#set document(title: "10.2 Linear Equations and Inequalities", 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")) == 10.2#h(0.6em)Linear Equations and Inequalities An #strong[equation] is just a mathematical statement that two expressions are equal. Equations relating two variables are particularly useful. If we know the value of one of the variables, we can find the corresponding value of the other variable by solving the equation. #examplebox("Example 1")[][ The equation #math.equation(block: false, alt: "w equals 6 h")[$w = 6 h$] gives Loren's wages, #math.equation(block: false, alt: "w")[$w$], in terms of the number of hours she works, #math.equation(block: false, alt: "h")[$h$]. How many hours does Loren need to work next week if she wants to earn \$#math.equation(block: false, alt: "225")[$225$]? #solutionbox[ We know that #math.equation(block: false, alt: "w equals 225")[$w = 225$], and we would like to know the value of #math.equation(block: false, alt: "h")[$h$]. We substitute the value for #math.equation(block: false, alt: "w")[$w$] into our equation and then solve for #math.equation(block: false, alt: "h")[$h$]. #math.equation(block: true, alt: "w, equals 6 h, Substitute 225 for w .; 225, equals 6 h, Divide both sides by 6.; the fraction 225 over 6, equals the fraction 6 h over 6, Simplify.; 375.5, equals h")[$w & = 6 h & & "Substitute 225 for " bold(italic(w)) . \ 225 & = 6 h & & "Divide both sides by " 6. \ frac(225, 6) & = frac(6 h, 6) & & "Simplify." \ 375.5 & = h$] Loren must work #math.equation(block: false, alt: "37.5")[$37.5$] hours in order to earn \$#math.equation(block: false, alt: "225")[$225$]. In reality, Loren will probably have to work for #math.equation(block: false, alt: "38")[$38$] hours, because most employers do not pay for portions of an hour's work. Thus, Loren needs to work for #math.equation(block: false, alt: "38")[$38$] hours. ] ] To solve an equation we can generate simpler equations that have the same solutions. Equations that have identical solutions are called #strong[equivalent equations]. For example, #math.equation(block: true, alt: "3 x minus 5 equals x plus 3")[$3 x − 5 = x + 3$] and #math.equation(block: true, alt: "2 x equals 8")[$2 x = 8$] are equivalent equations because the solution of each equation is #math.equation(block: false, alt: "4")[$4$]. Often we can find simpler equivalent equations by undoing in reverse order the operations performed on the variable. === Solving Linear Equations #strong[Linear], or first-degree, equations can be written so that every term is either a constant or a constant times the variable. The equations above are examples of linear equations. Recall the following rules for solving linear equations. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Generate Equivalent Equations] + We can add or subtract the #emph[same] number on #emph[both] sides of an equation. + We can multiply or divide #emph[both] sides of an equation by the #emph[same] number (except zero). ] Applying either of these rules produces a new equation equivalent to the old one and thus preserves the solution. We use the rules to isolate the variable on one side of the equation. #examplebox("Example 2")[][ Solve the equation #math.equation(block: false, alt: "3 x minus 5 equals x plus 3")[$" " " " 3 x − 5 = x + 3$]. #solutionbox[ We first collect all the variable terms on one side of the equation, and the constant terms on the other side. #math.equation(block: true, alt: "3 x minus 5 minus x, equals x plus 3 minus x, Subtract x from both sides.; 2 x minus 5, equals 3, Simplify.; 2 x minus 5 plus 5, equals 5 plus 5, Add 5 to both sides.; 2 x, equals 8, Simplify.; the fraction 2 x over 2, equals the fraction 8 over 2, Divide both sides by 2.; x, equals 4, Simplify.")