals, minus 1; minus 3 plus 2, equals ?, minus 1; minus 1, equals, minus 1 ✓, y, equals, x plus 5; 2, equals ?, minus 3 plus 5; 2, equals, 2 ✓")[$x + y & = & − 1 \ − 3 + 2 & limits(=)^(?) & − 1 \ − 1 & = & − 1 #h(0.2em) ✓ & & & y & = & x + 5 \ 2 & limits(=)^(?) & − 3 + 5 \ 2 & = & 2 #h(0.2em) ✓$]], [], [], [The solution is (−3, 2).], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ x plus y equals 6; y equals 3 x minus 2")[$\{ x + y = 6 \ y = 3 x − 2$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 , 4 close parenthesis")[$( 2 , 4 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ 2 x minus y equals 1; y equals −3 x minus 6")[$\{ 2 x − y = 1 \ y = −3 x − 6$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −1 , −3 close parenthesis")[$( −1 , −3 )$] ] ] If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for #emph[y]. #examplebox("Example 3")[][ Solve the system by substitution. #math.equation(block: false, alt: "{ 3 x plus y equals 5; 2 x plus 4 y equals −10")[$\{ 3 x + y = 5 \ 2 x + 4 y = −10$] #solutionbox[ We need to solve one equation for one variable. Then we will substitute that expression into the other equation. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([Solve for #emph[y]. #linebreak() #linebreak() Substitute into the other equation.], [#figure(figph[An image illustrating the substitution method for solving a system of linear equations. The first equation, 3x + y = 5, is rewritten as y = -3x + 5. This expression for y is then substituted into the second equation, 2x + 4y = -10, as indicated by the red arrow.], alt: "An image illustrating the substitution method for solving a system of linear equations. The first equation, 3x + y = 5, is rewritten as y = -3x + 5. This expression for y is then substituted into the second equation, 2x + 4y = -10, as indicated by the red arrow.", caption: none)]), [Replace the #emph[y] with −3#emph[x] + 5.], [#figure(figph[A mathematical equation is displayed against a white background: 2x + 4(-3x + 5) = -10. The numbers and variables are in a dark gray font, while the '-3x + 5' inside the parentheses is highlighted in red.], alt: "A mathematical equation is displayed against a white background: 2x + 4(-3x + 5) = -10. The numbers and variables are in a dark gray font, while the '-3x + 5' inside the parentheses is highlighted in red.", caption: none)], [Solve the resulting equation for #emph[x].], [#figure(figph[A mathematical equation is displayed on a white background: 2x - 12x + 20 = -10. The equation involves a variable 'x' and numerical constants, with subtraction, addition, and equality operations.], alt: "A mathematical equation is displayed on a white background: 2x - 12x + 20 = -10. The equation involves a variable 'x' and numerical constants, with subtraction, addition, and equality operations.", caption: none)], [], [#figure(figph[A mathematical equation is displayed with the expression '-10x + 20 = -10' in dark gray text against a plain white background.], alt: "A mathematical equation is displayed with the expression '-10x + 20 = -10' in dark gray text against a plain white background.", caption: none) #linebreak() #figure(figph[A mathematical equation displays '-10x = -30' against a white background.], alt: "A mathematical equation displays '-10x = -30' against a white background.", caption: none)], [Substitute #emph[x] = 3 into 3#emph[x] + #emph[y] = 5 to find #emph[y].], [#figure(figph[An image displays the algebraic expression 'x = 3' circled in red, with an arrow pointing from it to the equation '3x + y = 5,' indicating the substitution of the value of x into the equation.], alt: "An image displays the algebraic expression 'x = 3' circled in red, with an arrow pointing from it to the equation '3x + y = 5,' indicating the substitution of the value of x into the equation.", caption: none)], [], [#figure(figph[A mathematical equation is displayed on a white background, reading '3(3) + y = 5'. The first '3' is black, the second '3' inside the parentheses is red, and the rest of the equation is black. This simplifies to 9 + y = 5, so y = -4.], alt: "A mathematical equation is displayed on a white background, reading '3(3) + y = 5'. The first '3' is black, the second '3' inside the parentheses is red, and the rest of the equation is black. This simplifies to 9 + y = 5, so y = -4.", caption: none) #linebreak() #figure(figph[A mathematical equation is displayed on a white background: 9 + y = 5.], alt: "A mathematical equation is displayed on a white background: 9 + y = 5.", caption: none)], [The ordered pair is (3, −4).], [#figure(figph[The equation y = -4 is displayed in black text on a white background.], alt: "The equation y = -4 is displayed in black text on a white background.", caption: none)], [Check the ordered pair in both equations: #linebreak() #linebreak() #math.equation(block: false, alt: "3 x plus y, equals, 5; 3 times 3 plus open parenthesis minus 4 close parenthesis, equals ?, 5; 9 minus 4, equals ?, 5; 5, equals, 5 ✓, 2 x plus 4 y, equals, minus 10; 2 times 3 plus 4 open parenthesis minus 4 close parenthesis, equals, minus 10; 6 minus 16, equals ?, minus 10; minus 10, equals, minus 10 ✓")[$3 x + y & = & 5 \ 3 · 3 + ( − 4 ) & limits(=)^(?) & 5 \ 9 − 4 & limits(=)^(?) & 5 \ 5 & = & 5 #h(0.2em) ✓ & & & 2 x + 4 y & = & − 10 \ 2 · 3 + 4 ( − 4 ) & = & − 10 \ 6 − 16 & limits(=)^(?) & − 10 \ − 10 & = & − 10 #h(0.2em) ✓$]], [], [], [The solution is (3, −4).], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ 4 x plus y equals 2; 3 x plus 2 y equals −1")[$\{ 4 x + y = 2 \ 3 x + 2 y = −1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 1 , −2 close parenthesis")[$( 1 , −2 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ minus x plus y equals 4; 4 x minus y equals 2")[$\{ − x + y = 4 \ 4 x − y = 2$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 , 6 close parenthesis")[$( 2 , 6 )$] ] ] In it was easiest to solve for #emph[y] in the first equation because it had a coefficient of 1. In it will be easier to solve for #emph[x]. #examplebox("Example 4")[][ Solve the system by substitution. #math.equation(block: false, alt: "{ x minus 2 y equals −2; 3 x plus 2 y equals 34")[$\{ x − 2 y = −2 \ 3 x + 2 y = 34$] #solutionbox[ We will solve the first equation for #math.equation(block: false, alt: "x")[$x$] and then substitute the expression into the second equation. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A linear equation in two variables, x - 2y = -2, is displayed in black text on a white background.], alt: "A linear equation in two variables, x - 2y = -2, is displayed in black text on a white background.", caption: none)]), [Solve for #emph[x]. #linebreak() #linebreak() Substitute into the other equation.], [#figure(figph[This image illustrates the substitution method for solving a system of equations. The expression '2y - 2' from the equation 'x = 2y - 2' is shown being substituted into the equation '3x + 2y = 34'.], alt: "This image illustrates the substitution method for solving a system of equations. The expression '2y - 2' from the equation 'x = 2y - 2' is shown being substituted into the equation '3x + 2y = 34'.", caption: none)], [Replace the #emph[x] with 2#emph[y] − 2.], [#figure(figph[A mathematical equation is displayed against a white background: 3(2y - 2) + 2y = 34. The term '2y - 2' inside the parentheses is highlighted in red.], alt: "A mathematical equation is displayed against a white background: 3(2y - 2) + 2y = 34. The term '2y - 2' inside the parentheses is highlighted in red.", caption: none)], [Solve the resulting equation for #emph[y].], [#figure(figph[A mathematical equation displayed on a white background, which reads '6y - 6 + 2y = 34'.], alt: "A mathematical equation displayed on a white background, which reads '6y - 6 + 2y = 34'.", caption: none)], [#linebreak() Substitute #emph[y] = 5 into #emph[x] − 2#emph[y] = −2 to find #emph[x].], [#figure(figph[A mathematical equation, 8y - 6 = 34, is displayed on a white background.], alt: "A mathematical equation, 8y - 6 = 34, is displayed on a white background.", caption: none) #linebreak() #figure(figph[A mathematical equation is displayed on a white background, which reads '8y = 40' in black text.], alt: "A mathematical equation is displayed on a white background, which reads '8y = 40' in black text.", caption: none) #linebreak() #figure(figph[An image displays two mathematical expressions. The top expression, circled in red, is 'y = 5'. An arrow extends from this circled expression downwards to the bottom expression, which is 'x - 2y = -2'.], alt: "An image displays two mathematical expressions. The top expression, circled in red, is 'y = 5'. An arrow extends from this circled expression downwards to the bottom expression, which is 'x - 2y = -2'.", caption: none) #linebreak() #figure(figph[A mathematical equation is displayed, showing 'x - 2 \* 5 = -2'. The multiplication symbol and the number 5 are highlighted in red, suggesting the operation to be performed first.], alt: "A mathematical equation is displayed, showing 'x - 2 * 5 = -2'. The multiplication symbol and the number 5 are highlighted in red, suggesting the operation to be performed first.", caption: none) #linebreak() #figure(figph[A mathematical equation is displayed with a white background showing 'x - 10 = -2' in black characters. The equation represents a simple algebraic problem.], alt: "A mathematical equation is displayed with a white background showing 'x - 10 = -2' in black characters. The equation represents a simple algebraic problem.", caption: none) #linebreak() #figure(figph[The image displays a simple equation, 'X = 8', in a bold, sans-serif font centered on a plain white background.], alt: "The image displays a simple equation, 'X = 8', in a bold, sans-serif font centered on a plain white background.", caption: none)], [The ordered pair is (8, 5).], [], [Check the ordered pair in both equations: #linebreak() #linebreak() #math.equation(block: false, alt: "x minus 2 y, equals, minus 2; 8 minus 2 times 5, equals ?, minus 2; 8 minus 10, equals ?, minus 2; minus 2, equals, minus 2 ✓, 3 x plus 2 y, equals, 34; 3 times 8 plus 2 times 5, equals ?, 34; 24 plus 10, equals ?, 34; 34, equals, 34 ✓")[$x − 2 y & = & − 2 \ 8 − 2 · 5 & limits(=)^(?) & − 2 \ 8 − 10 & limits(=)^(?) & − 2 \ − 2 & = & − 2 #h(0.2em) ✓ & & & 3 x + 2 y & = & 34 \ 3 · 8 + 2 · 5 & limits(=)^(?) & 34 \ 24 + 10 & limits(=)^(?) & 34 \ 34 & = & 34 #h(0.2em) ✓$]], [], [], [The solution is (8, 5).], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ x minus 5 y equals 13; 4 x minus 3 y equals 1")[$\{ x − 5 y = 13 \ 4 x − 3 y = 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −2 , −3 close parenthesis")[$( −2 , −3 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ x minus 6 y equals −6; 2 x minus 4 y equals 4")[$\{ x − 6 y = −6 \ 2 x − 4 y = 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 6 , 2 close parenthesis")[$( 6 , 2 )$] ] ] When both equations are already solved for the same variable, it is easy to substitute! #examplebox("Example 5")[][ Solve the system by substitution. #math.equation(block: false, alt: "{ y equals −2 x plus 5; y equals the fraction 1 over 2 x")[$\{ y = −2 x + 5 \ y = frac(1, 2) x$] #solutionbox[ Since both equations are solved for #emph[y], we can substitute one into the other. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Substitute #math.equation(block: false, alt: "the fraction 1 over 2 x")[$frac(1, 2) x$] for #emph[y] in the first equation.], [#figure(figph[Equations y = (1/2)x and y = -2x + 5 are displayed. An arrow highlights the relationship between their slopes, 1/2 and -2, demonstrating they are negative reciprocals for perpendicular lines.], alt: "Equations y = (1/2)x and y = -2x + 5 are displayed. An arrow highlights the relationship between their slopes, 1/2 and -2, demonstrating they are negative reciprocals for perpendicular lines.", caption: none)]), [Replace the #emph[y] with #math.equation(block: false, alt: "the fraction 1 over 2 x .")[$frac(1, 2) x .$]], [#figure(figph[A mathematical equation is displayed: one-half x equals negative two x plus five (1/2x = -2x + 5).], alt: "A mathematical equation is displayed: one-half x equals negative two x plus five (1/2x = -2x + 5).", caption: none)], [Solve the resulting equation. Start #linebreak() by clearing the fraction.], [#figure(figph[A mathematical equation is displayed on a white background: 2(1/2x) = 2(-2x + 5). The numbers and variables are in black font.], alt: "A mathematical equation is displayed on a white background: 2(1/2x) = 2(-2x + 5). The numbers and variables are in black font.", caption: none)], [Solve for #emph[x].], [#figure(figph[A mathematical equation is displayed, showing 'x = -4x + 10' in bold, black text on a white background, representing an algebraic problem to solve for the value of x.], alt: "A mathematical equation is displayed, showing 'x = -4x + 10' in bold, black text on a white background, representing an algebraic problem to solve for the value of x.", caption: none)], [], [#figure(figph[A simple algebraic equation, 5x = 10, is displayed in black text on a white background.], alt: "A simple algebraic equation, 5x = 10, is displayed in black text on a white background.", caption: none)], [Substitute #emph[x] = 2 into #emph[y] = #math.equation(block: false, alt: "the fraction 1 over 2 x")[$frac(1, 2) x$] to find #emph[y].], [#figure(figph[A diagram illustrating the substitution of x=2 into the equation y = (1/2)x.], alt: "A diagram illustrating the substitution of x=2 into the equation y = (1/2)x.", caption: none) #linebreak() #figure(figph[A mathematical equation shows 'y = 1/2 ', with a multiplication dot followed by the number '2' highlighted in red.], alt: "A mathematical equation shows 'y = 1/2 ', with a multiplication dot followed by the number '2' highlighted in red.", caption: none) #linebreak() #figure(figph[The image shows a mathematical equation 'y=1' rendered in gray text against a plain white background, centrally positioned within the frame.], alt: "The image shows a mathematical equation 'y=1' rendered in gray text against a plain white background, centrally positioned within the frame.", caption: none)], [The ordered pair is (2,1).], [], [Check the ordered pair in both equations: #linebreak() #linebreak() #math.equation(block: false, alt: "y, equals, the fraction 1 over 2 x; 1, equals ?, the fraction 1 over 2 times 2; 1, equals, 1 ✓, y, equals, minus 2 x plus 5; 1, equals ?, minus 2 times 2 plus 5; 1, equals, minus 4 plus 5; 1, equals, 1 ✓")[$y & = & frac(1, 2) x \ 1 & limits(=)^(?) & frac(1, 2) · 2 \ 1 & = & 1 #h(0.2em) ✓ & & & y & = & − 2 x + 5 \ 1 & limits(=)^(?) & − 2 · 2 + 5 \ 1 & = & − 4 + 5 \ 1 & = & 1 #h(0.2em) ✓$]], [], [], [The solution is (2,1).], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ y equals 3 x minus 16; y equals the fraction 1 over 3 x")[$\{ y = 3 x − 16 \ y = frac(1, 3) x$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 6 , 2 close parenthesis")[$( 6 , 2 )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ y equals − x plus 10; y equals the fraction 1 over 4 x")[$\{ y = "−" x + 10 \ y = frac(1, 4) x$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 8 , 2 close parenthesis")[$( 8 , 2 )$] ] ] Be very careful with the signs in the next example. #examplebox("Example 6")[][ Solve the system by substitution. #math.equation(block: