#set document(title: "5.2 Linear Equations in One Variable with Applications", 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)Linear Equations in One Variable with Applications #figure(figph[A man is lifting weights at a gym.], alt: "A man is lifting weights at a gym.", caption: [Most gyms have a monthly membership fee.]) === Learning Objectives After completing this section, you should be able to: + Solve linear equations in one variable using properties of equations. + Construct a linear equation to solve applications. + Determine equations with no solution or infinitely many solutions. + Solve a formula for a given variable. In this section, we will study linear equations in one variable. There are several real-world scenarios that can be represented by linear equations: taxi rentals with a flat fee and a rate per mile; cell phone bills that charge a monthly fee plus a separate rate per text; gym memberships with a monthly fee plus a rate per class taken; etc. For example, if you join your local gym at \$10 per month and pay \$5 per class, how many classes can you take if your gym budget is \$75 per month? === Linear Equations and Applications Solving any equation is like discovering the answer to a puzzle. The purpose of solving an equation is to find the value or values of the variable that makes the equation a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle! There are many types of equations that we will learn to solve. In this section, we will focus on a #strong[linear equation], which is an equation in one variable that can be written as #math.equation(block: true, alt: "a x plus b equals 0")[$a x + b = 0$] where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are real numbers and #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$], such that #math.equation(block: false, alt: "a")[$a$] is the coefficient of #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "b")[$b$] is the constant. To solve a linear equation, it is a good idea to have an overall strategy that can be used to solve any linear equation. In the Example 1, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier. #examplebox("Example 1")[Solving a Linear Equation Using a General Strategy][ Solve #math.equation(block: false, alt: "7 open parenthesis n minus 3 close parenthesis minus 8 equals minus 15")[$7 ( n − 3 ) − 8 = − 15$] #solutionbox[ #figure(table( columns: 3, align: left, inset: 6pt, table.header([#strong[Step 1:] Simplify each side of the equation as much as possible.], [Use the Distributive Property. Notice that each side of the equation is now simplified as much as possible.], [#math.equation(block: false, alt: "7 open parenthesis n minus 3 close parenthesis minus 8, equals, minus 15; 7 n minus 21 minus 8, equals, minus 15; 7 n minus 29, equals, minus 15")[$7 ( n − 3 ) − 8 & = & − 15 \ 7 n − 21 − 8 & = & − 15 \ 7 n − 29 & = & − 15$]]), [#strong[Step 2:] Collect all variable terms on one side of the equation.], [Nothing to do; all #math.equation(block: false, alt: "n")[$n$]-terms are on the left side.], [#math.equation(block: false, alt: "7 n minus 29 equals minus 15")[$7 n − 29 = − 15$]], [#strong[Step 3:] Collect constant terms on the other side of the equation.], [To get constants only on the right, add 29 to each side. Simplify.], [#math.equation(block: false, alt: "7 n minus 29 plus 29, equals, minus 15 plus 29; 7 n, equals, 14")[$7 n − 29 + 29 & = & − 15 + 29 \ 7 n & = & 14$]], [#strong[Step 4:] Make the coefficient of the variable term equal to 1.], [Divide each side by 7. Simplify.], [#math.equation(block: