#### Solving Absolute Value Equations

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###### Exercises
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Exercises 1 Sometimes when solving an absolute value equation it is possible to encounter an extraneous solution. A solution is extraneous when it is accurately found through the solving process, but does not satisfy the original equation. Meaning, when it is substituted into the equation, it does not create a true statement.
Exercises 2 We have been given the following absolute value equation ∣4x−7∣=-1 and are told that it has no solution. Why? ∣4x−7∣≠-1 The absolute value of any number cannot be equal to a negative number. Because absolute values represent distance, they are always non-negative. Therefore, the equation has no solution.
Exercises 3 The absolute value of any number is the distance on a number line from that number to 0. We can use a number line to simplify ∣-9∣.From the number line, we can see that the distance between -9 and 0 is 9 units. ∣-9∣=9​
Exercises 4 The absolute value of any number is the distance on a number line from that number to 0. We can use a number line to simplify -∣15∣.From the number line, we can see that the distance between 15 and 0 is 15 units. Thus, ∣15∣=15. We will use this to simplify the given expression. -∣15∣∣15∣=15-(15)Remove parentheses-15
Exercises 5 To simplify the given expression completely, we will begin with the absolute values. The absolute value of any number is the distance on a number line from that number to 0. Therefore, when you remove the absolute value, it will always be positive. ∣14∣−∣-14∣∣14∣=1414−∣-14∣∣-14∣=1414−14Subtract terms0 The expression simplifies to 0.
Exercises 6 To simplify the given expression completely, we will begin with the absolute values. The absolute value of any number is the distance on a number line from that number to 0. Therefore, when you remove the absolute value, it will always be positive. ∣-3∣+∣3∣∣-3∣=33+∣3∣∣3∣=33+3Add terms6 The expression simplifies to 6.
Exercises 7 To simplify the given expression, we must follow the order of operations before we can apply the absolute value. -∣-5⋅(-7)∣-a(-b)=a⋅b-∣5⋅7∣Multiply-∣35∣ As 35 is a positive number, we can take away the absolute value without changing any sign. -∣35∣∣35∣=35-(35)Remove parentheses-35 The expression simplifies to -35.
Exercises 8 To simplify the given expression, we must follow the order of operations before we can apply the absolute value. ∣-0.8⋅10∣(-a)b=-ab∣-8∣ As -8 is a negative number we have to change the sign when we are taking away the absolute value. ∣-8∣∣-8∣=88 The expression simplifies to 8.
Exercises 9 To simplify the given expression, we must follow the order of operations before we can remove the absolute value. ∣∣∣∣​-327​∣∣∣∣​Put minus sign in front of fraction∣∣∣∣​-327​∣∣∣∣​Calculate quotient∣-9∣∣-9∣=99 The expression simplifies to 9.
Exercises 10 To simplify the given expression, we must follow the order of operations before we can remove the absolute value. ∣∣∣∣​-4-12​∣∣∣∣​Put minus sign in front of fraction∣∣∣∣​-(-412​)∣∣∣∣​-(-a)=a∣∣∣∣​412​∣∣∣∣​Calculate quotient∣3∣∣3∣=33 The expression simplifies to 3.
Exercises 11 An absolute value is always a non-negative number because it measures the number's distance from zero on a number line. In this exercise, the equation ∣w∣=6 means that the value of w is 6 steps from zero, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣w∣=6⇒w=-6w=-6​ Both w=6 and w=-6 are solutions to the given equation. When shown on a number line, we have the following.
Exercises 12 An absolute value represents distance and, thereby, is always non-negative. ∣x∣=x and ∣-x∣=x No matter what value you substitute for r, the equation will never equal -2. ∣r∣≠-2 Hence, there are no solutions to this equation.
Exercises 13 An absolute value represents distance and, thereby, is always non-negative. ∣x∣=x and ∣-x∣=x No matter what value you substitute for y, the equation will never equal -18. ∣y∣≠-18 Hence, there are no solutions to this equation.
Exercises 14 An absolute value is always a non-negative number because it measures the number's distance from zero on a number line. In this exercise, the equation ∣x∣=13 means that the value of x is 13 steps from zero, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣x∣=13⇒x=-13x=-13​ Both x=13 and x=-13 are solutions to the given equation. When shown on a number line, we have the following.
Exercises 15 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣m+3∣=7 means that the distance between m and -3 is 7, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣m+3∣=7⇒m+3=-7m+3=-7​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣m+3∣=7m+3≥0: m+3=7m+3<0: m+3=-7​(I)(II)​m+3=7m+3=-7​(I)(II)​(I), (II):  LHS−3=RHS−3m1​=4m2​=-10​ Both m=4 and m=-10 are solutions to the given equation. When shown on a number line, we have the following.
Exercises 16 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣q−8∣=14 means that the distance between q and 8 is 14, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣q−8∣=14⇒q−8=-14q−8=-14​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣q−8∣=14q−8≥0: q−8=14q−8<0: q−8=-14​(I)(II)​q−8=14q−8=-14​(I)(II)​(I), (II):  LHS+8=RHS+8q1​=22q2​=-6​ Both q=22 and q=-6 are solutions to the given equation. When shown on a number line, we have the following.