[$3 x − 5 − bold(italic(x)) & = x + 3 − bold(italic(x)) & & "Subtract " bold(italic(x)) " from both sides." \ 2 x − 5 & = 3 & & "Simplify." \ 2 x − 5 + 5 & = 5 + 5 & & "Add 5 to both sides." \ 2 x & = 8 & & "Simplify." \ frac(2 x, 2) & = frac(8, 2) & & "Divide both sides by 2." \ x & = 4 & & "Simplify."$] The solution is #math.equation(block: false, alt: "4")[$4$]. (You can check the solution by substituting #math.equation(block: false, alt: "4")[$4$] into the original equation to show that a true statement results.) ] ] The following steps should enable you to solve any linear equation. Of course, you may not need all the steps for a particular equation. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Solve a Linear Equation:] + Simplify each side of the equation separately. + Apply the distributive law to remove parentheses. + Collect like terms. + By adding or subtracting appropriate terms on both sides of the equation, get all the variable terms on one side and all the constant terms on the other. + Divide both sides of the equation by the coefficient of the variable. ] #examplebox("Example 3")[][ Solve #math.equation(block: false, alt: "3 open parenthesis 2 x minus 5 close parenthesis minus 4 x equals 2 x minus open parenthesis 6 minus 3 x close parenthesis")[$" " " " 3 ( 2 x − 5 ) − 4 x = 2 x − ( 6 − 3 x )$]. #solutionbox[ We begin by simplifying each side of the equation. #math.equation(block: true, alt: "3 open parenthesis 2 x minus 5 close parenthesis minus 4 x, equals 2 x minus open parenthesis 6 minus 3 x close parenthesis, Apply the distributive law.; 6 x minus 15 minus 4 x, equals 2 x minus 6 plus 3 x, Combine like terms on each side.; 2 x minus 15, equals 5 x minus 6")[$3 ( 2 x − 5 ) − 4 x & = 2 x − ( 6 − 3 x ) & & "Apply the distributive law." \ 6 x − 15 − 4 x & = 2 x − 6 + 3 x & & "Combine like terms on each side." \ 2 x − 15 & = 5 x − 6$] Next, we collect all the variable terms on the left side of the equation, and all the constant terms on the right side. #math.equation(block: true, alt: "2 x minus 15 minus 5 x plus 15, equals 5 x minus 6 minus 5 x plus 15, Add minus 5 x plus 15 to both sides.; minus 3 x, equals 9")[$2 x − 15 − 5 bold(italic(x)) + 15 & = 5 x − 6 − 5 bold(italic(x)) + 15 & & "Add " − 5 bold(italic(x)) + 15 " to both sides." \ − 3 x & = 9$] Finally, we divide both sides of the equation by the coefficient of the variable. #math.equation(block: true, alt: "minus 3 x, equals 9, Divide both sides by minus 3.; x, equals minus 3")[$− 3 x & = 9 & & "Divide both sides by " − 3. \ x & = − 3$] The solution is #math.equation(block: false, alt: "minus 3")[$− 3$]. ] ] === Formulas A #strong[formula] is an equation that relates several variables. For example, the equation #math.equation(block: true, alt: "P equals 2 l plus 2 w")[$P = 2 l + 2 w$] gives the perimeter of a rectangle in terms of its length and width. Suppose we have some wire fence to enclose an exercise area for rabbits, and we would like to see what dimensions are possible for different rectangles with that perimeter. In this case, it would be more useful to have a formula for the length of the rectangle in terms of its perimeter and its width. We can find such a formula by solving the perimeter formula for #math.equation(block: false, alt: "l")[$l$] in terms of #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "w")[$w$]. #math.equation(block: true, alt: "2 l plus 2 