false, alt: "{ 4 x plus 2 y equals 4; 6 x minus y equals 8")[$\{ 4 x + 2 y = 4 \ 6 x − y = 8$] #solutionbox[ We need to solve one equation for one variable. We will solve the first equation for #emph[y]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical equation is displayed, showing '4x + 2y = 4' in black text on a white background.], alt: "A mathematical equation is displayed, showing '4x + 2y = 4' in black text on a white background.", caption: none)]), [Solve the first equation for #emph[y].], [#figure(figph[A clear image of the linear equation 2y = -4x + 4, presented on a plain white background.], alt: "A clear image of the linear equation 2y = -4x + 4, presented on a plain white background.", caption: none)], [Substitute −2#emph[x] + 2 for #emph[y] in the second equation.], [#figure(figph[Illustration of solving a system of equations by substituting y = -2x + 2 into 6x - y = 8.], alt: "Illustration of solving a system of equations by substituting y = -2x + 2 into 6x - y = 8.", caption: none)], [Replace the #emph[y] with −2#emph[x] + 2.], [#figure(figph[A mathematical equation is displayed with the expression 6x - (-2x + 2) = 8, where the terms -2x and +2 are highlighted in red within the parentheses.], alt: "A mathematical equation is displayed with the expression 6x - (-2x + 2) = 8, where the terms -2x and +2 are highlighted in red within the parentheses.", caption: none)], [Solve the equation for #emph[x].], [#figure(figph[The image shows the mathematical equation 6x + 2x - 2 = 8.], alt: "The image shows the mathematical equation 6x + 2x - 2 = 8.", caption: none)], [], [#figure(figph[A mathematical equation is displayed on a white background, which reads '8x - 2 = 8'. The equation is presented clearly and centrally.], alt: "A mathematical equation is displayed on a white background, which reads '8x - 2 = 8'. The equation is presented clearly and centrally.", caption: none) #linebreak() #figure(figph[A simple mathematical equation '8x = 10' is displayed in black text on a plain white background.], alt: "A simple mathematical equation '8x = 10' is displayed in black text on a plain white background.", caption: none)], [#linebreak() #linebreak() Substitute #math.equation(block: false, alt: "x equals the fraction 5 over 4")[$x = frac(5, 4)$] into 4#emph[x] + 2#emph[y] = 4 to find #emph[y].], [#figure(figph[An equation 4x + 2y = 4 is shown, with an arrow pointing from the expression x = 5/4, which is circled in red, to the equation. This indicates substitution or a given value for x.], alt: "An equation 4x + 2y = 4 is shown, with an arrow pointing from the expression x = 5/4, which is circled in red, to the equation. This indicates substitution or a given value for x.", caption: none) #linebreak() #figure(figph[A mathematical equation is displayed against a white background: 4(5/4) + 2y = 4. The numerator and denominator of the fraction, 5 and 4 respectively, are highlighted in red.], alt: "A mathematical equation is displayed against a white background: 4(5/4) + 2y = 4. The numerator and denominator of the fraction, 5 and 4 respectively, are highlighted in red.", caption: none) #linebreak() #figure(figph[A mathematical equation is displayed on a white background, which reads '5 + 2y = 4'. The equation features the number 5, a plus sign, the number 2, the variable 'y', an equals sign, and the number 4.], alt: "A mathematical equation is displayed on a white background, which reads '5 + 2y = 4'. The equation features the number 5, a plus sign, the number 2, the variable 'y', an equals sign, and the number 4.", caption: none) #linebreak() #figure(figph[A simple algebraic equation, 2y = -1, is displayed on a white background, representing a basic mathematical problem to solve for the variable 'y'.], alt: "A simple algebraic equation, 2y = -1, is displayed on a white background, representing a basic mathematical problem to solve for the variable 'y'.", caption: none) #linebreak() #figure(figph[The equation y = -1/2 is displayed on a white background, representing a horizontal line at y equals negative one-half.], alt: "The equation y = -1/2 is displayed on a white background, representing a horizontal line at y equals negative one-half.", caption: none)], [The ordered pair is #math.equation(block: false, alt: "open parenthesis the fraction 5 over 4 , minus the fraction 1 over 2 close parenthesis .")[$( frac(5, 4) , − frac(1, 2) ) .$]], [], [Check the ordered pair in both equations. #linebreak() #linebreak() #math.equation(block: false, alt: "4 x plus 2 y, equals, 4; 4 open parenthesis the fraction 5 over 4 close parenthesis plus 2 open parenthesis minus the fraction 1 over 2 close parenthesis, equals ?, 4; 5 minus 1, equals ?, 4; 4, equals, 4 ✓, 6 x minus y, equals, 8; 6 open parenthesis the fraction 5 over 4 close parenthesis minus open parenthesis minus the fraction 1 over 2 close parenthesis, equals ?, 8; the fraction 15 over 4 minus open parenthesis minus the fraction 1 over 2 close parenthesis, equals ?, 8; the fraction 16 over 2, equals ?, 8; 8, equals, 8 ✓")[$4 x + 2 y & = & 4 \ 4 ( frac(5, 4) ) + 2 ( − frac(1, 2) ) & limits(=)^(?) & 4 \ 5 − 1 & limits(=)^(?) & 4 \ 4 & = & 4 #h(0.2em) ✓ & & & 6 x − y & = & 8 \ 6 ( frac(5, 4) ) − ( − frac(1, 2) ) & limits(=)^(?) & 8 \ frac(15, 4) − ( − frac(1, 2) ) & limits(=)^(?) & 8 \ frac(16, 2) & limits(=)^(?) & 8 \ 8 & = & 8 #h(0.2em) ✓$]], [], [], [The solution is #math.equation(block: false, alt: "open parenthesis the fraction 5 over 4 , minus the fraction 1 over 2 close parenthesis .")[$( frac(5, 4) , − frac(1, 2) ) .$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ x minus 4 y equals −4; −3 x plus 4 y equals 0")[$\{ x − 4 y = −4 \ −3 x + 4 y = 0$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 2 , the fraction 3 over 2 close parenthesis")[$( 2 , frac(3, 2) )$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ 4 x minus y equals 0; 2 x minus 3 y equals 5")[$\{ 4 x − y = 0 \ 2 x − 3 y = 5$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis minus the fraction 1 over 2 , −2 close parenthesis")[$( − frac(1, 2) , −2 )$] ] ] In , it will take a little more work to solve one equation for #emph[x] or #emph[y]. #examplebox("Example 7")[][ Solve the system by substitution. #math.equation(block: false, alt: "{ 4 x minus 3 y equals 6; 15 y minus 20 x equals −30")[$\{ 4 x − 3 y = 6 \ 15 y − 20 x = −30$] #solutionbox[ We need to solve one equation for one variable. We will solve the first equation for #emph[x]. #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#figure(figph[A mathematical equation is displayed on a white background, which reads 4x - 3y = 6.], alt: "A mathematical equation is displayed on a white background, which reads 4x - 3y = 6.", caption: none)]), [Solve the first equation for #emph[x].], [#figure(figph[A mathematical equation is displayed, showing '4x = 3y + 6' in a clear, digital font on a white background. This linear equation involves two variables, x and y, and constant terms.], alt: "A mathematical equation is displayed, showing '4x = 3y + 6' in a clear, digital font on a white background. This linear equation involves two variables, x and y, and constant terms.", caption: none)], [Substitute #math.equation(block: false, alt: "the fraction 3 over 4 y plus the fraction 3 over 2")[$frac(3, 4) y + frac(3, 2)$] for #emph[x] in the second equation.], [#figure(figph[Two mathematical equations are displayed: x = (3/4)y + 3/2, highlighted in a red oval, with an arrow pointing to the second equation, 15y - 20x = -30, indicating a relationship or substitution.], alt: "Two mathematical equations are displayed: x = (3/4)y + 3/2, highlighted in a red oval, with an arrow pointing to the second equation, 15y - 20x = -30, indicating a relationship or substitution.", caption: none)], [Replace the #emph[x] with #math.equation(block: false, alt: "the fraction 3 over 4 y plus the fraction 3 over 2 .")