#### Chapter Test

Find the solutions in the app
##### Sections
###### Exercises
Exercise name Free?
Exercises 1 We need to isolate x on one side to solve the equation. To do this, we will use the Addition Property of Equality to gather all the constant terms on the right-hand side of the equation and then simplify. x−7=15 LHS+7=RHS+7 LHS+7=RHS+7x−7+7=15+7Add terms x=22 We will check our solution by substituting this result back into the original equation and simplifying. x−7=15x=2222−7=?15Subtract term15=15 An identity occurs when the left-hand side and the right-hand side of an equation are equal to one another. Since substituting x=22 resulted in an identity, our solution is correct.
Exercises 2 To solve for x we need to isolate it by subtracting 5 from both sides of the equation. We want to subtract 5 because addition and subtraction are inverse operations - they cancel each other out. This is allowed by the Subtraction Property of Equality. 32​x+5=3 LHS−5=RHS−5 LHS−5=RHS−532​x+5−5=3−5Subtract terms 32​x=-2 Remember that multiplying a fraction by its inverse results in a product of 1. To eliminate the fraction coefficient of the variable term, we will use the Multiplication Property of Equality to distribute the fraction's inverse across the equation. 32​x=-2 LHS⋅23​=RHS⋅23​ LHS⋅23​=RHS⋅23​23​(32​x)=23​(-2)ba​⋅ab​=1x=23​(-2)a(-b)=-a⋅bx=-23​(2)Put minus sign in numeratorx=2-3​(2)2a​⋅2=a x=-3 We will check our solution by substituting this result back into the original equation and simplifying. 32​x+5=3x=-332​⋅(-3)+5=?3ca​⋅b=ca⋅b​3-6​+5=?3Calculate quotient-2+5=?3Add terms3=3 An identity occurs when the left-hand side and the right-hand side of an equation are equal to one another. Since substituting x=-3 resulted in an identity, our solution is correct.
Exercises 3 We need to isolate x on one side to solve the equation. We can see that we have a single x on the right-hand side so, by using the Subtraction Property of Equality, we can collect all the x-values on the left-hand side of the equation. 11x+1=-1+xLHS−x=RHS−x11x+1−x=-1+x−xSubtract term10x+1=-1 Now we can use the Subtraction Property of Equality again to collect all the constants on the right-hand side. 10x+1=-1LHS−1=RHS−110x+1−1=-1−1Subtract term10x=-2 Finally, we need to isolate x by dividing both sides by 10 using the Division Property of Equality. 10x=-2LHS/10=RHS/101010x​=10-2​ba​=b/2a/2​x=-51​ The solution is x=-51​.
Exercises 4 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint on a number line. First let's isolate the absolute value on the left-hand side: 2∣x−3∣−5=7LHS+5=RHS+52∣x−3∣−5+5=7+5Add terms2∣x−3∣=12LHS/2=RHS/2∣x−3∣=6 In this exercise, the equation ∣x−3∣=6 means that the value of the variable is 6 steps from the midpoint, in the positive direction and the negative direction. Thus, we can rewrite the given equation as follows. ∣x−3∣=6⇒x−3=-6x−3=-6​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣x−3∣=6x−3≥0: x−3=6x−3<0: x−3=-6​(I)(II)​x−3=6x−3=-6​(I)(II)​(I), (II):  LHS+3=RHS+3x1​=9x2​=-3​ Both x=9 and x=-3 are solutions to the given equation.
Exercises 5 When the absolute value of an expression is equal to an expression, either the expressions are equal or the opposite of the expressions are equal. For example, if we have the equation, ∣ax+b∣=cx+d​ then we can have either ax+b=cx+dorax+b=-cx−d.​ To solve the given absolute value equation, we will write two equations like above when we remove the absolute value. ∣2x−19∣=4x+12x−19≥0: 2x−19=(4x+1)2x−19<0: 2x−19=-(4x+1)​(I)(II)​2x−19=4x+12x−19=-4x−1​(I)(II)​ Solve for x (I), (II): LHS−2x=RHS−2x-19=2x+1-19=-6x−1​(I): LHS−1=RHS−1-20=2x-19=-6x−1​(II): LHS+1=RHS+1-20=2x-18=-6x​(I): LHS/2=RHS/2-10=x-18=-6x​(II): LHS/-6=RHS/-6-10=x3=x​(I), (II): Rearrange equation x=-10x=3​ After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made. ∣2x−19∣=4x+1x=-10∣2(-10)−19∣=?4(-10)+1 Simplify equation a(-b)=-a⋅b∣-20−19∣=?-40+1Add and subtract terms∣-39∣=?-39∣-39∣=39 39=-39 x=-10 is extraneous. ∣2x−19∣=4x+1x=3∣2⋅3−19∣=?4⋅3+1 Simplify equation Multiply∣6−19∣=?12+1Add and subtract terms∣-13∣=?13∣-13∣=13 13=13 x=3 is not extraneous.