Exercises 17 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣-3d∣=15 means that the distance between -3d and 0 is 15, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣-3d∣=15⇒-3d=-15-3d=-15​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣-3d∣=15-3d≥0: -3d=15-3d<0: -3d=-15​(I)(II)​-3d=15-3d=-15​(I)(II)​(I), (II):  LHS/(-3)=RHS/(-3)d1​=-5d2​=5​ Both d=-5 and d=5 are solutions to the given equation. When shown on a number line, we have the following.
Exercises 18 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣∣∣∣​2t​∣∣∣∣​=6 means that the value of the expression 2t​ is 6 steps from zero, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. 2t​=6and2t​=-6 We need to solve both of these cases so that we can find both values that satisfy the equation. 2t​=6LHS⋅2=RHS⋅2t=12 Now, let's look at the negative case. 2t​=-6LHS⋅2=RHS⋅2t=-12 Both t=12 and t=-12 are solutions to the given equation. When shown on a number line, we have the following.
Exercises 19 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣4b−5∣=19 means that the distance between 4b and 5 is 19, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣4b−5∣=19⇒4b−5=-194b−5=-19​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣4b−5∣=194b−5≥0: 4b−5=194b−5<0: 4b−5=-19​(I)(II)​4b−5=194b−5=-19​(I)(II)​(I), (II):  LHS+5=RHS+54b=244b=-14​(I), (II):  LHS/4=RHS/4b=6b=4-14​​ Write as Mixed Number (II): ba​=b/2a/2​b=6b=2-7​​(II): Put minus sign in front of fractionb=6b=-27​​(II): Rewrite 7 as 6+1b=6b=-26+1​​(II): Write as a sum of fractionsb=6b=-(26​+21​)​(II): Calculate quotientb=6b=-(3+21​)​Add terms b1​=6b2​=-321​​ Both b=6 and b=-321​ are solutions to the given equation. When shown on a number line, we have the following.
Exercises 20 Let's begin with simplifying the equation so that, on the left-hand side, we have only the absolute value expression. ∣x−1∣+5=2LHS−5=RHS−5∣x−1∣=-3 An absolute value is always a non-negative number because it measures a distance. No matter what value you substitute for x, the equation will never equal -3. ∣x−1∣≠-3 Hence, there are no solutions to this equation.
Exercises 21 Let's begin with simplifying the equation so that, on the left-hand side, we have only the absolute value expression. -4∣8−5n∣=13LHS/(-4)=RHS/(-4)∣9−5n∣=-413​ An absolute value is always a non-negative number because it measures a distance. No matter what value you substitute for n, the equation will never equal -413​. ∣9−5n∣≠-413​ Hence, there are no solutions to this equation.
Exercises 22 We will begin with isolating the absolute value on one side of the equation. -3∣∣∣∣​1−32​v∣∣∣∣​=-9LHS/(-3)=RHS/(-3)∣∣∣∣​1−32​v∣∣∣∣​=3 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣∣∣∣​1−32​v∣∣∣∣​=3 means that the distance between 1 and 32​v is 3, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. 1−32​v=3and1−32​v=-3 We need to solve both of these cases so that we can find both values that satisfy the equation. Let's solve the first one. 1−32​v=3LHS−1=RHS−1-32​v=2LHS⋅(-23​)=RHS⋅(-23​)-32​v(-23​)=2(-23​)ba​⋅ab​=1v=2(-23​)2a​⋅2=av=-3 Similarly, we can solve the second equation. 1−32​v=-3LHS−1=RHS−1-32​v=-4LHS⋅(-23​)=RHS⋅(-23​)-32​v(-23​)=-4(-23​)ba​⋅ab​=1v=-4(-23​)Put minus sign in numeratorv=-4(2-3​)a⋅cb​=ca⋅b​v=2-4(-3)​-a(-b)=a⋅bv=212​Calculate quotientv=6 Both v=-3 and v=6 are solutions to the given equation. When shown on a number line, we have the following.
Exercises 23 We will begin with isolating the absolute value on one side of the equation. 3=-2∣∣∣∣​41​s−5∣∣∣∣​+3LHS−3=RHS−30=-2∣∣∣∣​41​s−5∣∣∣∣​LHS/(-2)=RHS/(-2)0=∣∣∣∣​41​s−5∣∣∣∣​Rearrange equation∣∣∣∣​41​s−5∣∣∣∣​=0 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣∣∣∣​41​s−5∣∣∣∣​=0 means that the distance between 41​s and 5 is 0. We can write our equation as 41​s−5=0 We need to solve this for s. 41​s−5=0LHS+5=RHS+541​s=5LHS⋅4=RHS⋅4s=20 s=20 is the solution to given equation. When shown on a number line, we have the following.