[$frac(3, 4) y + frac(3, 2) .$]], [#figure(figph[A mathematical equation is displayed: 15y - 20(3/4y + 3/2) = -30. The fractions 3/4 and 3/2 are highlighted in red.], alt: "A mathematical equation is displayed: 15y - 20(3/4y + 3/2) = -30. The fractions 3/4 and 3/2 are highlighted in red.", caption: none)], [Solve for #emph[y].], [#figure(figph[An algebraic equation, '15y - 15y - 30 = -30,' is presented on a white background, which simplifies to -30 = -30, indicating that the equation holds true for any value of 'y' and has infinitely many solutions.], alt: "An algebraic equation, '15y - 15y - 30 = -30,' is presented on a white background, which simplifies to -30 = -30, indicating that the equation holds true for any value of 'y' and has infinitely many solutions.", caption: none)], [], [#figure(figph[The mathematical equation '0 - 30 = -30' is displayed in a simple, clear font on a white background, demonstrating a basic subtraction problem resulting in a negative number.], alt: "The mathematical equation '0 - 30 = -30' is displayed in a simple, clear font on a white background, demonstrating a basic subtraction problem resulting in a negative number.", caption: none)], [], [#figure(figph[The mathematical equation 0 = 0.], alt: "The mathematical equation 0 = 0.", caption: none)], )) Since 0 = 0 is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ 2 x minus 3 y equals 12; −12 y plus 8 x equals 48")[$\{ 2 x − 3 y = 12 \ −12 y + 8 x = 48$] #solutionbox[ infinitely many solutions ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ 5 x plus 2 y equals 12; −4 y minus 10 x equals −24")[$\{ 5 x + 2 y = 12 \ −4 y − 10 x = −24$] #solutionbox[ infinitely many solutions ] ] Look back at the equations in . Is there any way to recognize that they are the same line? Let’s see what happens in the next example. #examplebox("Example 8")[][ Solve the system by substitution. #math.equation(block: false, alt: "{ 5 x minus 2 y equals −10; y equals the fraction 5 over 2 x")[$\{ 5 x − 2 y = −10 \ y = frac(5, 2) x$] #solutionbox[ The second equation is already solved for #emph[y], so we can substitute for #emph[y] in the first equation. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Substitute #emph[x] for #emph[y] in the first equation.], [#figure(figph[An arrow points from the circled equation y = (5/2)x to 5x - 2y = -10, illustrating a transformation between two linear equation forms.], alt: "An arrow points from the circled equation y = (5/2)x to 5x - 2y = -10, illustrating a transformation between two linear equation forms.", caption: none)]), [Replace the #emph[y] with #math.equation(block: false, alt: "the fraction 5 over 2 x .")[$frac(5, 2) x .$]], [#figure(figph[A mathematical equation is displayed, reading '5x minus 2 multiplied by (5 over 2 multiplied by x) equals -10'. The fraction '5 over 2' and the 'x' immediately following it are colored red.], alt: "A mathematical equation is displayed, reading '5x minus 2 multiplied by (5 over 2 multiplied by x) equals -10'. The fraction '5 over 2' and the 'x' immediately following it are colored red.", caption: none)], [Solve for #emph[x].], [#figure(figph[A mathematical equation displays '5x - 5x = -10', which simplifies to '0 = -10', a contradiction indicating that there is no solution for x in this equation.], alt: "A mathematical equation displays '5x - 5x = -10', which simplifies to '0 = -10', a contradiction indicating that there is no solution for x in this equation.", caption: none)], [], [#figure(figph[Mathematical expression showing 0 does not equal -10.], alt: "Mathematical expression showing 0 does not equal -10.", caption: none)], )) Since 0 = −10 is a false statement the equations are inconsistent. The graphs of the two equation would be parallel lines. The system has no solutions. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ 3 x plus 2 y equals 9; y equals minus the fraction 3 over 2 x plus 1")[$\{ 3 x + 2 y = 9 \ y = − frac(3, 2) x + 1$] #solutionbox[ no solution ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the system by substitution. #math.equation(block: false, alt: "{ 5 x minus 3 y equals 2; y equals the fraction 5 over 3 x minus 4")[$\{ 5 x − 3 y = 2 \ y = frac(5, 3) x − 4$] #solutionbox[ no solution ] ] === Solve Applications of Systems of Equations by Substitution We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[How to use a problem solving strategy for systems of linear equations.] + #strong[Read] the problem. Make sure all the words and ideas are understood. + #strong[Identify] what we are looking for. + #strong[Name] what we are looking for. Choose variables to represent those quantities. + #strong[Translate] into a system of equations. + #strong[Solve] the system of equations using good algebra techniques. + #strong[Check] the answer in the problem and make sure it makes sense. + #strong[Answer] the question with a complete sentence. ] Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable? #examplebox("Example 9")[][ The sum of two numbers is zero. One number is nine less than the other. Find the numbers. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Step 1. Read] the problem.], []), [#strong[Step 2. Identify] what we are looking for.], [We are looking for two numbers.], [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "n equals")[$n =$] the first number #linebreak() Let #math.equation(block: false, alt: "m equals")[$m =$] the second number], [#strong[Step 4. Translate] into a system of equations.], [The sum of two numbers is zero.], [], [#figure(figph[A simple algebraic equation is displayed, showing 'n + m = 0' in black text against a plain white background.], alt: "A simple algebraic equation is displayed, showing 'n + m = 0' in black text against a plain white background.", caption: none)], [], [One number is nine less than the other.], [], [#figure(figph[A mathematical equation displays 'n = m - 9' in black text against a white background.], alt: "A mathematical equation displays 'n = m - 9' in black text against a white background.", caption: none)], [The system is:], [#figure(figph[A system of two linear equations is presented, with a light blue brace indicating they are grouped. The equations are: n + m = 0 and n = m - 9. The text is clear on a white background.], alt: "A system of two linear equations is presented, with a light blue brace indicating they are grouped. The equations are: n + m = 0 and n = m - 9. The text is clear on a white background.", caption: none)], [#strong[Step 5. Solve] the system of #linebreak() equations. We will use substitution #linebreak() since the second equation is solved #linebreak() for #emph[n].], [], [Substitute #emph[m] − 9 for #emph[n] in the first equation.], [#figure(figph[Two equations are displayed: n = m - 9 and n + m = 0. The expression 'm - 9' in the first equation is circled in red, and a red arrow points from it to the second equation.], alt: "Two equations are displayed: n = m - 9 and n + m = 0. The expression 'm - 9' in the first equation is circled in red, and a red arrow points from it to the second equation.", caption: none)], [Solve for #emph[m].], [#figure(figph[A mathematical equation is displayed on a white background, reading 'm - 9 + m = 0'. The 'm' and '- 9' are in a reddish-brown color, while the rest of the equation is in black.], alt: "A mathematical equation is displayed on a white background, reading 'm - 9 + m = 0'. The 'm' and '- 9' are in a reddish-brown color, while the rest of the equation is in