Exercises 6 To solve the equation we have to isolate x on one side. We will begin by moving all variable terms to the left-hand side and all constants to the right-hand side. -2+5x−7=3x−9+2xAdd and subtract terms-9+5x=5x−9LHS−5x=RHS−5x-9+5x−5x=5x−9−5xSubtract term-9=-9 The equation reduced to a statement that will always be true. This is called an identity. This means that the solution is all real numbers.
Exercises 7 On the left-hand side, we have a product of 3 and an expression in parentheses. To evaluate this product, we distribute 3 to each term inside the parentheses. 3(x+4)−1=-7 Distribute 3 Distribute 33(x)+3(4)−1=-7Multiply 3x+12−1=-7Subtract term3x+11=-7 Now we can continue solving for x using inverse operations. 3x+11=-7 LHS−11=RHS−11 LHS−11=RHS−113x+11−11=-7−11Subtract term 3x=-18 LHS/3=RHS/3 LHS/3=RHS/333x​=3-18​Put minus sign in front of fraction33x​=-318​Calculate quotient x=-6
Exercises 8 When the absolute values of two expressions are equal, either the expressions are equal or the opposite of the expressions are equal. For example, if we have the equation, ∣ax+b∣=∣cx+d∣​ then we can have either ax+b=cx+dorax+b=-(cx+d)​ To solve the given absolute value equation, we will write two equations like above when we remove the absolute value. ∣20+2x∣=∣4x+4∣20+2x≥0: 20+2x=(4x+4)20+2x<0: 20+2x=-(4x+4)​(I)(II)​20+2x=4x+420+2x=-(4x+4)​(I)(II)​ Solve for x (II): Distribute -120+2x=4x+420+2x=-4x−4​(I), (II): LHS−20=RHS−202x=4x−162x=-4x−24​(I): LHS−4x=RHS−4x-2x=-162x=-4x−24​(I): LHS/-2=RHS/-2x=82x=-4x−24​(II): LHS+4x=RHS+4xx=86x=-24​(II): LHS/6=RHS/6 x=8x=-4​ After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made. ∣20+2x∣=∣4x+4∣x=8∣20+2(8)∣=?∣4(8)+4∣ Simplify equation Multiply∣20+16∣=?∣32+4∣Add terms∣36∣=?∣36∣∣36∣=36 36=36 x=8 is not extraneous. ∣20+2x∣=∣4x+4∣x=-4∣20+2(-4)∣=?∣4(-4)+4∣ Simplify equation a(-b)=-a⋅b∣20−8∣=?∣-16+4∣Add and subtract terms∣12∣=?∣-12∣∣12∣=1212=?∣-12∣∣-12∣=12 12=12 x=-4 is not extraneous.
Exercises 9 To begin solving this equation, we can distribute the 31​ and -2 to each term in the parentheses on the left-hand side. Then we can continue solving by using the Properties of Equality. 31​(6x+12)−2(x−7)=19 Distribute Distribute 31​ & -26x(31​)+12(31​)+x(-2)−7(-2)=19Multiply36x​+312​−2x+14=19Calculate quotient 2x+4−2x+14=19Add and subtract terms18≠19 The statement 18=19 is never true, so the equation has no solution.
Exercises 10 The given equation is 3x−5=3x−c. When c=5 we have the following identity: 3x−5=3x−5. This equation holds true for all values of x as the left-hand side and right-hand side are identical. For example, if x=0 we have 3⋅0−5=3⋅0−5⇔-5=-5 However, if c≠5, then the equation has no solution because the left-hand side will never equal the right-hand side no matter what value we substitute for x. Let's pick c=1 and try to solve the equation. 3x−5=3x−cc=13x−5=3x−1LHS−3x=RHS−3x-5≠-1 We have reached a contradiction which means the two sides can never be equal. Simply, we can say that the equation has no solution if c≠5.
Exercises 11 We do not need to solve this equation. All we need to do is to think about the absolute value function itself. It takes any number as an input and gives a non-negative result as an output. In other words, for any number N we have ∣N∣≥0. It does not matter what kind of expression we have inside the absolute value because its result always obeys the above rule. Therefore, in our equation ∣x−7∣=c, we know that ∣x−7∣≥0. This restricts the values of c to be non-negative if we want our equation to have a solution. Therefore, the given equation has no solution if c<0.