Exercises 24 Let's begin with simplifying the equation so that on the left-hand side we have only the absolute value expression. 9∣4p+2∣+8=35LHS−8=RHS−89∣4p+2∣=27LHS/9=RHS/9∣4p+2∣=3 An absolute value is always a non-negative number because it measures the expression's distance from a midpoint. In this exercise, the equation ∣4p+2∣=3 means that the distance between 4p and -2 is 3, either in the positive direction or the negative direction. Thus, we can rewrite the given equation as follows. ∣4p+2∣=3⇒4p+2=-34p+2=-3​ We need to solve both of these cases so that we can find both values that satisfy the equation. ∣4p+2∣=34p+2≥0: 4p+2=34p+2<0: 4p+2=-3​(I)(II)​4p+2=34p+2=-3​(I)(II)​ First, we will solve Equation I. 4p+2=3LHS−2=RHS−24p=1LHS/4=RHS/4p=41​ In a similar process, we can solve Equation II. 4p+2=-3LHS−2=RHS−24p=-5LHS/4=RHS/4p=-45​Write fraction as a mixed numberp=-141​ Both p=41​ and p=-141​ are solutions to the given equation. When shown on a number line, we have the following.
Exercises 25 aWe can plot the maximum and minimum distances on the number line as the following.bAn absolute value equation takes on the following form. ∣x−midpoint∣=distance to midpoint​ To write an absolute value equation that models the situation, we can begin by thinking about the given minimum and maximum distance as solutions to the equation. Considering the number line that we formed, we can find the midpoint using the following equation. mid=2min+max​ Let's find it by substituting the maximum and minimum distances into the formula. mid=2min + max​min=91.4, max=94.5mid=291.4+94.5​Add termsmid=2185.9​Calculate quotientmid=92.95 The midpoint is 92.95 miles. The distance d from both values to the midpoint will be the same as the following. d=max−mid or d=mid−min Therefore, it will be enough to subtract the midpoint from the maximum value. Let's do it! d=max−midmax=94.5, mid=92.95d=94.5−92.95Subtract termd=1.55 The distance is 1.55 million miles. Let's illustrate this on the number line.As a result, we can write the equation using the form previously stated. ∣x−92.95∣=1.55​
Exercises 26 aWe can plot the maximum and minimum distances on the number line as the following.bAn absolute value equation takes on the following form. ∣x−midpoint∣=distance to midpoint​ To write an absolute value equation that models the situation, we can begin by thinking about the given minimum and maximum lengths as solutions to the equation. By plotting these points on a number line we can determine the midpoint and the distance from each point to the midpoint.From the number line, we can see that the midpoint between 10 and 15 is 12.5, and that the distance from both values to the midpoint is 2.5. We can then write the following equation. ∣x−12.5∣=2.5​
Exercises 27 If we have two points on a number line, we can use the midpoint between the points and the distance from each point to the midpoint to write an absolute value equation. ∣x−midpoint∣=distance to midpoint​ Notice that the left-hand side of the equation is in the form ∣x−a∣. Keeping that in mind, we can rewrite our equation. ∣x−(-2)∣=4​ Now, we can determine the midpoint as -2 and the distance from the midpoint as 4.This number line corresponds with option B.
Exercises 28 If we have two points on a number line, we can use the midpoint between the points and the distance from each point to the midpoint to write an absolute value equation as: ∣x−midpoint∣=distance to midpoint. In this exercise, we have: ∣x−4∣=2 Now we can determine the midpoint as 4 and the distance from the midpoint as 2.This number line corresponds with option D.
Exercises 29 If we have two points on a number line, we can use the midpoint between the points and the distance from each point to the midpoint to write an absolute value equation as: ∣x−midpoint∣=distance to midpoint. In this exercise, we have: ∣x−2∣=4 Now we can determine the midpoint as 2 and the distance from the midpoint as 4.This number line corresponds with option C.
Exercises 30 If we have two points on a number line, we can use the midpoint between the points and the distance from each point to the midpoint to write an absolute value equation as: ∣x−midpoint∣=distance to midpoint. Notice that the left-hand side of the equation is in the form ∣x−a∣. Keeping that in mind, we can rewrite our equation as ∣x−(-4)∣=2. Now, we can determine the midpoint as -4 and the distance from the midpoint as 2.This number line corresponds with option A.