black.", caption: none)], [], [#figure(figph[A mathematical equation is displayed on a white background, reading '2m - 9 = 0'.], alt: "A mathematical equation is displayed on a white background, reading '2m - 9 = 0'.", caption: none)], [], [#figure(figph[A mathematical equation '2m = 9' is displayed in black font on a plain white background, appearing as if handwritten or typed in a simple style.], alt: "A mathematical equation '2m = 9' is displayed in black font on a plain white background, appearing as if handwritten or typed in a simple style.", caption: none)], [Substitute #math.equation(block: false, alt: "m equals the fraction 9 over 2")[$m = frac(9, 2)$] into the second equation #linebreak() and then solve for #emph[n].], [#figure(figph[A white background displays two mathematical expressions. The top expression reads 'm = 9/2,' with the fraction '9/2' encircled in red. A red arrow points from this circled fraction to the variable 'm' in the lower expression, 'n = m - 9.' This visually indicates the substitution of the value of m into the second equation.], alt: "A white background displays two mathematical expressions. The top expression reads 'm = 9/2,' with the fraction '9/2' encircled in red. A red arrow points from this circled fraction to the variable 'm' in the lower expression, 'n = m - 9.' This visually indicates the substitution of the value of m into the second equation.", caption: none)], [], [#figure(figph[A mathematical equation is displayed, showing 'm' equals the fraction nine over two, minus nine. The fraction '9/2' is highlighted in red, while 'm =' and '- 9' are in black.], alt: "A mathematical equation is displayed, showing 'm' equals the fraction nine over two, minus nine. The fraction '9/2' is highlighted in red, while 'm =' and '- 9' are in black.", caption: none)], [], [#figure(figph[A mathematical equation displays 'm = 9/2 - 18/2' against a white background.], alt: "A mathematical equation displays 'm = 9/2 - 18/2' against a white background.", caption: none)], [], [#figure(figph[A mathematical expression shows 'n' equals negative nine-halves, written as 'n = -9/2' on a white background. The variable 'n' is isolated, indicating its value as a negative fraction or decimal.], alt: "A mathematical expression shows 'n' equals negative nine-halves, written as 'n = -9/2' on a white background. The variable 'n' is isolated, indicating its value as a negative fraction or decimal.", caption: none)], [#strong[Step 6. Check] the answer in the problem.], [Do these numbers make sense in #linebreak() the problem? We will leave this to you!], [#strong[Step 7. Answer] the question.], [The numbers are #math.equation(block: false, alt: "the fraction 9 over 2")[$frac(9, 2)$] and #math.equation(block: false, alt: "minus the fraction 9 over 2 .")[$− frac(9, 2) .$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The sum of two numbers is 10. One number is 4 less than the other. Find the numbers. #solutionbox[ The numbers are 3 and 7. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The sum of two number is −6. One number is 10 less than the other. Find the numbers. #solutionbox[ The numbers are 2 and −8. ] ] In the , we’ll use the formula for the perimeter of a rectangle, #emph[P] = 2#emph[L] + 2#emph[W]. #examplebox("Example 10")[][ The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and the width. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Step 1. Read] the problem.], [#figure(figph[A simple diagram depicts a rectangle with its width labeled 'W' on the left side and its length labeled 'L' below the bottom edge.], alt: "A simple diagram depicts a rectangle with its width labeled 'W' on the left side and its length labeled 'L' below the bottom edge.", caption: none)]), [#strong[Step 2. Identify] what you are looking for.], [We are looking for the length and width.], [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "L equals")[$L =$] the length #linebreak()   #math.equation(block: false, alt: "W equals")[$W =$] the width], [#strong[Step 4. Translate] into a system of equations.], [The perimeter of a rectangle is 88.], [], [2#emph[L] + 2#emph[W] = #emph[P] #linebreak() #figure(figph[The image shows the mathematical equation 2L + 2W = 88, which represents the formula for the perimeter of a rectangle, where L is the length and W is the width, equaling 88.], alt: "The image shows the mathematical equation 2L + 2W = 88, which represents the formula for the perimeter of a rectangle, where L is the length and W is the width, equaling 88.", caption: none)], [], [The length is five more than twice the width.], [], [#figure(figph[The image displays the mathematical equation L = 2W + 5, featuring uppercase L and W variables, the equals sign, multiplication, and addition.], alt: "The image displays the mathematical equation L = 2W + 5, featuring uppercase L and W variables, the equals sign, multiplication, and addition.", caption: none)], [The system is:], [#figure(figph[A system of two linear equations is presented, with the first equation being 2L + 2W = 88 and the second equation being L = 2W + 5, enclosed by a left curly brace.], alt: "A system of two linear equations is presented, with the first equation being 2L + 2W = 88 and the second equation being L = 2W + 5, enclosed by a left curly brace.", caption: none)], [#strong[Step 5. Solve] the system of equations. #linebreak() We will use substitution since the second #linebreak() equation is solved for #emph[L]. #linebreak() #linebreak() Substitute 2#emph[W] + 5 for #emph[L] in the first equation.], [#figure(figph[A red arrow points from the encircled expression '2W + 5' in the equation L = 2W + 5 to 'L' in the equation 2L + 2W = 88, illustrating the substitution method in algebra.], alt: "A red arrow points from the encircled expression '2W + 5' in the equation L = 2W + 5 to 'L' in the equation 2L + 2W = 88, illustrating the substitution method in algebra.", caption: none)], [Solve for #emph[W].], [#figure(figph[A mathematical equation is displayed: 2(2W + 5) + 2W = 88. The terms '2W' and '5' within the parentheses are highlighted in red, contrasting with the black color of the rest of the equation.], alt: "A mathematical equation is displayed: 2(2W + 5) + 2W = 88. The terms '2W' and '5' within the parentheses are highlighted in red, contrasting with the black color of the rest of the equation.", caption: none)], [], [#figure(figph[A mathematical equation is presented, reading '4W + 10 + 2W = 88' against a plain white background.], alt: "A mathematical equation is presented, reading '4W + 10 + 2W = 88' against a plain white background.", caption: none)], [], [#figure(figph[The image displays a mathematical equation: '6W + 10 = 88' on a white background. It represents a linear equation with one variable 'W' to be solved.], alt: "The image displays a mathematical equation: '6W + 10 = 88' on a white background. It represents a linear equation with one variable 'W' to be solved.", caption: none)], [], [#figure(figph[The image displays a mathematical equation '6W = 78' in black text on a white background, representing an algebraic expression where the variable W can be solved for.], alt: "The image displays a mathematical equation '6W = 78' in black text on a white background, representing an algebraic expression where the variable W can be solved for.", caption: none)], [Substitute #emph[W] = 13 into the second #linebreak() equation and then solve for #emph[L].], [#figure(figph[Mathematical expressions W=13 and L=2W+5 are shown. A red circle highlights the number 13, and a red arrow points from 13 to the W in the second equation, indicating substitution.], alt: "Mathematical expressions W=13 and L=2W+5 are shown. A red circle highlights the number 13, and a red arrow points from 13 to the W in the second equation, indicating substitution.", caption: none)], [], [#figure(figph[A mathematical equation is displayed, showing L = 2 \* 13 + 5. The number 13 is highlighted in red, indicating a specific focus on this value within the calculation.], alt: "A mathematical equation is displayed, showing L = 2 * 13 + 5. The number 13 is highlighted in red, indicating a specific focus on this value within the calculation.", caption: none)], [], [#figure(figph[The image displays the text 'L = 31' in a simple, clear font against a plain white background.], alt: "The image displays the text 'L = 31' in a simple, clear font against a plain white background.", caption: none)], [#strong[Step 6. Check] the answer in the problem.], [Does a rectangle with length 31 and width #linebreak() 13 have perimeter 88? Yes.], [#strong[Step 7. Answer] the equation.], [The length is 31 and the width is 13.