#set document(title: "10.5 Linear Systems in Two Variables", 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")) == 10.5#h(0.6em)Linear Systems in Two Variables A #math.equation(block: false, alt: "2 times 2")[$2 × 2$] #strong[system] of equations is a set of #math.equation(block: false, alt: "2")[$2$] equations in the same #math.equation(block: false, alt: "2")[$2$] variables. A #strong[solution] of a #math.equation(block: false, alt: "2 times 2")[$2 × 2$] system is an ordered pair that makes each equation in the system true. In this section, we review two algebraic methods for solving #math.equation(block: false, alt: "2 times 2")[$2 × 2$] linear systems: substitution and elimination. === Solving Systems by Substitution The basic strategy for the #strong[substitution] method can be described as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Steps for Solving a #math.equation(block: false, alt: "2 times 2")[$2 × 2$] System by Substitution] + Solve one of the equations for one of the variables in terms of the other. + Substitute this expression into the second equation; doing so yields an equation in one variable. + Solve the new equation. + Use the result of step 1 to find the other variable. ] #examplebox("Example 1")[][ Staci stocks two kinds of sleeping bags in her sporting goods store, a standard model and a down-filled model for colder temperatures. From past experience, she estimates that she will sell twice as many of the standard variety as of the down filled. She has room to stock 60 sleeping bags at a time. How many of each variety should Staci order? #solutionbox[ + #math.equation(block: true, alt: "Number of standard sleeping bags: x; Number of down-filled sleeping bags: y")[$& "Number of standard sleeping bags: " " " x \ & "Number of down-filled sleeping bags: " " " y$] + Write two equations about the variables. Staci needs twice as many standard model as down filled, so #math.equation(block: true, alt: "x, equals 2 y, open parenthesis 1 close parenthesis")[$x & = 2 y & & #hide($b l a n k e t y$) ( 1 )$] Also, the total number of sleeping bags is 60, so #math.equation(block: true, alt: "x plus y, equals 60, open parenthesis 2 close parenthesis")[$x + y & = 60 & & #hide($b l a n k$) ( 2 )$] + We will solve this system using substitution. Notice that Equation (1) is already solved for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "y")[$y$]: #math.equation(block: false, alt: "x equals 2 y")[$x = 2 y$]. Substitute #math.equation(block: false, alt: "2 y")[$2 bold(italic(y))$] for #math.equation(block: false, alt: "x")[$x$] in Equation (2) to obtain #math.equation(block: true, alt: "2 y plus y, equals 60; 3 y, equals 60")[$2 bold(italic(y)) + y & = 60 \ 3 y & = 60$] Solving for #math.equation(block: false, alt: "y")[$y$], we find #math.equation(block: false, alt: "y equals 20")[$y = 20$]. Finally, substitute this value into Equation (1) to find #math.equation(block: true, alt: "x equals 2 open parenthesis 20 close parenthesis equals 40")[$x = 2 ( 20 ) = 40$] The solution to the system is #math.equation(block: false, alt: "x equals 40 , y equals 20")[$x = 40 , y = 20$]. + Staci should order #math.equation(block: false, alt: "40")[$40$] standard sleeping bags and #math.equation(block: false, alt: "20")[$20$] down-filled bags. ] ] === Solving Systems by Elimination The method of substitution is convenient if one of the variables in the system has a coefficient of #math.equation(block: false, alt: "1")[$1$] or #math.equation(block: false, alt: "minus 1")[$− 1$], because it is easy to solve for that variable. If none of the coefficients is #math.equation(block: false, alt: "1")[$1$] or #math.equation(block: false, alt: "minus 1")[$− 1$], then a second method, called #strong[elimination], is usually more efficient. The method of elimination is based on the following properties of linear equations. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Properties of Linear Systems] + Multiplying a linear equation by a (nonzero) constant does not change its solutions. That is, any solution of the equation #math.equation(block: true, alt: "a x plus b y equals c")[$a x + b y = c$] is also a solution of the equation #math.equation(block: true, alt: "k a x plus k b y equals k c")[$k a x + k b y = k c$] + Adding (or subtracting) two linear equations does not change their common solutions. That is, any solution of the system #math.equation(block: true, alt: "a sub 1 x plus b sub 1 y, equals c sub 1; a sub 2 x plus b sub 2 y, equals c sub 2")[$a_(1) x + b_(1) y & = c_(1) \ a_(2) x + b_(2) y & = c_(2)$] is also a solution of the equation #math.equation(block: true, alt: "open parenthesis a sub 1 plus a sub 2 close parenthesis x plus open parenthesis b sub 1 plus b sub 2 close parenthesis y equals c sub 1 plus c sub 2")[$( a_(1) + a_(2) ) x + ( b_(1) + b_(2) ) y = c_(1) + c_(2)$] ] #examplebox("Example 2")[][ Solve the system by the method of elimination. #math.equation(block: true, alt: "2 x, plus, 3 y, equals, 8, open parenthesis 1 close parenthesis; 3 x, minus, 4 y, equals, minus 5, open parenthesis 2 close parenthesis")[$2 x & + & 3 y & = & 8 #hide($b l a n k b l a n k$) & ( 1 ) \ 3 x & − & 4 y & = & − 5 #hide($b l a n k b l a n k$) & ( 2 )$] #solutionbox[ We first decide which variable to eliminate, #math.equation(block: false, alt: "x")[$x$] or #math.equation(block: false, alt: "y")[$y$]. We can choose whichever looks easiest. In this problem, we choose to eliminate #math.equation(block: false, alt: "x")[$x$]. We next look for the smallest number that both coefficients, #math.equation(block: false, alt: "2")[$2$] and #math.equation(block: false, alt: "3")[$3$], divide into evenly. This number is #math.equation(block: false, alt: "6")[$6$]. We want the coefficients of #math.equation(block: false, alt: "x")[$x$] to become #math.equation(block: false, alt: "6")[$6$] and #math.equation(block: false, alt: "minus 6")[$− 6$], so we will multiply Equation (1) by #math.equation(block: false, alt: "3")[$3$] and Equation (2) by #math.equation(block: false, alt: "minus 2")[$− 2$] to obtain #math.equation(block: true, alt: "6 x, plus, 9 y, equals, 24, open parenthesis 1 a close parenthesis; minus 6 x, plus, 8 y, equals, 10, open parenthesis 2 a close parenthesis")[$6 x & + & 9 y & = & 24 #hide($b l a n k b l a n k$) & ( 1 "a" ) \ − 6 x & + & 8 y & = & 10 #hide($b l a n k b l a n k$) & ( 2 "a" )$] Now we add the corresponding terms of (1a) and (2a). The #math.equation(block: false, alt: "x")[$x$]-terms are eliminated, yielding an equation in one variable. #math.equation(block: true, alt: "6 x, plus, 9 y, equals, 24, open parenthesis 1 a close parenthesis; minus 6 x, plus, 8 y, equals, 10, open parenthesis 2 a close parenthesis; 17 y, equals, 34, open parenthesis 3 close parenthesis")[$6 x & + & 9 y & = & 24 #hide($b l a n k b l a n k$) & ( 1 "a" ) \ − 6 x & + & 8 y & = & 10 #hide($b l a n k b l a n k$) & ( 2 "a" ) \ & & 17 y & = & 34 #hide($b l a n k b l a n k$) & ( 3 )$] We solve this equation for #math.equation(block: false, alt: "y")[$y$] to find #math.equation(block: false, alt: "y equals 2")[$y = 2$]. We can substitute this value of #math.equation(block: false, alt: "y")[$y$] into any of our equations involving both #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. If we choose Equation (1), then #math.equation(block: true, alt: "2 x plus 3 open parenthesis 2 close parenthesis equals 8")[$2 x + 3 ( 2 ) = 8$] and solving this equation yields #math.equation(block: false, alt: "x equals 1")[$x = 1$]. The ordered pair #math.equation(block: false, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$] is a solution to the system. You should verify that these values satisfy both original equations. ] ] We summarize the strategy for solving a linear system by elimination. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Steps for Solving a #math.equation(block: false, alt: "2 times 2")[$2 × 2$] Linear System by Elimination] + Choose one of the variables to eliminate. Multiply each equation by a suitable factor so that the coefficients of that variable are opposites. + Add the two new equations termwise. + Solve the resulting equation for the remaining variable. + Substitute the value found in step 3 into either of the original equations and solve for the other variable. ] In Example, we added #math.equation(block: false, alt: "3")[$3$] times the first equation to #math.equation(block: false, alt: "minus 2")[$− 2$] times the second equation. The result from adding a constant multiple of one equation to a constant multiple of another equation is called a #strong[linear combination] of the two equations. The method of elimination is also called the method of linear combinations. If either equation in a system has fractional coefficients, it is helpful to clear the fractions before applying the method of linear combinations. #examplebox("Example 3")[][ Solve the system by linear combinations. #math.equation(block: true, alt: "the fraction 2 over 3 x, minus, y, equals, 2, open parenthesis 1 close parenthesis; x, plus, the fraction 1 over 2 y, equals, 7, open parenthesis 2 close parenthesis")[$frac(2, 3) x & − & y & = & 2 #hide($b l a n k b l a n k$) & ( 1 ) \ x & + & frac(1, 2) y & = & 7 #hide($b l a n k b l a n k$) & ( 2 )$] #solutionbox[ Multiply each side of Equation (1) by #math.equation(block: false, alt: "3")[$3$] and each side of Equation (2) by #math.equation(block: false, alt: "2")[$2$] to clear the fractions: #math.equation(block: true, alt: "2 x, minus, 3 y, equals, 6, open parenthesis 1 a close parenthesis; 2 x, plus, y, equals, 14, open parenthesis 2 a close parenthesis")[$2 x & − & 3 y & = & 6 #hide($b l a n k b l a n k$) & ( 1 "a" ) \ 2 x & + & y & = & 14 #hide($b l a n k b l a n k$) & ( 2 "a" )$] To eliminate the variable #math.equation(block: false, alt: "x")[$x$], multiply Equation (2a) by #math.equation(block: false, alt: "minus 1")[$− 1$] and add the result to Equation (1a) to get #math.equation(block: true, alt: "minus 4 y, equals minus 8, Divide both sides by -4.; y, equals 2")[$− 4 y & = − 8 & & "Divide both sides by -4." \ y & = 2$] Substitute #math.equation(block: false, alt: "2")[$2$] for #math.equation(block: false, alt: "y")[$y$] in one of the original equations and solve for #math.equation(block: false, alt: "x")[$x$]. We use Equation (2). #math.equation(block: true, alt: "x plus the fraction 1 over 2 open parenthesis 2 close parenthesis, equals 7, Subtract 1 from both sides.; x, equals 6")[$x + frac(1, 2) ( 2 ) & = 7 & & "Subtract 1 from both sides." \ x & = 6$] Verify that #math.equation(block: false, alt: "x equals 6")[$x = 6$] and #math.equation(block: false, alt: "y equals 2")[$y = 2$] satisfy both Equations (1) and (2). The solution to the system is the ordered pair #math.equation(block: false, alt: "open parenthesis 6 , 2 close parenthesis")[$( 6 , 2 )$]. ] ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - System of equations - Dependent - Linear combination - Inconsistent - Elimination - Solution of a system - Substitution ==== SKILLS Practice each skill in the exercises listed. + Solve a system by substitution: \#1–4 + Solve a system by elimination: \#5–8 + Choose a method and solve the system: \#9–18, 29–32 + Solve problems by writing and solving a system: \#19–28 === Exercises A.5 For Problems 1-4, solve the system by substitution. #math.equation(block: true, alt: "a, plus, 2 b, equals minus 6; 2 a, minus, 3 b, equals 16")[$a & + & 2 b & = − 6 \ 2 a & − & 3 b & = 16$] #math.equation(block: true, alt: "open parenthesis 2 , minus 4 close parenthesis")[$( 2 , − 4 )$] #math.equation(block: true, alt: "7 r, minus, 4 s, equals 1; 3 r, plus, s, equals 14")[$7 r & − & 4 s & = 1 \ 3 r & + & s & = 14$] #math.equation(block: true, alt: "2 x, minus, 3 y, equals 6; x, plus, 3 y, equals 3")[$2 x & − & 3 y & = 6 \ x & + & 3 y & = 3$] #math.equation(block: true, alt: "open parenthesis 3 , 0 close parenthesis")[$( 3 , 0 )$] #math.equation(block: true, alt: "2 r, equals s plus 7; 2 s, equals 14 minus 3 r")[$2 r & = s + 7 \ 2 s & = 14 − 3 r$] For Problems 5-8, solve the system by elimination. #math.equation(block: true, alt: "5 x, minus, 2 y, equals minus 4; minus 6 x, plus, 3 y, equals 5")[$5 x & − & 2 y & = − 4 \ − 6 x & + & 3 y & = 5$] #math.equation(block: true, alt: "open parenthesis the fraction minus 2 over 3 , the fraction 1 over 3 close