Exercises 31 To write an absolute value equation, we can begin by thinking about the solutions to the equation as points on a number line. We can use the number line to determine the midpoint and the distance from each point to the midpoint. Our equation will take the following form. ∣x−midpoint∣=distance to midpoint​ Let's plot the given solutions on a number line and determine the midpoint.From the number line, we can see that the midpoint between 8 and 18 is 13, and that the distance from both values to the midpoint is 5. Now we can write our equation. ∣x−midpoint∣∣x−13∣​=distance to midpoint=5​ We can solve the equation we've created to ensure it has the desired solutions. ∣x−13∣=5x−13≥0: x−13=5x−13<0: x−13=-5​(I)(II)​x−13=5x−13=-5​(I)(II)​(I), (II):  LHS+13=RHS+13x1​=18x2​=8​
Exercises 32 To write an absolute value equation, we can begin by thinking about the solutions to the equation as points on a number line. We can use the number line to determine the midpoint and the distance from each point to the midpoint. Our equation will take the following form ∣x−midpoint∣=distance to midpoint.​ Let's plot the given solutions on a number line and determine the midpoint.From the number line, we can see that the midpoint between -6 and 10 is 2, and that the distance from both values to the midpoint is 8. Now we can write our equation as ∣x−midpoint∣∣x−2∣​=distance to midpoint=8​ We can solve the equation we've created to ensure it has the desired solutions. ∣x−2∣=8x−2≥0: x−2=8x−2<0: x−2=-8​(I)(II)​x−2=8x−2=-8​(I)(II)​(I), (II):  LHS+2=RHS+2x1​=10x2​=-6​
Exercises 33 To write an absolute value equation, we can begin by thinking about the solutions to the equation as points on a number line. We can use the number line to determine the midpoint and the distance from each point to the midpoint. Our equation will take the following form ∣x−midpoint∣=distance to midpoint.​ Let's plot the given solutions on a number line and determine the midpoint.From the number line, we can see that the midpoint between 2 and 9 is 5.5, and that the distance from both values to the midpoint is 3.5. Now we can write our equation as ∣x−midpoint∣∣x−5.5∣​=distance to midpoint=3.5​ We can solve the equation we've created to ensure it has the desired solutions. ∣x−5.5∣=3.5x−5.5≥0: x−5.5=3.5x−5.5<0: x−5.5=-3.5​(I)(II)​x−5.5=3.5x−5.5=-3.5​(I)(II)​(I), (II):  LHS+5.5=RHS+5.5x1​=9x2​=2​
Exercises 34 To write an absolute value equation, we can begin by thinking about the solutions to the equation as points on a number line. We can use the number line to determine the midpoint and the distance from each point to the midpoint. Our equation will take the following form ∣x−midpoint∣=distance to midpoint.​ Let's plot the given solutions on a number line and determine the midpoint.From the number line, we can see that the midpoint between -10 and -5 is -7.5, and that the distance from both values to the midpoint is 2.5. Now we can write our equation as ∣x−midpoint∣∣x−(-7.5)∣∣x+7.5∣​=distance to midpoint=2.5=2.5​ We can solve the equation we've created to ensure it has the desired solutions. ∣x+7.5∣=2.5x+7.5≥0: x+7.5=2.5x+7.5<0: x+7.5=-2.5​(I)(II)​x+7.5=2.5x+7.5=-2.5​(I)(II)​(I), (II):  LHS−7.5=RHS−7.5x1​=-5x2​=-10​
Exercises 35 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. ∣4n−15∣=∣n∣4n−15≥0: 4n−15=n4n−15<0: 4n−15=-n​(I)(II)​4n−15=n4n−15=-n​(I)(II)​ Solve for n (I), (II): LHS+15=RHS+154n=n+154n=-n+15​(I): LHS−n=RHS−n3n=154n=-n+15​(II): LHS+n=RHS+n3n=155n=15​(I): LHS/3=RHS/3n=55n=15​(II): LHS/5=RHS/5 n1​=5n2​=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. ∣4n−15∣=∣n∣n=5∣4⋅5−15∣=?∣5∣ Simplify equation Multiply∣20−15∣=?∣5∣Subtract term∣5∣=?∣5∣∣5∣=5 5=5 n=5 is not extraneous. ∣4n−15∣=∣n∣n=3∣4⋅3−15∣=?∣3∣ Simplify equation Multiply∣12−15∣=?∣3∣Subtract term∣-3∣=?∣3∣∣-3∣=33=?∣3∣∣3∣=3 3=3 n=3 is also not extraneous.
Exercises 36 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. ∣2c+8∣=∣10c∣2c+8≥0: 2c+8=10c2c+8<0: 2c+8=-10c​(I)(II)​2c+8=10c2c+8=-10c​(I)(II)​ Solve for c (I), (II): LHS−8=RHS−82c=10c−82c=-10c−8​(I): LHS−10c=RHS−10c-8c=-82c=-10c−8​(I): LHS/(-8)=RHS/(-8)c=12c=-10c−8​(II): LHS+10c=RHS+10cc=112c=-8​(II): LHS/12=RHS/12c=1c=-128​​ba​=b/4a/4​ c1​=1c2​=-32​​ 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. ∣2c+8∣=∣10c∣c=1∣2⋅1+8∣=?∣10⋅1∣ Simplify equation Multiply∣2+8∣=?∣10∣Add terms∣10∣=?∣10∣∣10∣=10 10=10 c=1 is not extraneous. ∣2c+8∣=∣10c∣c=-32​∣∣∣∣​2(-32​)+8∣∣∣∣​=?∣∣∣∣​10(-32​)∣∣∣∣​ Simplify equation a⋅cb​=ca⋅b​∣∣∣∣​-34​+8∣∣∣∣​=?∣∣∣∣​-320​∣∣∣∣​Rewrite 8 as 324​∣∣∣∣​-34​+324​∣∣∣∣​=?∣∣∣∣​-320​∣∣∣∣​Add fractions∣∣∣∣​320​∣∣∣∣​=?∣∣∣∣​-320​∣∣∣∣​∣∣∣∣​320​∣∣∣∣​=320​320​=?∣∣∣∣​-320​∣∣∣∣​∣∣∣∣​-320​∣∣∣∣​=320​ 320​=320​ c=-32​ is also not extraneous.