false, alt: "the fraction 7 n over 7 equals the fraction 14 over 7")[$frac(7 n, 7) = frac(14, 7)$] #math.equation(block: false, alt: "n equals 2")[$n = 2$]], [#strong[Step 5:] Check the solution.], [Let #math.equation(block: false, alt: "n equals 2")[$n = 2$] Subtract.], [Check: #math.equation(block: false, alt: "7 open parenthesis n minus 3 close parenthesis minus 8, equals, minus 15; 7 open parenthesis 2 minus 3 close parenthesis minus 8, equals ?, minus 15; 7 open parenthesis minus 1 close parenthesis minus 8, equals ?, minus 15; minus 7 minus 8, equals ?, minus 15; minus 15, equals, minus 15 ✓")[$7 ( n − 3 ) − 8 & = & − 15 \ 7 ( 2 − 3 ) − 8 & limits(=)^(?) & − 15 \ 7 ( − 1 ) − 8 & limits(=)^(?) & − 15 \ − 7 − 8 & limits(=)^(?) & − 15 \ − 15 & = & − 15 ✓$]], )) ] ] In Example 1, we used both the addition and division property of equations. All the properties of equations are summarized in table below. Basically, what you do to one side of the equation, you must do to the other side of the equation to preserve equality. #figure(table( columns: 3, align: left, inset: 6pt, table.header([Operation], [Property], [Example]), [Addition], [If #math.equation(block: false, alt: "a equals b")[$a = b$] Then #math.equation(block: false, alt: "a plus c equals b plus c")[$a + c = b + c$]], [#math.equation(block: false, alt: "2, equals, 2; 2 plus 3, equals, 2 plus 3; 5, equals, 5")[$2 & = & 2 \ 2 + 3 & = & 2 + 3 \ 5 & = & 5$]], [Subtraction], [If #math.equation(block: false, alt: "a equals b")[$a = b$] Then #math.equation(block: false, alt: "a minus c equals b minus c")[$a − c = b − c$]], [#math.equation(block: false, alt: "5, equals, 5; 5 minus 2, equals, 5 minus 2; 3, equals, 3")[$5 & = & 5 \ 5 − 2 & = & 5 − 2 \ 3 & = & 3$]], [Multiplication], [If #math.equation(block: false, alt: "a equals b")[$a = b$] Then #math.equation(block: false, alt: "a times c equals b times c")[$a ⋅ c = b ⋅ c$]], [#math.equation(block: false, alt: "3, equals, 3; 3 times 4, equals, 3 times 4; 12, equals, 12")[$3 & = & 3 \ 3 ⋅ 4 & = & 3 ⋅ 4 \ 12 & = & 12$]], [Division], [If #math.equation(block: false, alt: "a equals b")[$a = b$] Then #math.equation(block: false, alt: "a divided by c equals b divided by c")[$a ÷ c = b ÷ c$] for #math.equation(block: false, alt: "c not equal to 0")[$c ≠ 0$]], [#math.equation(block: false, alt: "8, equals, 8; 8 divided by 2, equals, 8 divided by 2; 4, equals, 4")[$8 & = & 8 \ 8 ÷ 2 & = & 8 ÷ 2 \ 4 & = & 4$]], )) #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Be careful to multiply and divide every term on each side of the equation. For example,] #math.equation(block: false, alt: "2 plus x equals the fraction x over 3")[$2 + x = frac(x, 3)$] #emph[is solved by multiplying BOTH sides of the equation by] 3 #emph[to get] #math.equation(block: false, alt: "3 open parenthesis 2 plus x close parenthesis equals 3 open parenthesis the fraction x over 3 close parenthesis")[$3 ( 2 + x ) = 3 ( frac(x, 3) )$] #emph[which gives] #math.equation(block: false, alt: "6 plus 3 x equals x")[$6 + 3 x = x$]. #emph[Using parentheses will help you remember to use the distributive property! A division example, such as] #math.equation(block: false, alt: "3 open parenthesis x plus 2 close parenthesis equals 6 x plus 9")[$3 ( x + 2 ) = 6 x + 9$], #emph[can be solved by dividing BOTH sides of the equation by] 3 #emph[to get] #math.equation(block: false, alt: "the fraction 3 open parenthesis x plus 2 close parenthesis over 3 equals the fraction 6 x plus 9 over 3 ,")[$frac(3 ( x + 2 ), 3) = frac(6 x + 9, 3) ,$] #emph[which then