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The perimeter of a rectangle is 40. The length is 4 more than the width. Find the length and width of the rectangle. #solutionbox[ The length is 12 and the width is 8. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width of the rectangle. #solutionbox[ The length is 23 and the width is 6. ] ] For we need to remember that the sum of the measures of the angles of a triangle is 180 degrees and that a right triangle has one 90 degree angle. #examplebox("Example 11")[][ The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles. #solutionbox[ We will draw and label a figure. #linebreak() #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Step 1. Read] the problem.], [#figure(figph[A right-angled triangle is shown with angle 'a' at the top vertex and angle 'b' at the bottom right vertex. A square symbol indicates the 90-degree angle.], alt: "A right-angled triangle is shown with angle 'a' at the top vertex and angle 'b' at the bottom right vertex. A square symbol indicates the 90-degree angle.", caption: none)]), [#strong[Step 2. Identify] what you are looking for.], [We are looking for the measures of the angles.], [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "a equals")[$a =$] the measure of the 1#super[st] angle #linebreak() #math.equation(block: false, alt: "b equals")[$#h(1.5em) b =$] the measure of the 2#super[nd] angle], [#strong[Step 4. Translate] into a system of equations.], [The measure of one of the small angles #linebreak() of a right triangle is ten more than three #linebreak() times the measure of the other small angle.], [], [#figure(figph[A mathematical equation shows 'a = 3b + 10' in a simple, clear font against a white background.], alt: "A mathematical equation shows 'a = 3b + 10' in a simple, clear font against a white background.", caption: none)], [], [The sum of the measures of the angles of #linebreak() a triangle is 180.], [], [#figure(figph[The image displays the mathematical equation 'a + b + 90 = 180' in black text against a white background.], alt: "The image displays the mathematical equation 'a + b + 90 = 180' in black text against a white background.", caption: none)], [The system is:], [#figure(figph[A system of two linear equations is presented, with the equations being a = 3b + 10 and a + b + 90 = 180.], alt: "A system of two linear equations is presented, with the equations being a = 3b + 10 and a + b + 90 = 180.", caption: none)], [#strong[Step 5. Solve] the system of equations. #linebreak() We will use substitution since the first #linebreak() equation is solved for #emph[a].], [#figure(figph[Two algebraic equations, 'a = 3b + 10' and 'a + b + 90 = 180', are displayed on a white background, with a red arrow pointing from the first (circled) to the second.], alt: "Two algebraic equations, 'a = 3b + 10' and 'a + b + 90 = 180', are displayed on a white background, with a red arrow pointing from the first (circled) to the second.", caption: none)], [Substitute 3#emph[b] + 10 for #emph[a] in the #linebreak() second equation.], [#figure(figph[A mathematical equation is displayed: (3b + 10) + b + 90 = 180. The term '3b + 10' is highlighted in red, indicating it is a specific angle measurement in a geometric problem.], alt: "A mathematical equation is displayed: (3b + 10) + b + 90 = 180. The term '3b + 10' is highlighted in red, indicating it is a specific angle measurement in a geometric problem.", caption: none)], [Solve for #emph[b].], [#figure(figph[A mathematical equation is displayed against a white background, which reads '4b + 100 = 180'.], alt: "A mathematical equation is displayed against a white background, which reads '4b + 100 = 180'.", caption: none)], [], [#figure(figph[A close-up shot of a white background with a mathematical equation in the center. The equation reads '4b = 80' in a simple black font.], alt: "A close-up shot of a white background with a mathematical equation in the center. The equation reads '4b = 80' in a simple black font.", caption: none)], [], [#figure(figph[An algebraic problem illustrating substitution: given b=20, substitute this value into a=3b+10 to solve for a.], alt: "An algebraic problem illustrating substitution: given b=20, substitute this value into a=3b+10 to solve for a.", caption: none)], [Substitute #emph[b] = 20 into the first #linebreak() equation and then solve for #emph[a].], [#figure(figph[A mathematical equation, 'a = 3 \* 20 + 10', is displayed on a white background. Performing the multiplication first (3 \* 20 = 60) and then the addition (60 + 10 = 70) reveals that 'a' equals 70.], alt: "A mathematical equation, 'a = 3 * 20 + 10', is displayed on a white background. Performing the multiplication first (3 * 20 = 60) and then the addition (60 + 10 = 70) reveals that 'a' equals 70.", caption: none) #linebreak() #figure(figph[The equation a = 70 is displayed on a white background.], alt: "The equation a = 70 is displayed on a white background.", caption: none)], [#strong[Step 6. Check] the answer in the problem.], [We will leave this to you!], [#strong[Step 7. Answer] the question.], [The measures of the small angles are #linebreak() 20 and 70.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles. #solutionbox[ The measure of the angles are 22 degrees and 68 degrees. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles. #solutionbox[ The measure of the angles are 36 degrees and 54 degrees. ] ] #examplebox("Example 12")[][ Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her \$25,000 plus \$15 for each training session. Option B would pay her \$10,000 + \$40 for each training session. How many training sessions would make the salary options equal? #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([#strong[Step 1. Read] the problem.], []), [#strong[Step 2. Identify] what you are looking for.], [We are looking for the number of training sessions #linebreak() that would make the pay equal.], [#strong[Step 3. Name] what we are looking for.], [Let #math.equation(block: false, alt: "s equals")[$s =$] Heather’s salary. #linebreak() #math.equation(block: false, alt: "n equals")[$#h(1.5em) n =$] the number of training sessions], [#strong[Step 4. Translate] into a system of equations.], [Option A would pay her \$25,000 plus \$15 #linebreak() for each training session.], [], [#figure(figph[The image shows the equation s = 25,000 + 15n, presented in a black font against a white background.], alt: "The image shows the equation s = 25,000 + 15n, presented in a black font against a white background.", caption: none)], [], [Option B would pay her \$10,000 + \$40 #linebreak() for each training session], [], [#figure(figph[A mathematical equation is