Exercises 37 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. ∣2b−9∣=∣b−6∣2b−9≥0: 2b−9=(b−6)2b−9<0: 2b−9=-(b−6)​(I)(II)​2b−9=b−62b−9=-(b−6)​(I)(II)​ Solve for b (II):  Distribute -12b−9=b−62b−9=-b+6​(I), (II): LHS+9=RHS+92b=b+32b=-b+15​(I): LHS−b=RHS−bb=32b=-b+15​(II): LHS+b=RHS+bb=33b=15​(II): LHS/3=RHS/3 b1​=3b2​=5​ 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. ∣2b−9∣=∣b−6∣b=3∣2⋅3−9∣=?∣3−6∣ Simplify equation Multiply∣6−9∣=?∣3−6∣Subtract terms∣-3∣=?∣-3∣∣-3∣=3 3=3 b=3 is not extraneous. ∣2b−9∣=∣b−6∣b=5∣2⋅5−9∣=?∣5−6∣ Simplify equation Multiply∣10−9∣=?∣5−6∣Subtract terms∣1∣=?∣-1∣∣1∣=11=?∣-1∣∣-1∣=1 1=1 b=5 is also not extraneous.
Exercises 38 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. ∣3k−2∣=2∣k+2∣3k−2≥0: 3k−2=2(k+2)3k−2<0: 3k−2=-2(k+2)​(I)(II)​3k−2=2(k+2)3k−2=-2(k+2)​(I)(II)​ Solve for k (I): Distribute 23k−2=2k+43k−2=-2(k+2)​(I)(II)​(II): Distribute -23k−2=2k+43k−2=-2k−4​(I)(II)​(I), (II): LHS+2=RHS+23k=2k+63k=-2k−2​(I): LHS−2k=RHS−2kk=63k=-2k−2​(II): LHS+2k=RHS+2kk=65k=-2​(II): LHS/5=RHS/5 k1​=6k2​=-52​​ 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. ∣3k−2∣=2∣k+2∣k=6∣3⋅6−2∣=?2∣6+2∣ Simplify equation Multiply∣18−2∣=?2∣6+2∣Add and subtract terms∣16∣=?2∣8∣∣8∣=816=?2⋅8Multiply 16=16 k=6 is not extraneous. ∣3k−2∣=2∣k+2∣k=-52​∣∣∣∣​3(-52​)−2∣∣∣∣​=?2∣∣∣∣​-52​+2∣∣∣∣​ Simplify equation a⋅cb​=ca⋅b​∣∣∣∣​-56​−2∣∣∣∣​=?2∣∣∣∣​-52​+2∣∣∣∣​Rewrite 2 as 510​∣∣∣∣​-56​−510​∣∣∣∣​=?2∣∣∣∣​-52​+510​∣∣∣∣​Add and subtract fractions∣∣∣∣​-516​∣∣∣∣​=?2∣∣∣∣​58​∣∣∣∣​∣∣∣∣​-516​∣∣∣∣​=516​516​=?2∣∣∣∣​58​∣∣∣∣​∣∣∣∣​58​∣∣∣∣​=58​516​=?2⋅58​a⋅cb​=ca⋅b​ 516​=516​ k=-52​ is not extraneous.
Exercises 39 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 when we remove the absolute value. For 4∣p−3∣=∣2p+8∣ we receive 4(p−3)=2p+8or4(p−3)=-(2p+8) 4(p−3)=2p+84(p−3)=-(2p+8)​ Solve for p (I): Distribute 44p−12=2p+84(p−3)=-(2p+8)​(II): Distribute 4 & -14p−12=2p+84p−12=-2p−8​(I), (II): LHS+12=RHS+124p=2p+204p=-2p+4​(I): LHS−2p=RHS−2p2p=204p=-2p+4​(I): LHS/2=RHS/2p=104p=-2p+4​(II): LHS+2p=RHS+2pp=106p=4​(II): LHS/6=RHS/6p=10p=64​​(II): ba​=b/2a/2​ p=10p=32​​ 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. 4∣p−3∣=∣2p+8∣p=104∣10−3∣=?∣2⋅10+8∣ Simplify equation Multiply4∣10−3∣=?∣20+8∣Add and subtract terms4∣7∣=?∣28∣∣a∣=a4⋅7=?28Multiply 28=28 p=10 is not extraneous. 4∣p−3∣=∣2p+8∣p=32​4∣∣∣∣​32​−3∣∣∣∣​=?∣∣∣∣​2⋅32​+8∣∣∣∣​ Simplify equation a⋅cb​=ca⋅b​4∣∣∣∣​32​−3∣∣∣∣​=?∣∣∣∣​34​+8∣∣∣∣​Rewrite 3 as 39​4∣∣∣∣​32​−39​∣∣∣∣​=?∣∣∣∣​34​+8∣∣∣∣​Rewrite 8 as 324​4∣∣∣∣​32​−39​∣∣∣∣​=?∣∣∣∣​34​+324​∣∣∣∣​Add and subtract fractions4∣∣∣∣​-37​∣∣∣∣​=?∣∣∣∣​328​∣∣∣∣​∣∣∣∣​-37​∣∣∣∣​=37​4⋅37​=?∣∣∣∣​328​∣∣∣∣​a⋅cb​=ca⋅b​328​=?∣∣∣∣​328​∣∣∣∣​∣∣∣∣​328​∣∣∣∣​=328​ 328​=328​ p=32​ is also not extraneous.