will lead to] #math.equation(block: false, alt: "x plus 2 equals 2 x plus 3")[$x + 2 = 2 x + 3$]. ] #examplebox("Example 2")[Solving a Linear Equation Using Properties of Equations][ Solve #math.equation(block: false, alt: "9 open parenthesis y minus 2 close parenthesis minus y equals 16 plus 7 y")[$9 ( y − 2 ) − y = 16 + 7 y$]. #solutionbox[ #strong[Step 1:] Simplify each side. #math.equation(block: true, alt: "9 open parenthesis y minus 2 close parenthesis minus y, equals, 16 plus 7 y; 9 y minus 18 minus y, equals, 16 plus 7 y; 8 y minus 18, equals, 16 plus 7 y")[$9 ( y − 2 ) − y & = & 16 + 7 y \ 9 y − 18 − y & = & 16 + 7 y \ 8 y − 18 & = & 16 + 7 y$] #strong[Step 2:] Collect all variables on one side. #math.equation(block: true, alt: "8 y minus 18 minus 7 y, equals, 16 plus 7 y minus 7 y; y minus 18, equals, 16")[$8 y − 18 − 7 bold(italic(y)) & = & 16 + 7 y − 7 bold(italic(y)) \ y − 18 & = & 16$] #strong[Step 3:] Collect constant terms on one side. #math.equation(block: true, alt: "y minus 18 plus 18, equals, 16 plus 18; y, equals, 34")[$y − 18 + 18 & = & 16 + 18 \ y & = & 34$] #strong[Step 4:] Make the coefficient of the variable 1. Already done! #linebreak() #strong[Step 5:] Check. #math.equation(block: true, alt: "9 open parenthesis 34 close parenthesis minus 18 minus open parenthesis 34 close parenthesis, equals ?, 16 plus 7 open parenthesis 34 close parenthesis; 306 minus 18 minus 34, equals ?, 16 plus 238; 288 minus 34, equals ?, 254; 254, equals, 254 ✓")[$9 ( 34 ) − 18 − ( 34 ) & limits(=)^(?) & 16 + 7 ( 34 ) \ 306 − 18 − 34 & limits(=)^(?) & 16 + 238 \ 288 − 34 & limits(=)^(?) & 254 \ 254 & = & 254 ✓$] ] ] #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[ #emph[Who Invented the Symbol for Equals ?] Before the creation of a symbol for equality, it was usually expressed with a word that meant equals, such as #emph[aequales] (Latin), #emph[esgale] (French), or #emph[gleich] (German). Welsh mathematician and physician Robert Recorde is given credit for inventing the modern sign. It first appears in writing in #emph[The Whetstone of Witte], a book Recorde wrote about algebra, which was published in 1557. In this book, Recorde states, "I will set as I do often in work use, a pair of parallels, or Gemowe (twin) lines of one length, thus: ===, because no two things can be more equal." Although his version of the sign was a bit longer than the one we use today, his idea stuck and "=" is used throughout the world to indicate equality in mathematics. ] In Algebraic Expressions, you translated an English sentence into an equation. In this section, we take that one step further and translate an English paragraph into an equation, and then we solve the equation. We can go back to the opening question in this section: #emph[If you join your local gym at \$10 per month and pay \$5 per class, how many classes can you take if your gym budget is \$75 per month?] We can create an equation for this scenario and then solve the equation (see Example 4). #examplebox("Example 3")[Constructing a Linear Equation to Solve an Application][ The Beaudrie family has two cats, Basil and Max. Together, they weigh 23 pounds. Basil weighs 16 pounds. How much does Max weigh? #solutionbox[ Let #math.equation(block: false, alt: "b")[$b$] = Basil’s weight and #math.equation(block: false, alt: "m")[$m$] = Max’s weight. #math.equation(block: true, alt: "b plus m equals 23")[$b + m = 23$] We also know that Basil weighs 16 pounds so: #strong[Steps 1 and 2:] #math.equation(block: false, alt: "16 plus m equals 23")[$16 + m = 23$] Since both