displayed on a white background, reading 's = 10,000 + 40n'.], alt: "A mathematical equation is displayed on a white background, reading 's = 10,000 + 40n'.", caption: none)], [The system is:], [#figure(figph[A system of two linear equations is presented. The first equation is s = 25,000 + 15n, and the second equation is s = 10,000 + 40n, where 's' and 'n' are variables.], alt: "A system of two linear equations is presented. The first equation is s = 25,000 + 15n, and the second equation is s = 10,000 + 40n, where 's' and 'n' are variables.", caption: none)], [#strong[Step 5. Solve] the system of equations. #linebreak() We will use substitution.], [#figure(figph[Two equations are presented: s = 25,000 + 15n, circled in red, and s = 10,000 + 40n. An arrow from the circled equation points to the second.], alt: "Two equations are presented: s = 25,000 + 15n, circled in red, and s = 10,000 + 40n. An arrow from the circled equation points to the second.", caption: none)], [Substitute 25,000 + 15#emph[n] for #emph[s] in the second equation.], [#figure(figph[A mathematical equation is displayed: 25,000 + 15n = 10,000 + 40n.], alt: "A mathematical equation is displayed: 25,000 + 15n = 10,000 + 40n.", caption: none)], [Solve for #emph[n].], [#figure(figph[A mathematical equation is displayed with a white background and dark gray text. The equation reads: 25,000 = 10,000 + 25n.], alt: "A mathematical equation is displayed with a white background and dark gray text. The equation reads: 25,000 = 10,000 + 25n.", caption: none)], [], [#figure(figph[A mathematical equation is displayed, showing 15,000 equals 25n.], alt: "A mathematical equation is displayed, showing 15,000 equals 25n.", caption: none)], [], [#figure(figph[The image displays a mathematical equation written in a simple, clear font on a white background, stating '600 = n'.], alt: "The image displays a mathematical equation written in a simple, clear font on a white background, stating '600 = n'.", caption: none)], [#strong[Step 6. Check] the answer.], [Are 600 training sessions a year reasonable? #linebreak() Are the two options equal when #emph[n] = 600?], [#strong[Step 7. Answer] the question.], [The salary options would be equal for 600 training sessions.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Geraldine has been offered positions by two insurance companies. The first company pays a salary of \$12,000 plus a commission of \$100 for each policy sold. The second pays a salary of \$20,000 plus a commission of \$50 for each policy sold. How many policies would need to be sold to make the total pay the same? #solutionbox[ There would need to be 160 policies sold to make the total pay the same. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Kenneth currently sells suits for company A at a salary of \$22,000 plus a \$10 commission for each suit sold. Company B offers him a position with a salary of \$28,000 plus a \$4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal? #solutionbox[ Kenneth would need to sell 1,000 suits. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access these online resources for additional instruction and practice with solving systems of equations by substitution. - #link("https://www.openstax.org/l/25SolvingLinear")[Instructional Video-Solve Linear Systems by Substitution] - #link("https://www.openstax.org/l/25Substitution")[Instructional Video-Solve by Substitution] ] === Key Concepts - #strong[Solve a system of equations by substitution] + Solve one of the equations for either variable. + Substitute the expression from Step 1 into the other equation. + Solve the resulting equation. + Substitute the solution in Step 3 into one of the original equations to find the other variable. + Write the solution as an ordered pair. + Check that the ordered pair is a solution to both original equations. ==== Practice Makes Perfect #strong[Solve a System of Equations by Substitution] In the following exercises, solve the systems of equations by substitution. #math.equation(block: true, alt: "{ 2 x plus y equals −4; 3 x minus 2 y equals −6")[$\{ 2 x + y = −4 \ 3 x − 2 y = −6$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −2 , 0 close parenthesis")[$( −2 , 0 )$] ] #math.equation(block: true, alt: "{ 2 x plus y equals −2; 3 x minus y equals 7")[$\{ 2 x + y = −2 \ 3 x − y = 7$] #math.equation(block: true, alt: "{ x minus 2 y equals −5; 2 x minus 3 y equals −4")[$\{ x − 2 y = −5 \ 2 x − 3 y = −4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 7 , 6 close parenthesis")[$( 7 , 6 )$] ] #math.equation(block: true, alt: "{ x minus 3 y equals −9; 2 x plus 5 y equals 4")[$\{ x − 3 y = −9 \ 2 x + 5 y = 4$] #math.equation(block: true, alt: "{ 5 x minus 2 y equals −6; y equals 3 x plus 3")[$\{ 5 x − 2 y = −6 \ y = 3 x + 3$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 0 , 3 close parenthesis")[$( 0 , 3 )$] ] #math.equation(block: true, alt: "{ −2 x plus 2 y equals 6; y equals −3 x plus 1")[$\{ −2 x + 2 y = 6 \ y = −3 x + 1$] #math.equation(block: true, alt: "{ 2 x plus 3 y equals 3; y equals − x plus 3")[$\{ 2 x + 3 y = 3 \ y = "−" x + 3$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 6 , −3 close parenthesis")[$( 6 , −3 )$] ] #math.equation(block: true, alt: "{ 2 x plus 5 y equals −14; y equals −2 x plus 2")[$\{ 2 x + 5 y = −14 \ y = −2 x + 2$] #math.equation(block: true, alt: "{ 2 x plus 5 y equals 1; y equals the fraction 1 over 3 x minus 2")[$\{ 2 x + 5 y = 1 \ y = frac(1, 3) x − 2$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 3 , −1 close parenthesis")[$( 3 , −1 )$] ] #math.equation(block: true, alt: "{ 3 x plus 4 y equals 1; y equals minus the fraction 2 over 5 x plus 2")[$\{ 3 x + 4 y = 1 \ y = − frac(2, 5) x + 2$] #math.equation(block: true, alt: "{ 3 x minus 2 y equals 6; y equals the fraction 2 over 3 x plus 2")[$\{ 3 x − 2 y = 6 \ y = frac(2, 3) x + 2$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 6 , 6 close parenthesis")[$( 6 , 6 )$] ] #math.equation(block: true, alt: "{ −3 x minus 5 y equals 3; y equals the fraction 1 over 2 x minus 5")[$\{ −3 x − 5 y = 3 \ y = frac(1, 2) x − 5$] #math.equation(block: true, alt: "{ 2 x plus y equals 10; minus x plus y equals −5")[$\{ 2 x + y = 10 \ − x + y = −5$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 , 0 close parenthesis")[$( 5 , 0 )$] ] #math.equation(block: true, alt: "{ −2 x plus y equals 10; minus x plus 2 y equals 16")[$\{ −2 x + y = 10 \ − x + 2 y = 16$] #math.equation(block: true, alt: "{ 3 x plus y equals 1; −4 x plus y equals 15")[$\{ 3 x + y = 1 \ −4 x + y = 15$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −2 , 7 close parenthesis")[$( −2 , 7 )$] ] #math.equation(block: true, alt: "{ x plus y equals 0; 2 x plus 3 y equals −4")[$\{ x + y = 0 \ 2 x + 3 y = −4$] #math.equation(block: true, alt: "{ x plus 3 y equals 1; 3 x plus 5 y equals −5")[$\{ x + 3 y = 1 \ 3 x + 5 y = −5$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −5 , 2 close parenthesis")[$( −5 , 2 )$] ] #math.equation(block: true, alt: "{ x plus 2 y equals −1; 2 x plus 3 y equals 1")[$\{ x + 2 y = −1 \ 2 x + 3 y = 1$] #math.equation(block: true, alt: "{ 2 x plus y equals 5; x minus 2 y equals −15")[$\{ 2 x + y = 5 \ x − 2 y = −15$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −1 , 7 close parenthesis")[$( −1 , 7 )$] ] #math.equation(block: true, alt: "{ 4 x plus y equals 10; x minus 2 y equals −20")[$\{ 4 x + y = 10 \ x − 2 y = −20$] #math.equation(block: true, alt: "{ y equals −2 x minus 1; y equals minus the fraction 1 over 3 x plus 4")[$\{ y = −2 x − 1 \ y = − frac(1, 3) x + 4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis −3 , 5 close parenthesis")[$( −3 , 5 )$] ] #math.equation(block: true, alt: "{ y