Exercises 40 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 the examples above when we remove the absolute value symbols. 2∣4w−1∣=3∣4w+2∣2∣4w−1∣≥0: 2∣4w−1∣=3∣4w+2∣2∣4w−1∣<0: 2∣4w−1∣=-3∣4w+2∣​(I)(II)​2(4w−1)=3(4w+2)2(4w−1)=-3(4w+2)​(I)(II)​(I), (II): Distribute 28w−2=3(4w+2)8w−2=-3(4w+2)​(I), (II):  Distribute 38w−2=12w+68w−2=-(12w+6)​(II):  Remove parentheses and change signs8w−2=12w+68w−2=-12w−6​ (I), (II):  Solve for w (I):  LHS−12w=RHS−12w-4w−2=68w−2=-12w−6​(I):  LHS+2=RHS+2-4w=88w−2=-12w−6​(II):  LHS/(-4)=RHS/(-4)w=-28w−2=-12w−6​(II):  LHS+12w=RHS+12ww=-220w−2=-6​(II):  LHS+2=RHS+2w=-220w=-4​(II):  LHS/20=RHS/20w=-2w=-204​​(II):  ba​=b/4a/4​ w1​=-2w2​=-51​​ After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the original equation and determine if the statement is true. 2∣4w−1∣=3∣4w+2∣w=-22∣4(-2)−1∣=?3∣4(-2)+2∣ Simplify equation a(-b)=-a⋅b2∣-8−1∣=?3∣-8+2∣Add and subtract terms2∣-9∣=?3∣-6∣∣-9∣=92⋅9=?3∣-6∣∣-6∣=62⋅9=?3⋅6Multiply 18=18 w=-2 is not extraneous. 2∣4w−1∣=3∣4w+2∣w=-51​2∣∣∣∣​4(-51​)−1∣∣∣∣​=?3∣∣∣∣​4(-51​)+2∣∣∣∣​ Simplify equation a(-b)=-a⋅b2∣∣∣∣​-4⋅51​−1∣∣∣∣​=?3∣∣∣∣​-4⋅51​+2∣∣∣∣​a⋅b1​=ba​2∣∣∣∣​5-4​−1∣∣∣∣​=?3∣∣∣∣​5-4​+2∣∣∣∣​Write as a fraction2∣∣∣∣​5-4​−55​∣∣∣∣​=?3∣∣∣∣​5-4​+510​∣∣∣∣​Add and subtract fractions2∣∣∣∣​5-9​∣∣∣∣​=?3∣∣∣∣​56​∣∣∣∣​∣∣∣∣​-59​∣∣∣∣​=59​2⋅59​=?3∣∣∣∣​56​∣∣∣∣​∣∣∣∣​56​∣∣∣∣​=56​2⋅59​=?3⋅56​a⋅cb​=ca⋅b​ 518​=518​ w=-51​ is not extraneous.
Exercises 41 When the absolute values 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​ then we can have either ax+b=cxorax+b=-cx​ To solve the given absolute value equation, we will write two equations like above when we remove the absolute value. ∣3h+1∣=7h3h+1≥0: 3h+1=7h3h+1<0: 3h+1=-7h​(I)(II)​3h+1=7h3h+1=-7h​(I)(II)​ Let's solve the equations one at time, starting with Equation I. 3h+1=7h Solve for h LHS−3h=RHS−3h1=4hLHS/4=RHS/441​=hRearrange equation h=41​ Now, we can solve Equation II. 3h+1=-7h Solve for h LHS−3h=RHS−3h1=-10hLHS/(-10)=RHS/(-10)-101​=hRearrange equation h=-101​ 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. ∣3h+1∣=7hh=41​∣∣∣∣​3⋅41​+1∣∣∣∣​=?7⋅41​ Simplify equation a⋅b1​=ba​∣∣∣∣​43​+1∣∣∣∣​=?47​Rewrite 1 as 44​∣∣∣∣​43​+44​∣∣∣∣​=?47​Add fractions∣∣∣∣​47​∣∣∣∣​=?47​∣∣∣∣​47​∣∣∣∣​=47​ 47​=47​ h=41​ is not extraneous. ∣3h+1∣=7hh=-101​∣∣∣∣​3(-101​)+1∣∣∣∣​=?7(-101​) Simplify equation a⋅b1​=ba​∣∣∣∣​-103​+1∣∣∣∣​=?-107​Rewrite 1 as 1010​∣∣∣∣​-103​+1010​∣∣∣∣​=?-107​Add fractions∣∣∣∣​107​∣∣∣∣​=?-107​∣∣∣∣​107​∣∣∣∣​=107​ 107​≠-107​ h=-101​ is extraneous. It is not a valid solution to the equation.