w, equals P, Subtact 2 w from both sides.; 2 l, equals P minus 2 w, Divide both sides by 2.; l, equals the fraction P minus 2 w over 2")[$2 l + 2 w & = P & & "Subtact " 2 bold(italic(w)) " from both sides." \ 2 l & = P − 2 w & & "Divide both sides by 2." \ l & = frac(P − 2 w, 2)$] The result is a new formula that gives the length of a rectangle in terms of its perimeter and its width. #examplebox("Example 4")[][ The formula #math.equation(block: false, alt: "5 F equals 9 C plus 160")[$" " 5 F = 9 C + 160 " "$] relates the temperature in degrees Fahrenheit, #math.equation(block: false, alt: "F")[$F$], to the temperature in degrees Celsius, #math.equation(block: false, alt: "C")[$C$]. Solve the formula for #math.equation(block: false, alt: "C")[$C$] in terms of #math.equation(block: false, alt: "F")[$F$]. #solutionbox[ We begin by isolating the term that contains #math.equation(block: false, alt: "C")[$C$]. #math.equation(block: true, alt: "5 F, equals 9 C plus 160, Subtract 160 from both sides.; 5 F minus 160, equals 9 C, Divide both sides by 9.; the fraction 5 F minus 160 over 9, equals C")[$5 F & = 9 C + 160 & & "Subtract 160 from both sides." \ 5 F − 160 & = 9 C & & "Divide both sides by 9." \ frac(5 F − 160, 9) & = C$] We can also write the formula for #math.equation(block: false, alt: "C")[$C$] in terms of #math.equation(block: false, alt: "F")[$F$] as #math.equation(block: false, alt: "C equals the fraction 5 over 9 F minus the fraction 160 over 9")[$" " C = display(frac(5, 9)) F − display(frac(160, 9))$]. ] ] #examplebox("Example 5")[][ Solve #math.equation(block: false, alt: "3 x minus 5 y equals 40")[$" " 3 x − 5 y = 40 " "$] for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$]. #solutionbox[ We isolate #math.equation(block: false, alt: "y")[$y$] on one side of the equation. #math.equation(block: true, alt: "3 x minus 5 y, equals 40, Subtract 3 x from both sides.; minus 5 y, equals 40 minus 3 x, Divide both sides by minus 5.; the fraction minus 5 y over minus 5, equals the fraction 40 minus 3 x over minus 5, Simplify both sides.; y, equals minus 8 plus the fraction 3 over 5 x")[$3 x − 5 y & = 40 & & "Subtract " 3 bold(italic(x)) " from both sides." \ − 5 y & = 40 − 3 x & & "Divide both sides by " − 5. \ frac(− 5 y, − 5) & = frac(40 − 3 x, − 5) & & "Simplify both sides." \ y & = − 8 + frac(3, 5) x$] ] ] === Linear Inequalities The symbol #math.equation(block: false, alt: "greater than")[$>$] is called an #strong[inequality symbol], and the statement #math.equation(block: false, alt: "a greater than b")[$a > b$] is called an #strong[inequality]. There are four inequality symbols: #math.equation(block: true, alt: "greater than, is greater than; less than, is less than; greater than or equal to, is greater than or equal to; less than or equal to, is less than or equal to")[$& > & & "is greater than" \ & < & & "is less than" \ & ≥ & & "is greater than or equal to" \ & ≤ & & "is less than or equal to"$] Inequalities that include the symbols #math.equation(block: false, alt: "greater than")[$>$] or #math.equation(block: false, alt: "less than or equal to")[$≤$] are called #strong[strict inequalities]; those that include #math.equation(block: false, alt: "greater than or equal to")[$≥$] or #math.equation(block: false, alt: "less than or equal to")[$≤$] are called #strong[nonstrict]. If we multiply or divide both sides of an inequality by a negative number, the direction of the inequality must be reversed. For example, if we multiply both sides of the inequality #math.equation(block: true, alt: "2 less than 5")[$2 < 5$] by #math.equation(block: false, alt: "minus 3")[$− 3$], we get #math.equation(block: true, alt: "minus 3 open parenthesis 2 close parenthesis, greater than minus 3 open parenthesis 5 close parenthesis, Change inequality symbol from less than to greater than .; minus 6, greater than minus 15")[$− 3 ( 2 ) & > − 3 ( 5 ) & & "Change inequality symbol from " < " to " > . \ − 6 & > − 15$] Because of this property, the rules for solving linear equations must be revised slightly for solving linear inequalities. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Solve a Linear Inequality:] + We may add or subtract the same number to both sides of an inequality without changing its solutions. + We may multiply or divide both sides of an inequality by a #emph[positive] number without changing its solutions. + If we multiply or divide both sides of an inequality by a #emph[negative] number, we must #emph[reverse the direction of the inequality symbol]. ] #examplebox("Example 6")[][ Solve the inequality #math.equation(block: false, alt: "4 minus 3 x greater than or equal to minus 17")[$" " 4 − 3 x ≥ − 17$]. #solutionbox[ Use the rules above to isolate #math.equation(block: false, alt: "x")[$x$] on one side of the inequality. #math.equation(block: true, alt: "4 minus 3 x, greater than or equal to minus 17, Subtract 4 from both sides.; minus 3 x, greater than or equal to minus 21, Divide both sides by minus 3.; x, less than or equal to 7")[$4 − 3 x & ≥ − 17 & & "Subtract 4 from both sides." \ − 3 x & ≥ − 21 & & "Divide both sides by " − 3. \ x & ≤ 7$] Notice that we reversed the direction of the inequality when we divided by #math.equation(block: false, alt: "minus 3")[$− 3$]. Any number less than or equal to #math.equation(block: false, alt: "7")[$7$] is a solution of the inequality. ] ] A #strong[compound inequality] involves two inequality symbols. #examplebox("Example 7")[][ Solve #math.equation(block: false, alt: "4 less than or equal to 3 x plus 10 less than or equal to 16")[$" " 4 ≤ 3 x + 10 ≤ 16$]. #solutionbox[ We isolate #math.equation(block: false, alt: "x")[$x$] by performing the same operations on all three sides of the inequality. #math.equation(block: true, alt: "4, less than or equal to 3 x, plus, 10, less than or equal to 16, Subtract 10.; minus 6, less than or equal to, 3 x, less than or equal to 6, Divide by 3.; minus 2, less than or equal to, x, less than or equal to 2")[$4 & ≤ 3 x & + & 10 & ≤ 16 #hide($b l a n k$) & "Subtract " 10. \ − 6 & ≤ & 3 x & & ≤ 6 #hide($b l a n k$) & "Divide by " 3. \ − 2 & ≤ & x & & ≤ 2 #hide($b l a n k$) &$] The solutions are all numbers between #math.equation(block: false, alt: "minus 2")[$− 2$] and #math.equation(block: false, alt: "2")[$2$], inclusive. ] ] === Interval Notation The solutions of the inequality in Example form an interval. An #strong[interval] is a set that consists of all the real numbers between two numbers #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$]. The set #math.equation(block: false, alt: "minus 2 less than or equal to x less than or equal to 2")[$− 2 ≤ x ≤ 2$] includes its endpoints #math.equation(block: false, alt: "minus 2")[$− 2$] and #math.equation(block: false, alt: "2")[$2$], so we call it a #strong[closed interval], and we denote it by #math.equation(block: false, alt: "open bracket minus 2 , 2 close bracket")[$[ − 2 , 2 ]$]. Its graph is shown in figure (a). The square brackets tell us that the endpoints are included in the interval. An interval that does not include its endpoints, such as #math.equation(block: false, alt: "minus 2 less than x less than 2")[$− 2 < x < 2$], is