sides are simplified, the variable is on one side of the equation, we start in Step 3 and collect the constants on one side: #strong[Step 3:] #math.equation(block: true, alt: "16 plus m minus 16, equals, 23 minus 16; m, equals, 7")[$16 + m − 16 & = & 23 − 16 \ m & = & 7$] #strong[Step 4:] is already done so we go to Step 5: #strong[Step 5:] #math.equation(block: true, alt: "16 plus 7, equals ?, 23; 23, equals, 23 ✓")[$16 + 7 & limits(=)^(?) & 23 \ 23 & = & 23 ✓$] Basil weighs 16 pounds and Max weighs 7 pounds. ] ] #examplebox("Example 4")[Constructing a Linear Equation to Solve Another Application][ If you join your local gym at \$10 per month and pay \$5 per class, how many classes can you take if your gym budget is \$75 per month? #solutionbox[ If we let #math.equation(block: false, alt: "x")[$x$] = number of classes, the expression #math.equation(block: false, alt: "5 x plus 10")[$5 x + 10$] would represent what you pay per month if each class is \$5 and there’s a \$10 monthly fee per class. \$10 is your constant. If you want to know how many classes you can take if you have a \$75 monthly gym budget, set the equation equal to 75. Then solve the equation #math.equation(block: false, alt: "5 x plus 10 equals 75")[$5 x + 10 = 75$] for #math.equation(block: false, alt: "x")[$x$]. #strong[Steps 1 and 2:] #math.equation(block: true, alt: "5 x plus 10 equals 75")[$5 x + 10 = 75$] #strong[Step 3:] #math.equation(block: true, alt: "5 x plus 10 minus 10, equals, 75 minus 10; 5 x, equals, 65")[$5 x + 10 − 10 & = & 75 − 10 \ 5 x & = & 65$] #strong[Step 4:] #math.equation(block: true, alt: "the fraction 5 x over 5, equals, the fraction 65 over 5; x, equals, 13")[$frac(5 x, 5) & = & frac(65, 5) \ x & = & 13$] #strong[Step 5:] #math.equation(block: true, alt: "5 open parenthesis 13 close parenthesis plus 10, equals ?, 75; 65 plus 10, equals ?, 75; 75, equals, 75 ✓")[$5 ( 13 ) + 10 & limits(=)^(?) & 75 \ 65 + 10 & limits(=)^(?) & 75 \ 75 & = & 75 ✓$] The solution is 13 classes. You can take 13 classes on a \$75 monthly gym budget. ] ] #examplebox("Example 5")[Constructing an Application from a Linear Equation][ Write an application that can be solved using the equation #math.equation(block: false, alt: "50 x plus 35 equals 185")[$50 x + 35 = 185$]. Then solve your application. #solutionbox[ Answers will vary. Let’s say you want to rent a snowblower for a huge winter storm coming up. If #math.equation(block: false, alt: "x")[$x$] = the number of days you rent a snowblower, then the expression #math.equation(block: false, alt: "50 x plus 35")[$50 x + 35$] represents what you pay if, for each day, it costs \$50 to rent the snowblower and there is a \$35 flat rental fee. \$35 is the constant. To find out how many days you can rent a snowblower for \$185, set the expression equal to 185. Then solve the equation #math.equation(block: false, alt: "50 x plus 35 equals 185")[$50 x + 35 = 185$] for #math.equation(block: false, alt: "x")[$x$]. #strong[Steps 1 and 2:] #math.equation(block: true, alt: "50 x plus 35 equals 185")[$50 x + 35 = 185$] #strong[Step 3:] #math.equation(block: true, alt: "50 x plus 35 minus 35, equals, 185 minus 35; 50 x, equals, 150")[$50 x + 35 − 35 & = & 185 − 35 \ 50 x & = & 150$] #strong[Step 4:] #math.equation(block: true, alt: "the fraction 50 x over 50, equals, the fraction 150 over 50; x, equals, 3")[$frac(50 x, 50) & = & frac(150, 50) \ x & = & 3$] #strong[Step 5:] #math.equation(block: true, alt: "50 open parenthesis 3 close parenthesis plus 35, equals ?, 185; 150 plus 35, equals ?, 185; 185, equals, 