Exercises 42 When the absolute values 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​ then we can have either ax+b=cxorax+b=-cx​ To solve the given absolute value equation, we will write two equations like above when we remove the absolute value. ∣6a−5∣=4a6a−5≥0: 6a−5=4a6a−5<0: 6a−5=-4a​(I)(II)​6a−5=4a6a−5=-4a​(I)(II)​ Let's solve these one at a time, starting with Equation I. 6a−5=4a Solve for a LHS−6a=RHS−6a-5=-2aLHS/(-2)=RHS/(-2)25​=aRearrange equation a=25​ Now, we can solve Equation II. 6a−5=-4a Solve for a LHS−6a=RHS−6a-5=-10aLHS/(-10)=RHS/(-10)105​=aba​=b/5a/5​21​=aRearrange equation a=21​ 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. ∣6a−5∣=4aa=25​∣∣∣∣​6⋅25​−5∣∣∣∣​=?4⋅25​ Simplify equation a⋅cb​=ca⋅b​∣∣∣∣​230​−5∣∣∣∣​=?220​Calculate quotient∣15−5∣=?10Subtract term∣10∣=?10∣10∣=10 10=10 a=25​ is not extraneous. ∣6a−5∣=4aa=21​∣∣∣∣​6⋅21​−5∣∣∣∣​=?4⋅21​ Simplify equation a⋅b1​=ba​∣∣∣∣​26​−5∣∣∣∣​=?24​Calculate quotient∣3−5∣=?2Subtract term∣-2∣=?2∣-2∣=2 2=2 a=21​ is not extraneous.
Exercises 43 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. ∣f−6∣=∣f+8∣f−6≥0: f−6=(f+8)f−6<0: f−6=-(f+8)​(I)(II)​f−6=f+8f−6=-(f+8)​(I)(II)​ Solve for f (II):  Distribute -1f−6=f+8f−6=-f−8​(I): LHS−f=RHS−f-6≠8f−6=-f−8​(II): LHS+f=RHS+f-6≠82f−6=-8​(II): LHS+6=RHS+6-6≠82f=-2​(II): LHS/2=RHS/2 -6≠8f=-1​ Notice that, when solving the first equation, we encountered a contradiction. Thus, the first equation has no solution. The solution to Equation II is f=-1. We need to check if it is an extraneous solution. To do this, we substitute the found solution into the given equation and determine if a true statement is made. ∣f−6∣=∣f+8∣f=-1∣-1−6∣=?∣-1+8∣ Simplify equation Add and subtract terms∣-7∣=?∣7∣∣-7∣=77=?∣7∣∣7∣=7 7=7 f=-1 is not extraneous.
Exercises 44 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. ∣3x−4∣=∣3x−5∣3x−4≥0: 3x−4=(3x−5)3x−4<0: 3x−4=-(3x−5)​(I)(II)​3x−4=3x−53x−4=-(3x−5)​(I)(II)​ Solve for x (II):  Distribute -13x−4=3x−53x−4=-3x+5​(I): LHS−3x=RHS−3x-4≠-53x−4=-3x+5​(II): LHS+3x=RHS+3x-4≠-56x−4=5​(II): LHS+4=RHS+4-4≠-56x=9​(II): LHS/6=RHS/6-4≠-5x=69​​(II): ba​=b/3a/3​ -4≠-5x=23​​ Notice that, when solving the first equation, we encountered a contradiction. Thus, the first equation has no solution. The solution to Equation II is x=23​. We need to check for if it is an extraneous solution. To do this, we substitute the found solution into the given equation and determine if a true statement is made. ∣3x−4∣=∣3x−5∣x=23​∣∣∣∣​3⋅23​−4∣∣∣∣​=?∣∣∣∣​3⋅23​−5∣∣∣∣​ Simplify equation a⋅cb​=ca⋅b​∣∣∣∣​29​−4∣∣∣∣​=?∣∣∣∣​29​−5∣∣∣∣​Rewrite 4 as 28​∣∣∣∣​29​−28​∣∣∣∣​=?∣∣∣∣​29​−5∣∣∣∣​Rewrite 5 as 210​∣∣∣∣​29​−28​∣∣∣∣​=?∣∣∣∣​29​−210​∣∣∣∣​Subtract fractions∣∣∣∣​21​∣∣∣∣​=?∣∣∣∣​-21​∣∣∣∣​∣∣∣∣​21​∣∣∣∣​=21​21​=?∣∣∣∣​-21​∣∣∣∣​∣∣∣∣​-21​∣∣∣∣​=21​ 21​=21​ x=23​ is not extraneous.
Exercises 45 To identify the various parts of this equation, we need to first recognize that this equation is an adaptation of the standard distance formula, d=r⋅t where d is distance, r is the rate, and t is the time. With this in mind, we can recognize that 48 must be our rate, which is measured in feet per second. However, inside the absolute value we have: 300−48t. We can know that 300 is the initial distance away from you because it is constant in an equation about distance. We can solve for the time(s) when the car is 60 ft away from you by substituting d=60 into our equation. d=∣300−48t∣⇒60=∣300−48t∣​ When solving an absolute value equation, we need to remember to look at both the positive and negative cases. ​Case 1: -60=300−48tCase 2: -60=300−48t​ Now, we can solve these cases separately. Let's start with the positive case. 60=300−48tLHS−300=RHS−300-240=-48tLHS⋅(-1)=RHS⋅(-1)240=48tLHS/48=RHS/4848240​=tCalculate quotient5=tRearrange equationt=5 Our first solution is that the car will be 60 feet away from you when 5 seconds have passed. Let's take a look at the negative case. -60=300−48tLHS−300=RHS−300-360=-48tLHS⋅(-1)=RHS⋅(-1)360=48tLHS/48=RHS/4848360​=tba​=b/24a/24​215​=t Write as a Mixed Number Rewrite 15 as 14+1214+1​=tWrite as a sum of fractions214​+21​=tCalculate quotient7+21​=tRearrange equationt=7+21​Rewrite 7+21​ as 721​ t=721​Rewrite 721​ as 7.5t=7.5 Our second solution is that the car will be 60 feet away from you again when 7.5 seconds have passed.