called an #strong[open interval], and we denote it with round brackets, #math.equation(block: false, alt: "open parenthesis minus 2 , 2 close parenthesis")[$( − 2 , 2 )$]. Its graph is shown in figure (b). #figure(figph[closed interval and open interva], alt: "closed interval and open interva", caption: none) #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Do not confuse the open interval #math.equation(block: false, alt: "open parenthesis minus 2 , 2 close parenthesis")[$( − 2 , 2 )$] with the point #math.equation(block: false, alt: "open parenthesis minus 2 , 2 close parenthesis")[$( − 2 , 2 )$]! The notation is the same, so you must decide from the context whether an interval or a point is being discussed. ] We can also discuss #strong[infinite intervals], such as #math.equation(block: false, alt: "x less than 3")[$" " x < 3 " "$] and #math.equation(block: false, alt: "x greater than or equal to minus 1")[$" " x ≥ − 1 " "$], shown in the figure below. We denote the interval #math.equation(block: false, alt: "x less than 3")[$" " x < 3 " "$] by #math.equation(block: false, alt: "open parenthesis minus ∞ , 3 close parenthesis")[$( − upright(∞) , 3 )$], and the interval #math.equation(block: false, alt: "x greater than or equal to minus 1")[$" " x ≥ − 1 " "$] by #math.equation(block: false, alt: "open bracket minus 1 , ∞ close parenthesis")[$[ − 1 , upright(∞) )$]. The symbol #math.equation(block: false, alt: "∞")[$upright(∞)$], for infinity, does not represent a specific real number; it indicates that the interval continues forever along the real line. #figure(figph[infinite intervals], alt: "infinite intervals", caption: none) Finally, we can combine two or more intervals into a larger set. For example, the set consisting of #math.equation(block: false, alt: "x less than minus 1")[$x < − 1$] or #math.equation(block: false, alt: "x greater than 2")[$x > 2$], shown below, is the #strong[union] of two intervals and is denoted by #math.equation(block: false, alt: "open parenthesis minus ∞ , minus 2 close parenthesis union open parenthesis 2 , ∞ close parenthesis")[$( − upright(∞) , − 2 ) ∪ ( 2 , upright(∞) )$]. #figure(figph[number line with two disjoint infinite intervals], alt: "number line with two disjoint infinite intervals", caption: none) Many solutions of inequalities are intervals or unions of intervals. #examplebox("Example 8")[][ Write each of the solution sets with interval notation and graph the solution set on a number line. + #math.equation(block: false, alt: "3 less than or equal to x less than 6")[$3 ≤ x < 6$] + #math.equation(block: false, alt: "x greater than or equal to minus 9")[$x ≥ − 9$] + #math.equation(block: false, alt: "x less than or equal to 1 or x greater than 4")[$x ≤ 1 " " " " " or " " " " " x > 4$] + #math.equation(block: false, alt: "minus 8 less than x less than or equal to minus 5 or minus 1 less than or equal to x less than 3")[$− 8 < x ≤ − 5 " " " " " or " " " " " − 1 ≤ x < 3$] #solutionbox[ + #math.equation(block: false, alt: "open bracket 3 , 6 close parenthesis")[$[ 3 , 6 )$]. This is called a #strong[half-open] or #strong[half-closed] interval. #figure(figph[half-open interval], alt: "half-open interval", caption: none) + #math.equation(block: false, alt: "open bracket minus 9 , ∞ close parenthesis")[$[ − 9 , upright(∞) )$]. We always use round brackets next to the symbol #math.equation(block: false, alt: "∞")[$upright(∞)$] because #math.equation(block: false, alt: "∞")[$upright(∞)$] is not a specific number and is not included in the set. #figure(figph[half-closed infinite interval], alt: "half-closed infinite interval", caption: none) + #math.equation(block: false, alt: "open parenthesis minus ∞ , 1 close bracket union open parenthesis 4 , ∞ close parenthesis")[$( − upright(∞) , 1 ] ∪ ( 4 , upright(∞) )$]. The word #emph[or] describes the union of two sets. #figure(figph[disjoint infinite intervals], alt: "disjoint infinite intervals", caption: none) + #math.equation(block: false, alt: "open parenthesis minus 8 , minus 5 close bracket union open bracket minus 1 , 3 close parenthesis")[$( − 8 , − 5 ] ∪ [ − 1 , 3 )$]. #figure(figph[disjoint finite intervals], alt: "disjoint finite intervals", caption: none) ] ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Equation - Closed interval - Compound inequality - Inequality - Linear equation - Equivalent equation - Open interval - Interval - Strict inequality - Formula - Solve an equation - Union ==== SKILLS Practice each skill in the exercises listed. + Solve a linear equation: \#1–10 + Solve a formula for one variable in terms of the others: \#11–28 + Solve a linear inequality: \#26–34 + Solve a compound inequality: \#35–41 + Write solutions to inequalities in interval notation: \#41–50 === Exercises A.2 For Problems 1-10, solve the linear equation. #math.equation(block: true, alt: "3 x plus 5 equals 26")[$3 x + 5 = 26$] #math.equation(block: true, alt: "7")[$7$] #math.equation(block: true, alt: "2 plus 5 x equals 37")[$2 + 5 x = 37$] #math.equation(block: true, alt: "3 open parenthesis z plus 2 close parenthesis equals 37")[$3 ( z + 2 ) = 37$] #math.equation(block: true, alt: "the fraction 31 over 3")[$display(frac(31, 3))$] #math.equation(block: true, alt: "2 open parenthesis z minus 3 close parenthesis equals 15")[$2 ( z − 3 ) = 15$] #math.equation(block: true, alt: "3 y minus 2 open parenthesis y minus 4 close parenthesis equals 12 minus 5 y")[$3 y − 2 ( y − 4 ) = 12 − 5 y$] #math.equation(block: true, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] #math.equation(block: true, alt: "5 y minus 3 open parenthesis y plus 1 close parenthesis equals 14 plus 2 y")[$5 y − 3 ( y + 1 ) = 14 + 2 y$] #math.equation(block: true, alt: "6")[$6$] #math.equation(block: true, alt: "0.8 w minus 2.6 equals 1.4 w plus 0.3")[$0.8 w − 2.6 = 1.4 w + 0.3$] #math.equation(block: true, alt: "minus 4.83 bar")[$− 4.8 limits(3)^(―)$] #math.equation(block: true, alt: "4.8 minus 1.3 w equals 0.7 w plus 2.1")[$4.8 − 1.3 w = 0.7 w + 2.1$] #math.equation(block: true, alt: "0.25 t plus 0.10 open parenthesis t minus 4 close parenthesis equals 11.60")[$0.25 t + 0.10 ( t − 4 ) = 11.60$] #math.equation(block: true, alt: "34.29")[$34.29$] #math.equation(block: true, alt: "0.12 t plus 0.08 open parenthesis t plus 10 , 000 close parenthesis equals 12 , 000")[$0.12 t + 0.08 ( t + 10 , 000 ) = 12 , 000$] For problems 11-20, solve for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "4 x plus 3 y equals minus 2")[$4 x + 3 y = − 2$] #math.equation(block: true, alt: "y equals the fraction minus 2 over 3 minus the fraction 4 x over 3")[$y = display(frac(− 2, 3)) − display(frac(4 x, 3))$] #math.equation(block: true, alt: "x minus 2 y equals minus 7")[$x − 2 y = − 7$] #math.equation(block: true, alt: "the fraction x over 8 minus the fraction y over 2 equals 1")[$display(frac(x, 8)) − display(frac(y, 2)) = 1$] #math.equation(block: true, alt: "y equals the fraction x over 4 minus 2")[$y = display(frac(x, 4)) − 2$] #math.equation(block: true, alt: "the fraction x over 5 plus the fraction y over 7 equals 1")[$display(frac(x, 5)) + display(frac(y, 7)) = 1$] #math.equation(block: true, alt: "3 x plus the fraction 2 over 