equals x minus 6; y equals minus the fraction 3 over 2 x plus 4")[$\{ y = x − 6 \ y = − frac(3, 2) x + 4$] #math.equation(block: true, alt: "{ y equals 2 x minus 8; y equals the fraction 3 over 5 x plus 6")[$\{ y = 2 x − 8 \ y = frac(3, 5) x + 6$] #solutionbox[ (10, 12) ] #math.equation(block: true, alt: "{ y equals − x minus 1; y equals x plus 7")[$\{ y = "−" x − 1 \ y = x + 7$] #math.equation(block: true, alt: "{ 4 x plus 2 y equals 8; 8 x minus y equals 1")[$\{ 4 x + 2 y = 8 \ 8 x − y = 1$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 , 3 close parenthesis")[$( frac(1, 2) , 3 )$] ] #math.equation(block: true, alt: "{ minus x minus 12 y equals −1; 2 x minus 8 y equals −6")[$\{ − x − 12 y = −1 \ 2 x − 8 y = −6$] #math.equation(block: true, alt: "{ 15 x plus 2 y equals 6; −5 x plus 2 y equals −4")[$\{ 15 x + 2 y = 6 \ −5 x + 2 y = −4$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 , minus the fraction 3 over 4 close parenthesis")[$( frac(1, 2) , − frac(3, 4) )$] ] #math.equation(block: true, alt: "{ 2 x minus 15 y equals 7; 12 x plus 2 y equals −4")[$\{ 2 x − 15 y = 7 \ 12 x + 2 y = −4$] #math.equation(block: true, alt: "{ y equals 3 x; 6 x minus 2 y equals 0")[$\{ y = 3 x \ 6 x − 2 y = 0$] #solutionbox[ Infinitely many solutions ] #math.equation(block: true, alt: "{ x equals 2 y; 4 x minus 8 y equals 0")[$\{ x = 2 y \ 4 x − 8 y = 0$] #math.equation(block: true, alt: "{ 2 x plus 16 y equals 8; minus x minus 8 y equals −4")[$\{ 2 x + 16 y = 8 \ − x − 8 y = −4$] #solutionbox[ Infinitely many solutions ] #math.equation(block: true, alt: "{ 15 x plus 4 y equals 6; −30 x minus 8 y equals −12")[$\{ 15 x + 4 y = 6 \ −30 x − 8 y = −12$] #math.equation(block: true, alt: "{ y equals −4 x; 4 x plus y equals 1")[$\{ y = −4 x \ 4 x + y = 1$] #solutionbox[ No solution ] #math.equation(block: true, alt: "{ y equals minus the fraction 1 over 4 x; x plus 4 y equals 8")[$\{ y = − frac(1, 4) x \ x + 4 y = 8$] #math.equation(block: true, alt: "{ y equals the fraction 7 over 8 x plus 4; −7 x plus 8 y equals 6")[$\{ y = frac(7, 8) x + 4 \ −7 x + 8 y = 6$] #solutionbox[ No solution ] #math.equation(block: true, alt: "{ y equals minus the fraction 2 over 3 x plus 5; 2 x plus 3 y equals 11")[$\{ y = − frac(2, 3) x + 5 \ 2 x + 3 y = 11$] #strong[Solve Applications of Systems of Equations by Substitution] In the following exercises, translate to a system of equations and solve. The sum of two numbers is 15. One number is 3 less than the other. Find the numbers. #solutionbox[ The numbers are 6 and 9. ] The sum of two numbers is 30. One number is 4 less than the other. Find the numbers. The sum of two numbers is −26. One number is 12 less than the other. Find the numbers. #solutionbox[ The numbers are −7 and −19. ] The perimeter of a rectangle is 50. The length is 5 more than the width. Find the length and width. The perimeter of a rectangle is 60. The length is 10 more than the width. Find the length and width. #solutionbox[ The length is 20 and the width is 10. ] The perimeter of a rectangle is 58. The length is 5 more than three times the width. Find the length and width. The perimeter of a rectangle is 84. The length is 10 more than three times the width. Find the length and width. #solutionbox[ The length is 34 and the width is 8. ] The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles. The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles. #solutionbox[ The measures are 16° and 74°. ] The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles. The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles. #solutionbox[ The measures are 45° and 45°. ] Maxim has been offered positions by two car dealers. The first company pays a salary of \$10,000 plus a commission of \$1,000 for each car sold. The second pays a salary of \$20,000 plus a commission of \$500 for each car sold. How many cars would need to be sold to make the total pay the same? Jackie has been offered positions by two cable companies. The first company pays a salary of \$ 14,000 plus a commission of \$100 for each cable package sold. The second pays a salary of \$20,000 plus a commission of \$25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same? #solutionbox[ 80 cable packages would need to be sold. ] Amara currently sells televisions for company A at a salary of \$17,000 plus a \$100 commission for each television she sells. Company B offers her a position with a salary of \$29,000 plus a \$20 commission for each television she sells. How many televisions would Amara need to sell for the options to be equal? Mitchell currently sells stoves for company A at a salary of \$12,000 plus a \$150 commission for each stove he sells. Company B offers him a position with a salary of \$24,000 plus a \$50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal? #solutionbox[ Mitchell would need to sell 120 stoves. ] ==== Everyday Math When Gloria spent 15 minutes on the elliptical trainer and then did circuit training for 30 minutes, her fitness app says she burned 435 calories. When she spent 30 minutes on the elliptical trainer and 40 minutes circuit training she burned 690 calories. Solve the system #math.equation(block: false, alt: "{ 15 e plus 30 c equals 435; 30 e plus 40 c equals 690")[$\{ 15 e + 30 c = 435 \ 30 e + 40 c = 690$] for #math.equation(block: false, alt: "e")[$e$], the number of calories she burns for each minute on the elliptical trainer, and #math.equation(block: false, alt: "c")[$c$], the number of calories she burns for each minute of circuit training. Stephanie left Riverside, California, driving her motorhome north on Interstate 15 towards Salt Lake City at a speed of 56 miles per hour. Half an hour later, Tina left Riverside in her car on the same route as Stephanie, driving 70 miles per hour. Solve the system #math.equation(block: false, alt: "{ 56 s equals 70 t; s equals t plus the fraction 1 over 2")[$\{ 56 s = 70 t \ s = t + frac(1, 2)$]. + ⓐ for #math.equation(block: false, alt: "t")[$t$] to find out how long it will take Tina to catch up to Stephanie. + ⓑ what is the value of #math.equation(block: false, alt: "s")[$s$], the number of hours Stephanie will have driven before Tina catches up to her? #solutionbox[ ⓐ #math.equation(block: false, alt: "t equals 2")[$t = 2$] hours ⓑ #math.equation(block: false, alt: "s equals 2 the fraction 1 over 2")[$s = 2 frac(1, 2)$] hours ] ==== Writing Exercises Solve the system of equations #linebreak() #math.equation(block: false, alt: "{ x plus y equals 10; x minus y equals 6")[$\{ x + y = 10 \ x − y = 6$] ⓐ by graphing. ⓑ by substitution. ⓒ Which method do you prefer? Why? Solve the system of equations #linebreak() #math.equation(block: false, alt: "{ 3 x plus y equals 12; x equals y minus 8")[$\{ 3 x + y = 12 \ x = y − 8$] by substitution and explain all your steps in words. #solutionbox[ Answers will vary. ] ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This figure shows a table with three rows and four columns. The columns are labeled, “I can…,” “Confidently.” “With some help.” and “No - I don’t get it.” The only column with filled in cells below it is labeled “I can…” It reads, “solve a system of equations by substitution.” “solve applications of systems of equations by substitution.”], alt: "This figure shows a table with three rows and four columns. The columns are labeled, “I can…,” “Confidently.” “With some help.” and “No - I don’t get it.” The only column with filled in cells below it is labeled “I can…” It reads, “solve a system of equations by substitution.” “solve applications of systems of equations by substitution.”", caption: none) ⓑ After reviewing this checklist, what will you do to become confident for all objectives?