Exercises 46 Unfortunately, our friend is incorrect. They are on the right track, but have misunderstood a key piece of the idea. An absolute value is a distance. Hence, it must always be positive. If we encounter the absolute value of an expression and it is set equal to a negative number, the equation has no solution. For example, we can conclude, without solving, that the equation ∣x+2∣=-3 has no solution because it shows the absolute value of an expression equal to a negative number. However, notice that on the left-hand side, the absolute value is isolated. We have to make sure this is the case before concluding there is no solution.The given equation is ∣3x+8∣−9=-5. Although the left-hand side is set equal to a negative number, the absolute value is not isolated. Thus, we cannot conclude that there is no solution. This is the error our friend made. To isolate the absolute value, we can add 9 to both sides. This gives ∣3x+8∣=4. Now we can see that absolute value is not equal to a negative number. We'll need to solve to actually determine the solutions.
Exercises 47 aLooking at the graph, we can tell that the percent of teens who are in favor of year-round school is 32%. The actual percent of students who are in favor cannot differ from this by more than 5%. The difference between the actual percent, x%, and the percent displayed can be expressed algebraically as an absolute value. ∣x−32∣ We must use an absolute value because the value of x can differ in either the positive or negative direction. Because the error must be not more than 5%, we should use an inequality to show this. ∣x−3∣≤5 To solve for x, we need to remove the absolute value. This creates a compound inequality where the argument must be greater than or equal to -5 and less than or equal to 5. -5≤x−32≤5 Let's solve for x by looking at our compound inequality in two separate cases. ​Case 1:   -5≤x−32Case 2:   -5≤x−32≤5​ Let's solve these cases one at a time.Case 1 -5≤x−32LHS+32≤RHS+3227≤x The percent of students in favor cannot be less than 27%.Case 2 x−32≤5LHS+32≤RHS+32x≤37 The percent of the students in favor cannot be greater than 37%.Conclusion The percent of students who are in favor of year-round school could be between 27% to 37%.bTo see if the claim will fit in the known range, we need to convert the fraction to a percentage. 31​Use a calculator≈33% 33% is greater than 27% but less than 37%. Therefore, it does not conflict with the survey data.
Exercises 48 aThe recommended weight of a soccer ball is 430 grams. The difference between the recommended weight and the actual weight, x grams, can be expressed algebraically as the following absolute value. ∣x−430∣ We must use an absolute value because the weight can differ in either the positive or negative direction. Because our weight must not be more than 20 grams off, we should use an inequality to show this. ∣x−430∣≤20 To solve for x, we need to remove the absolute value. This creates a compound inequality where the argument must be greater than or equal to -20 and less than or equal to 20. -20≤x−430≤20 Let's solve for x by looking at our compound inequality in two separate cases: ​Case 1:   -20≤x−430Case 2:   -20≤x−430≤20​ Let's solve these cases one at a time.Case 1 -20≤x−430LHS+430≤RHS+430410≤x The weight cannot be less than 410 grams.Case 2 x−430≤20LHS+430≤RHS+430x≤450 The weight cannot be greater than 450 gramsConclusion The minimum acceptable weight is 410 grams and the maximum acceptable weight is 450 grams.bTo find out whether or not the weight is acceptable, we have to determine if it falls between 410 grams and 450 grams. We know that the weight of a ball that was originally 423 grams decreased by 16 grams. 423−16=407 The ball now weighs 407 grams. We know the minimum acceptable weight is 410 grams. 407≤410 Therefore, the ball weighs less than the minimum acceptable weight and is not acceptable.
Exercises 49 The given equation is: ∣2x−1∣=-9. Upon analyzing the equation, we can see that the absolute value expression is isolated on one side, and it is set equal to a negative number. Since absolute value represents a distance, it can never be negative. In other words, there are no values for x that will make the given equation true. It has no solution.
Exercises 50 To find the error in the work provided, it can be helpful to analyze the work line by line and think through how we would solve the equation. Looking at the provided work, no errors have been committed. The absolute value equation was written as two equations correctly, and inverse operations were applied correctly. However, the solutions were not checked. When solving absolute value equations, we have to check that the solutions actually satisfy the original equations or if they are extraneous.Check x=-2 ∣5x+8∣=xx=-2∣5(-2)+8∣=?-2a(-b)=-a⋅b∣-5⋅2+8∣=?-2Multiply∣-10+8∣=?-2Add terms∣-2∣=?-2∣-2∣=22≠-2 x=-2 is an extraneous solution.Check x=-34​ ∣5x+8∣=xx=-34​∣∣∣∣​5(-34​)+8∣∣∣∣​=?-34​a(-b)=-a⋅b∣∣∣∣​-5⋅34​+8∣∣∣∣​=?-34​Multiply∣∣∣∣​-320​+8∣∣∣∣​=?-34​Write as a fraction∣∣∣∣​-320​+324​∣∣∣∣​=?-34​Add terms∣∣∣∣​-34​∣∣∣∣​=?-34​∣∣∣∣​-34​∣∣∣∣​=34​34​≠-34​ x=-34​ is an extraneous solution.
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