7 y equals 1")[$3 x + display(frac(2, 7)) y = 1$] #math.equation(block: true, alt: "y equals the fraction 7 over 2 minus the fraction 21 x over 2")[$y = display(frac(7, 2)) − display(frac(21 x, 2))$] #math.equation(block: true, alt: "the fraction 5 over 6 x plus 8 y equals 1")[$display(frac(5, 6)) x + 8 y = 1$] #math.equation(block: true, alt: "minus open parenthesis x minus 1 close parenthesis equals 6 open parenthesis y minus 3 close parenthesis")[$− ( x − 1 ) = 6 ( y − 3 )$] #math.equation(block: true, alt: "y equals the fraction 19 over 6 minus the fraction x over 6")[$y = display(frac(19, 6)) − display(frac(x, 6))$] #math.equation(block: true, alt: "2 y minus 4 equals 3 open parenthesis x plus 5 close parenthesis")[$2 y − 4 = 3 ( x + 5 )$] #math.equation(block: true, alt: "the fraction y plus 8 over x minus 1 equals the fraction minus 7 over 4")[$display(frac(y + 8, x − 1)) = display(frac(− 7, 4))$] #math.equation(block: true, alt: "y equals the fraction minus 25 over 4 minus the fraction 7 x over 4")[$y = display(frac(− 25, 4)) − display(frac(7 x, 4))$] #math.equation(block: true, alt: "the fraction 2 over 3 equals the fraction y minus 5 over x plus 2")[$display(frac(2, 3)) = display(frac(y − 5, x + 2))$] For Problems 21-28, solve the formula for the specified variable. #math.equation(block: true, alt: "v equals k plus g t , for t")[$v = k + g t , " " " " " for " t$] #math.equation(block: true, alt: "t equals the fraction v minus k over g")[$t = display(frac(v − k, g))$] #math.equation(block: true, alt: "S equals 3 π d plus π a , for d")[$S = 3 π d + π a , " " " " " for " d$] #math.equation(block: true, alt: "S equals 2 w open parenthesis w plus 2 h close parenthesis , for h")[$S = 2 w ( w + 2 h ) , " " " " " for " h$] #math.equation(block: true, alt: "h equals the fraction S minus 2 w squared over 4 w")[$h = display(frac(S − 2 w^(2), 4 w))$] #math.equation(block: true, alt: "A equals P open parenthesis 1 plus r t close parenthesis , for r")[$A = P ( 1 + r t ) , " " " " " for " r$] #math.equation(block: true, alt: "P equals a plus open parenthesis n minus 1 close parenthesis d , for n")[$P = a + ( n − 1 ) d , " " " " " for " n$] #math.equation(block: true, alt: "n equals the fraction P minus a plus d over d")[$n = display(frac(P − a + d, d))$] #math.equation(block: true, alt: "R equals 2 d plus h open parenthesis a plus b close parenthesis , for b")[$R = 2 d + h ( a + b ) , " " " " " for " b$] #math.equation(block: true, alt: "A equals π r h plus π r squared , for h")[$A = π r h + π r^(2) , " " " " " for " h$] #math.equation(block: true, alt: "h equals the fraction A minus π r squared over π r")[$h = display(frac(A − π r^(2), π r))$] #math.equation(block: true, alt: "A equals 2 w squared plus 4 l w , for l")[$A = 2 w^(2) + 4 l w , " " " " " for " l$] For Problems 29-40, solve the inequality. #math.equation(block: true, alt: "3 x minus 2 greater than 1 plus 2 x")[$3 x − 2 > 1 + 2 x$] #math.equation(block: true, alt: "x greater than 3")[$x > 3$] #math.equation(block: true, alt: "2 x plus 3 less than or equal to x minus 1")[$2 x + 3 ≤ x − 1$] #math.equation(block: true, alt: "the fraction minus 2 x minus 6 over minus 3 greater than 2")[$display(frac(− 2 x − 6, − 3)) > 2$] #math.equation(block: true, alt: "x greater than 0")[$x > 0$] #math.equation(block: true, alt: "the fraction minus 2 x minus 3 over 2 less than or equal to minus 5")[$display(frac(− 2 x − 3, 2)) ≤ − 5$] #math.equation(block: true, alt: "the fraction 2 x minus 3 over 3 less than or equal to the fraction 3 x over minus 2")[$display(frac(2 x − 3, 3)) ≤ display(frac(3 x, − 