Exercises 12 An absolute value equation takes on the following form: ∣x−midpoint∣=distance to midpoint​ To write an absolute value equation that models the situation, we should begin by thinking about the given minimum and maximum heights as solutions to the equation. By plotting these points on a number line, we can determine the midpoint as well as the distance from each point to the midpoint.From the number line, we can see that the midpoint between 30 and 38 is 34 and that the distance from both values to the midpoint is 4. We can now substitute these values into the formula. ​∣x−midpoint∣=distance to midpoint  ∣x−34∣=4​ ⧼ebox-type-checking-our-solution⧽ Checking Our Solution We can solve the equation we've created to make sure it has the desired solutions. ∣x−34∣=4x−34≥0: x−34=4x−34<0: x−34=-4​(I)(II)​x−34=4x−34=-4​(I)(II)​(I), (II):  LHS+34=RHS+34x=38x=30​ Since our created equation yielded the values 30 and 38, we can conclude that it is correct.
Exercises 13 aLet's isolate w on left-hand side. P=2l+2wLHS−2l=RHS−2lP−2l=2wLHS/2=RHS/22P​−l=wba​=b1​⋅a21​P−l=wRearrange equationw=21​P−lbIn the figure, we are given that the perimeter is 330 yards and the length is 100 yards. Let's insert these values into the rewritten equation. w=21​P−lP=330, l=100w=21​⋅330−100b1​⋅a=ba​w=2330​−100Calculate quotientw=165−100Subtract termw=65 The width of the field is 65 yards.cWe can use the percent proportion to solve this problem: WholePart​=100Percent​. In the equation, the part is the circle's area. We can calculate this since we know its radius: AC​=π⋅102=100π yd2. In the denominator, the whole represents the entire area of the rectangular field which is simply its width times its length: AR​=65⋅100=6500. Now, let's plug them into our formula. WholePart​=100Percent​Part=100π, Whole=65006500100π​=100Percent​LHS⋅100=RHS⋅100650010000π​=PercentRearrange equationPercent=650010000π​Use a calculatorPercent=4.8332…Percent=4.83 About 4.83% of the field is inside the center circle.
Exercises 14 aThe total amount we will pay for repairs at each location has two components:The cost of the parts, which is a flat fee. The hourly labor cost charged for the duration of the repair process. We can model these costs with algebraic expressions. Let's write the expressions for the total cost for each location separately.Dealership The cost of the parts is given to be \$24 and the hourly labor cost is \$99. If we let the number of hours worked be x, then we can write the total cost expression as: 99x+24.Local Mechanic The cost of the parts is given to be \$45 and the hourly labor cost is \$89. If we let the number of hours worked be x, then we can write the total cost expression as: 89x+45.When are costs the same? To determine the number of hours worked when we will have paid the same total amount, we need to know when these two expressions will be equal to one another. We can do that by equating them and solving for x. 99x+24=89x+45 Solve for x LHS−89x=RHS−89x10x+24=45LHS−24=RHS−2410x=21LHS/10=RHS/10 x=2.1 After 2.1 hours, the total costs will be identical.bSince 2.1 hours is the point of intersection, we can find out when the repairs will cost less at the dealership or at the local mechanic by checking which one is more expensive before and after 2.1 hours. Let's graph the lines and compare.As we can see on the graph, the blue line has lower values for y before the point of intersection and the red line has lower values afterwards. This means that the dealership costs less before 2.1 hours and the local mechanic costs less after 2.1 hours have passed. Notice, we could have also figured this out using logic. The dealership charges far less than the local mechanic for the parts. Therefore, it is cheaper to use the dealership until the point that the hourly wage makes up for the cheaper parts.
Exercises 15 The left-hand side of the equation is an absolute value and, therefore, its result will always be non-negative. The right-hand side is a single term with x and a coefficient of positive 6. If we substitute x=-2 into the equation, the right-hand side will equal a negative number, the product of a positive and a negative number is always negative. This means we have a contradiction, an absolute value can't equal a negative number. Therefore, the solution must be extraneous.
Exercises 16 When an equation is reduced into the number=number form, the solution to the equation does not depend on its variable(s). It means the equation works for any value of the variable. This happens when we have the same expression or quantity on both sides. For example: 0=0or2x=2x.​ These are completely true regardless of our choice of x. There is no difference in terms of validity between 0=0 and 2x=2x. When we have this case, we say that there are infinitely many solutions, because any value for the variable will work.