185 ✓")[$50 ( 3 ) + 35 & limits(=)^(?) & 185 \ 150 + 35 & limits(=)^(?) & 185 \ 185 & = & 185 ✓$] The equation is #math.equation(block: false, alt: "50 x plus 35 equals 185")[$50 x + 35 = 185$] and the solution is 3 days. You can rent a snowblower for 3 days on a \$185 budget. ] ] === Linear Equations with No Solutions or Infinitely Many Solutions Every linear equation we have solved thus far has given us one numerical solution. Now we'll look at linear equations for which there are no solutions or infinitely many solutions. #examplebox("Example 6")[Solving a Linear Equation with No Solution][ Solve #math.equation(block: false, alt: "3 open parenthesis x plus 4 close parenthesis equals 4 x plus 8 minus x")[$3 ( x + 4 ) = 4 x + 8 − x$]. #solutionbox[ #strong[Step 1:] Simplify each side. #math.equation(block: false, alt: "3 open parenthesis x plus 4 close parenthesis equals 4 x plus 8 minus x")[$3 ( x + 4 ) = 4 x + 8 − x$] #math.equation(block: true, alt: "3 x plus 12 minus 3 x equals 3 x plus 8 minus 3 x")[$3 x + 12 − 3 x = 3 x + 8 − 3 x$] #strong[Step 2:] Collect all variables to one side. #math.equation(block: false, alt: "3 x plus 12 minus 3 x equals 3 x plus 8 minus 3 x")[$3 x + 12 − 3 bold(italic(x)) = 3 x + 8 − 3 bold(italic(x))$] #math.equation(block: true, alt: "12 equals 8")[$12 = 8$] The variable #math.equation(block: false, alt: "x")[$x$] disappeared! When this happens, you need to examine what remains. In this particular case, we have #math.equation(block: false, alt: "12 equals 8")[$12 = 8$], which is not a true statement. When you have a false statement, then you know the equation has no solution; there does not exist a value for #math.equation(block: false, alt: "x")[$x$] that can be put into the equation that will make it true. ] ] #examplebox("Example 7")[Solving a Linear Equation with Infinitely Many Solutions][ Solve #math.equation(block: false, alt: "2 open parenthesis x plus 5 close parenthesis equals 4 open parenthesis x plus 3 close parenthesis minus 2 x minus 2")[$2 ( x + 5 ) = 4 ( x + 3 ) − 2 x − 2$]. #solutionbox[ #strong[Step 1:] #math.equation(block: true, alt: "2 open parenthesis x plus 5 close parenthesis, equals, 4 open parenthesis x plus 3 close parenthesis minus 2 x minus 2; 2 x plus 10, equals, 4 x plus 12 minus 2 x minus 2; 2 x plus 10, equals, 2 x plus 10")[$2 ( x + 5 ) & = & 4 ( x + 3 ) − 2 x − 2 \ 2 x + 10 & = & 4 x + 12 − 2 x − 2 \ 2 x + 10 & = & 2 x + 10$] #strong[Step 2:] #math.equation(block: true, alt: "2 x plus 10 minus 2 x, equals, 2 x plus 10 minus 2 x; 10, equals, 10")[$2 x + 10 − 2 bold(italic(x)) & = & 2 x + 10 − 2 bold(italic(x)) \ 10 & = & 10$] As with the previous example, the variable disappeared. In this case, however, we have a true statement (#math.equation(block: false, alt: "10 equals 10")[$10 = 10$]). When this occurs we say there are infinitely many solutions; any value for #math.equation(block: false, alt: "x")[$x$] will make this statement true. ] ] === Solving a Formula for a Given Variable You are probably familiar with some geometry formulas. A #strong[formula] is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be able to manipulate formulas and solve for specific variables. To solve a formula for a specific variable means to isolate that variable on one side of the equal sign with a coefficient of 1. All other variables and constants are on the other side of the equal sign. To see how to solve a formula for a specific variable, we will start with the distance, rate, and time formula. #examplebox("Example 8")[Solving for a Given Variable with