2))$] #math.equation(block: true, alt: "x less than or equal to the fraction 6 over 13")[$x ≤ display(frac(6, 13))$] #math.equation(block: true, alt: "the fraction 3 x minus 4 over minus 2 greater than the fraction minus 2 x over 5")[$display(frac(3 x − 4, − 2)) > display(frac(− 2 x, 5))$] #math.equation(block: true, alt: "minus 6 less than 4 x plus 10 less than 20")[$− 6 < 4 x + 10 < 20$] #math.equation(block: true, alt: "minus 4 less than x less than the fraction 5 over 2")[$− 4 < x < display(frac(5, 2))$] #math.equation(block: true, alt: "3 less than minus 2 x minus 5 less than 15")[$3 < − 2 x − 5 < 15$] #math.equation(block: true, alt: "minus 9 less than or equal to minus 3 x plus 6 less than 2")[$− 9 ≤ − 3 x + 6 < 2$] #math.equation(block: true, alt: "the fraction 4 over 3 less than or equal to x less than or equal to 5")[$display(frac(4, 3)) ≤ x ≤ 5$] #math.equation(block: true, alt: "4 less than 8 x plus 12 less than or equal to 16")[$4 < 8 x + 12 ≤ 16$] #math.equation(block: true, alt: "5 less than the fraction 8 minus 2 x over 4 less than or equal to 7")[$5 < display(frac(8 − 2 x, 4)) ≤ 7$] #math.equation(block: true, alt: "minus 10 less than or equal to x less than or equal to minus 6")[$− 10 ≤ x ≤ − 6$] #math.equation(block: true, alt: "minus 1 less than or equal to the fraction 4 x minus 6 over minus 3 less than or equal to 0")[$− 1 ≤ display(frac(4 x − 6, − 3)) ≤ 0$] For Problems 41-50, write the set with interval notation, and graph the set on a number line. #math.equation(block: true, alt: "minus 5 less than x less than or equal to 3")[$− 5 < x ≤ 3$] #math.equation(block: true, alt: "open parenthesis minus 5 , 3 close bracket")[$( − 5 , 3 ]$] #figure(figph[half-open interval], alt: "half-open interval", caption: none) #math.equation(block: true, alt: "0 less than or equal to x less than 4")[$0 ≤ x < 4$] #math.equation(block: true, alt: "0 greater than or equal to x greater than or equal to minus 4")[$0 ≥ x ≥ − 4$] #math.equation(block: true, alt: "open bracket minus 4 , 0 close bracket")[$[ − 4 , 0 ]$] #figure(figph[closed interval], alt: "closed interval", caption: none) #math.equation(block: true, alt: "8 greater than x greater than 5")[$8 > x > 5$] #math.equation(block: true, alt: "x greater than minus 6")[$x > − 6$] #math.equation(block: true, alt: "open parenthesis minus 6 , ∞ close parenthesis")[$( − 6 , upright(∞) )$] #figure(figph[half-infinite open interval], alt: "half-infinite open interval", caption: none) #math.equation(block: true, alt: "x less than or equal to 1")[$x ≤ 1$] #math.equation(block: true, alt: "x less than minus 3 or x greater than or equal to minus 1")[$x < − 3 " " " or " " " x ≥ − 1$] #math.equation(block: true, alt: "open parenthesis minus ∞ , minus 3 close parenthesis union open bracket minus 1 , ∞ close parenthesis")[$( − upright(∞) , − 3 ) ∪ [ − 1 , upright(∞) )$] #figure(figph[disjoint intervals], alt: "disjoint intervals", caption: none) #math.equation(block: true, alt: "x greater than or equal to 3 or x less than or equal to minus 3")[$x ≥ 3 " " " or " " " x ≤ − 3$] #math.equation(block: true, alt: "minus 6 less than or equal to x less than minus 4 or minus 2 less than x less than or equal to 0")[$− 6 ≤ x < − 4 " " " or " " " − 2 < x ≤ 0$] #math.equation(block: true, alt: "open bracket minus 6 , minus 4 close parenthesis union open parenthesis minus 2 , 0 close bracket")[$[ − 6 , − 4 ) ∪ ( − 2 , 0 ]$] #figure(figph[disjoint intervals], alt: "disjoint intervals", caption: none) #math.equation(block: true, alt: "x less than 2 or 2 less than x less than 3")[$x < 2 " " " or " " " 2 < x < 3$]