parenthesis")[$( display(frac(− 2, 3)) , display(frac(1, 3)) )$] #math.equation(block: true, alt: "2 p, plus, 3 q, equals 38; 6 p, minus, 5 q, equals 2")[$2 p & + & 3 q & = 38 \ 6 p & − & 5 q & = 2$] #math.equation(block: true, alt: "3 x, minus, 4 y, equals minus 11; 2 x, plus, 6 y, equals minus 3")[$3 x & − & 4 y & = − 11 \ 2 x & + & 6 y & = − 3$] #math.equation(block: true, alt: "open parenthesis minus 3 , the fraction 1 over 2 close parenthesis")[$( − 3 , display(frac(1, 2)) )$] #math.equation(block: true, alt: "2 u, minus, 3 v, equals minus 4; 5 u, plus, 2 v, equals 9")[$2 u & − & 3 v & = − 4 \ 5 u & + & 2 v & = 9$] For Problems 9–12, solve the system by substitution or by linear combinations. #math.equation(block: true, alt: "3 m, plus n equals 7; 2 m, equals 5 n minus 1")[$3 m & + n = 7 \ 2 m & = 5 n − 1$] #math.equation(block: true, alt: "open parenthesis 2 , 1 close parenthesis")[$( 2 , 1 )$] #math.equation(block: true, alt: "3 x, plus, 5 y, equals 1; 2 x, minus, 3 y, equals 7")[$3 x & + & 5 y & = 1 \ 2 x & − & 3 y & = 7$] #math.equation(block: true, alt: "3 y, equals 2 x minus 8; 4 y, plus 11 equals 3 x")[$3 y & = 2 x − 8 \ 4 y & + 11 = 3 x$] #math.equation(block: true, alt: "open parenthesis 1 , minus 2 close parenthesis")[$( 1 , − 2 )$] #math.equation(block: true, alt: "4 L minus 3, equals 3 W; 25 plus 5 L, equals minus 2 W")[$4 L − 3 & = 3 W \ 25 + 5 L & = − 2 W$] In Problems 13–18, clear the fractions in each equation first, then solve the system by substitution or by linear combinations. #math.equation(block: true, alt: "the fraction 2 over 3 A, minus, B, equals 4; A, minus, the fraction 3 over 4 B, equals 6")[$display(frac(2, 3)) A & − & B & = 4 \ A & − & display(frac(3, 4)) B & = 6$] #math.equation(block: true, alt: "open parenthesis 6 , 0 close parenthesis")[$( 6 , 0 )$] #math.equation(block: true, alt: "the fraction 1 over 8 w, minus, the fraction 3 over 8 z, equals 1; the fraction 1 over 2 w, minus, the fraction 1 over 4 z, equals minus 1")[$display(frac(1, 8)) w & − & display(frac(3, 8)) z & = 1 \ display(frac(1, 2)) w & − & display(frac(1, 4)) z & = − 1$] #math.equation(block: true, alt: "the fraction M over 4, equals the fraction N over 3 minus the fraction 5 over 12; the fraction N over 5, equals the fraction 1 over 2 minus the fraction M over 10")[$display(frac(M, 4)) & = display(frac(N, 3)) − display(frac(5, 12)) \ display(frac(N, 5)) & = display(frac(1, 2)) − display(frac(M, 10))$] #math.equation(block: true, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$] #math.equation(block: true, alt: "the fraction R over 3, equals the fraction S over 3 plus 2; the fraction S over 3, equals the fraction R over 6 minus 1")[$display(frac(R, 3)) & = display(frac(S, 3)) + 2 \ display(frac(S, 3)) & = display(frac(R, 6)) − 1$] #math.equation(block: true, alt: "the fraction s over 2, equals the fraction 7 over 6 minus the fraction t over 3; the fraction s over 4, equals the fraction 3 over 4 minus the fraction t over 4")[$display(frac(s, 2)) & = display(frac(7, 6)) − display(frac(t, 3)) \ display(frac(s, 4)) & = display(frac(3, 4)) − display(frac(t, 4))$] #math.equation(block: true, alt: "open parenthesis 1 , 2 close parenthesis")[$( 1 , 2 )$] #math.equation(block: true, alt: "the fraction 2 p over 3, plus the fraction 8 q over 9 equals the fraction 4 over 3; the fraction p over 3, equals 2 plus the fraction q over 2")[$display(frac(2 p, 3)) & + display(frac(8 q, 9)) = display(frac(4, 3)) \ display(frac(p, 3)) & = 2 + display(frac(q, 2))$] In Problems 19–28, write a system of equations for the problem, then solve algebraically. Francine has \$2000, part of it invested in bonds paying 10%, and the rest in a certificate account at 8%. Her annual income from the two investments is \$184. How much did Francine invest at each rate? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Principal], [Interest rate], [Interest]), [Bonds], [], [], [], [Certificate], [], [], [], [Total], [], [——], [], )) + Write one equation about the amount Francine invested. + Write a second equation about Francine's annual interest. + Solve the system and answer the question in the problem. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Principal], [Interest rate], [Interest]), [Bonds], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.10")[$0.10$]], [#math.equation(block: false, alt: "0.10 x")[$0.10 x$]], [Certificate], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0.08")[$0.08$]], [#math.equation(block: false, alt: "0.08 y")[$0.08 y$]], [Total], [#math.equation(block: false, alt: "x plus y")[$x + y$]], [——], [#math.equation(block: false, alt: "0.10 x plus 0.08 y")[$0.10 x + 0.08 y$]], )) + #math.equation(block: false, alt: "x plus y equals 2000")[$x + y = 2000$] + #math.equation(block: false, alt: "0.10 x plus 0.08 y equals 184")[$0.10 x + 0.08 y = 184$] + \$800 at 8%, \$1200 at 10% Carmella has \$1200 invested in two stocks; one returns 8% per year, and the other returns 12% per year. The income from the 8% stock is \$3 more than the income from the 12% stock. How much did Carmella invest in each stock? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Principal], [Interest rate], [Interest]), [First stock], [], [], [], [Second stock], [], [], [], [Total], [], [——], [——], )) + Write one equation about the amount Carmella invested. + Write a second equation about Carmella's annual interest. + Solve the system and answer the question in the problem. Paul needs 40 pounds of 48% silver alloy to finish a collection of jewelry. How many pounds of 45% silver alloy should he melt with 60% silver alloy to obtain the alloy he needs? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Pounds], [% silver], [Amount of silver]), [First alloy], [], [], [], [Second alloy], [], [], [], [Mixture], [], [], [], )) + Write one equation about the amount of alloy Paul needs. + Write a second equation about the amount of silver in the alloys. + Solve the system and answer the question in the problem. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Pounds], [% silver], [Amount of silver]), [First alloy], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.45")[$0.45$]], [#math.equation(block: false, alt: "0.45 x")[$0.45 x$]], [Second alloy], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0.60")[$0.60$]], [#math.equation(block: false, alt: "0.60 y")[$0.60 y$]], [Mixture], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "0.48")[$0.48$]], [#math.equation(block: false, alt: "0.48 open parenthesis 40 close parenthesis")[$0.48 ( 40 )$]], )) + #math.equation(block: false, alt: "x plus y equals 40")[$x + y = 40$] + #math.equation(block: false, alt: "0.45 x plus 0.60 y equals 19.2")[$0.45 x + 0.60 y = 19.2$] + 32 lb Amal plans to make 10 liters of a 17% acid solution by mixing a 20% acid solution with a 15% acid solution. How much of each should she use? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Liters], [% acid], [Amount of acid]), [First solution], [], [], [], [Second solution], [], [], [], [Mixture], [], [], [], )) + Write one equation about the amount of solution Amal needs. + Write a second equation about the acid in the solution. + Solve the system and answer the question in the problem. Delbert answered 13 true-false and 9 fill-in questions correctly on his last test and got a score of 71. If he had answered 9 true-false and 13 fill-ins correctly, he would have made an 83. How many points was each type of problem worth? True-false: 2 points; fill-ins: 5 points In a recent election, 7179 votes were cast for the two candidates. If 6 votes had been switched from the winner to the loser, the loser would have won by 1 vote. How many votes were cast for each candidate? Because of prevailing winds, a flight from Detroit to Denver, a distance of 1120 miles, takes 4 hours on Econoflite, while the return trip takes 3.5 hours. What were the speed of the airplane and the speed of the wind? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Rate], [Time], [Distance]), [Detroit to Denver], [], [], [], [Denver to Detroit], [], [], [], )) + Write one equation about the trip from Detroit to Denver. + Write a second equation about the return trip. + Solve the system and answer the question in the problem. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Rate], [Time], [Distance]), [Detroit to Denver], [#math.equation(block: false, alt: "x plus y")[$x + y$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "1120")[$1120$]], [Denver to Detroit], [#math.equation(block: false, alt: "x plus y")[$x + y$]], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "1120")[$1120$]], )) + #math.equation(block: false, alt: "4 open parenthesis x minus y close parenthesis equals 1120")[$4 ( x − y ) = 1120$] + #math.equation(block: false, alt: "3.5 open parenthesis x plus y close parenthesis equals 1120")[$3.5 ( x + y ) = 1120$] + Airplane: 300 mph; wind: 20 mph On a breezy day, Bonnie propelled her human-powered aircraft 100 meters in 15 seconds going into the wind and made the return trip in 10 seconds with the wind. What were the speed of the wind and Bonnie's speed in still air? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Rate], [Time], [Distance]), [Against the wind], [], [], [], [With the wind], [], [], [], )) + Write one equation about Bonnie's initial flight. + Write a second equation about Bonnie's return trip. + Solve the system and answer the question in the problem. A cup of rolled oats provides 310 calories. A cup of rolled wheat flakes provides 290 calories. A new breakfast cereal combines wheat and oats to provide 302 calories per cup. How much of each grain does 1 cup of the cereal include? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Cups], [Calories per cup], [Calories]), [Oat flakes], [], [], [], [Wheat flakes], [], [], [], [Mixture], [], [——], [], )) + Write one equation about the amounts of each grain. + Write a second equation about the number of calories. + Solve the system and answer the question in the problem. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Cups], [Calories per cup], [Calories]), [Oat flakes], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "310")[$310$]], [#math.equation(block: false, alt: "310 x")[$310 x$]], [Wheat flakes], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "290")[$290$]], [#math.equation(block: false, alt: "290 y")[$290 y$]], [Mixture], [#math.equation(block: false, alt: "x plus y")[$x + y$]], [——], [#math.equation(block: false, alt: "310 x plus 290 y")[$310 x + 290 y$]], )) + #math.equation(block: false, alt: "x plus y equals 1")[$x + y = 1$] + #math.equation(block: false, alt: "310 x plus 290 y equals 302")[$310 x + 290 y = 302$] + 0.6 cup oats, 0.4 cup wheat Acme Motor Company is opening a new plant to produce chassis for two of its models, a sports coupe and a wagon. Each sports coupe requires a riveter for 3 hours and a welder for 4 hours; each wagon requires a riveter for 4 hours and a welder for 5 hours. The plant has available 120 hours of riveting and 155 hours of welding per day. How many of each model of chassis can it produce in a day? + Choose variables for the unknown quantities, and fill in the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([], [Sports coupes], [Wagons], [Total]), [Hours of riveting], [], [], [], [Hours of welding], [], [], [], )) + Write one equation about the hours of riveting. + Write a second equation about the hours of welding. + Solve the system and answer the question in the problem. For Problems 19-28, use a calculator to solve the system. #math.equation(block: true, alt: "4.8 x, minus, 3.5 y, equals 5.44; 2.7 x, plus, 1.3 y, equals 8.29")[$4.8 x & − & 3.5 y & = 5.44 \ 2.7 x & + & 1.3 y & = 8.29$] #math.equation(block: true, alt: "open parenthesis 2.3 , 1.6 close parenthesis")[$( 2.3 , 1.6 )$] #math.equation(block: true, alt: "6.4 x, plus, 2.3 y, equals minus 14.09; minus 5.2 x, minus, 3.7 y, equals minus 25.37")[$6.4 x & + & 2.3 y & = − 14.09 \ − 5.2 x & − & 3.7 y & = − 25.37$] #math.equation(block: true, alt: "0.9 x equals 25.78, plus 1.03 y; 0.25 x plus 0.3 y, equals 85.7")[$0.9 x = 25.78 & + 1.03 y \ 0.25 x + 0.3 y & = 85.7$] #math.equation(block: true, alt: "open parenthesis 182 , 134 close parenthesis")[$( 182 , 134 )$] #math.equation(block: true, alt: "0.02 x, equals 0.6 y minus 78.72; 1.1 y, equals minus 0.4 x plus 108.3")[$0.02 x & = 0.6 y − 78.72 \ 1.1 y & = − 0.4 x + 108.3$]