Distance, Rate, and Time][ Solve the formula #math.equation(block: false, alt: "d equals r t")[$d = r t$] for #math.equation(block: false, alt: "t")[$t$]. This is the distance formula where #math.equation(block: false, alt: "d")[$d$] = distance, #math.equation(block: false, alt: "r")[$r$] = rate, and #math.equation(block: false, alt: "t")[$t$] = time. #solutionbox[ Divide both sides by #math.equation(block: false, alt: "r")[$r$]: #math.equation(block: false, alt: "d / r equals r t / r")[$d / bold(italic(r)) = r t / bold(italic(r))$] #math.equation(block: true, alt: "d / r equals t")[$d / r = t$] ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Solving_for_a_variable")[Solving for a Variable in an Equation] ] #examplebox("Example 9")[Solving for a Given Variable in the Area Formula for a Triangle][ Solve the formula #math.equation(block: false, alt: "A equals ½")[$A = ½$] #math.equation(block: false, alt: "bh")[$italic("bh")$] for #math.equation(block: false, alt: "h")[$h$]. This is the area formula of a triangle where #math.equation(block: false, alt: "A")[$A$] = area, #math.equation(block: false, alt: "b")[$b$] = base, and #math.equation(block: false, alt: "h")[$h$] = height. #solutionbox[ #strong[Step 1:] Multiply both sides by 2. #math.equation(block: true, alt: "2 A equals 2 open parenthesis ½ b h close parenthesis; 2 A equals b h")[$2 A = 2 ( ½ #h(0.28em) b h ) \ 2 A = b h$] #strong[Step 2:] Divide both sides by #math.equation(block: false, alt: "b")[$b$]. #math.equation(block: true, alt: "the fraction 2 A over b, equals, the fraction b h over b; the fraction 2 A over b, equals, h; h, equals, the fraction 2 A over b")[$frac(2 A, b) & = & frac(b h, b) \ frac(2 A, b) & = & h \ h & = & frac(2 A, b)$] ] ] #notebox("Work It Out", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ #emph[Using Algebra to Understand Card Tricks] You will need to perform this card trick with another person. Before you begin, the two people must first decide which of the two will be the #emph[Dealer] and which will be the #emph[Partner], as each will do something different. Once you have decided upon that, follow the steps here: #strong[Step 1:] Dealer and Partner: Take a regular deck of 52 cards, and remove the face cards and the 10s. #strong[Step 2:] Dealer and Partner: Shuffle the remaining cards #strong[Step 3:] Dealer and Partner: Select one card each, but keep them face down and don’t look at them yet. #strong[Step 4:] Dealer: Look at your card (just the Dealer!). Multiply its value by 2 (Aces = 1). #strong[Step 5:] Dealer: Add 2 to this result. #strong[Step 6:] Dealer: Multiply your answer by 5. #strong[Step 7:] Partner: Look at your card. #strong[Step 8:] Partner: Calculate: 10 - your card, and tell this information to the dealer. #strong[Step 9:] Dealer: Subtract the value the Partner tells you from your total to get a final answer. #strong[Step 10:] Dealer: verbally state the final answer. #strong[Step 11:] Dealer and Partner: Turn over your cards. Now, answer the following questions + Did the trick work? How do you know? + Why did this occur? In other words, how does this trick work? ] === Key Terms - linear equation === Key Concepts - Solving linear equations means discovering what the value of the variable in a linear equation represents in the given conditions. - When solving a linear equation, most often you will have one solution; however, a linear equation may have no solutions or infinitely many solutions. === Videos - #link("https://openstax.org/r/Solving_for_a_variable")[Solving for a Variable in an Equation]