Solving Simple Equations

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Exercises 1 Inverse operations are operations that undo each other. They are used in solving equations to isolate a variable.Adding and subtracting Addition and subtraction are inverse operations. Consider the equation below. On the left-hand side we have x plus 2. Subtracting 2 from both sides "undoes" the addition of 2 to x. x+2x+2 − 2x​=4=4 − 2=2​Multiplication and division Similarly, multiplication and division are inverse operations. Consider the equation below. On the left-hand side we have x multiplied by 4. Dividing by 4 on both sides "undoes" the multiplication of x by 4. 4x44x​x​=12=412​=3​
Exercises 2 Two equations are "equivalent" when they have the same solution. To determine if the given equations are equivalent we will solve each one individually and then compare their solutions.Equation 1 -2x=10LHS/(-2)=RHS/(-2)x=-5 The solution to Equation 1 is x=-5.Equation 2 -5x=25LHS/(-5)=RHS/(-5)x=-5 The solution to Equation 1 is also x=-5. Since both equations have the same solution, we can conclude that the equations are equivalent.
Exercises 3 To solve an equation, we must isolate the variable. We do this by applying inverse operations. Let's consider the given equation. 14x=56 Notice that the x is being multiplied by 14. We must undo the multiplication by dividing by 14 on both sides. Thus, we will use the Division Property of Equality to solve.
Exercises 4 To begin, we will name the equations.NameEquation A8=2x​ B3=x÷4 Cx−6=5 D3x​=9 We notice that Equations A and D contain a fraction, Equation B contains division and Equation C contains subtraction. Fractions are one way to represent division. 2x​=x divided by 2​ This means that Equations A and D actually contain division. Since Equation C is the only equation that does not contain division, it does not belong.
Exercises 5 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. x+5=8LHS−5=RHS−5x+5−5=8−5Subtract termsx=3 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. x+5=8x=33+5=?8Add terms8=8 Since substituting x=3 resulted in an identity, our solution is correct.
Exercises 6 To solve for m, we need to isolate it by subtracting 9 from both sides of the equation. We want to subtract 9 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Subtraction Property of Equality. m+9=2LHS−9=RHS−9m+9−9=2−9Subtract termsm=-7 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. m+9=2m=-7-7+9=?2Add terms2=2 Since substituting m=-7 resulted in an identity, our solution is correct.
Exercises 7 To solve for y, we need to isolate it by adding 4 to both sides of the equation. We want to add 4 because subtraction and addition are inverse operations — they cancel each other out. This is allowed by the Addition Property of Equality. y−4=3LHS+4=RHS+4y−4+4=3+4Add termsy=7 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. y−4=3y=77−4=?3Subtract term3=3 Since substituting y=7 resulted in an identity, our solution is correct.
Exercises 8 To solve for s, we need to isolate it by adding 2 from both sides of the equation. We want to add 2 because subtraction and addition are inverse operations — they cancel each other out. This is allowed by the Addition Property of Equality. s−2=1LHS+2=RHS+2s−2+2=1+2Add termss=3 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. s−2=1s=33−2=?1Subtract term1=1 Since substituting s=3 resulted in an identity, our solution is correct.
Exercises 9 To solve for w, we need to isolate it by subtracting 3 from both sides of the equation. We want to subtract 3 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Subtraction Property of Equality. w+3=-4LHS−3=RHS−3w+3−3=-4−3Subtract termsw=-7 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. w+3=-4w=-7-7+3=?-4Add terms-4=-4 Since substituting w=-7 resulted in an identity, our solution is correct.
Exercises 10 To solve for n, we need to isolate it by adding 6 to both sides of the equation. We want to add 6 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Addition Property of Equality. n−6=-7LHS+6=RHS+6n−6+6=-7+6Add termsn=-1 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. n−6=-7n=-1-1−6=?-7Subtract term-7=-7 Since substituting n=-1 resulted in an identity, our solution is correct.
Exercises 11 To solve for p, we need to isolate it by adding 11 to both sides of the equation. We want to add 11 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Addition Property of Equality. -14=p−11LHS+11=RHS+11-14+11=p−11+11Add terms-3=pRearrange equationp=-3 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. -14=p−11p=-3-14=?-3−11Subtract term-14=-14 Since substituting p=-3 resulted in an identity, our solution is correct.
Exercises 12 To solve for q, we need to isolate it by subtracting 4 from both sides of the equation. We want to subtract 4 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Subtraction Property of Equality. 0=4+qLHS−4=RHS−40−4=4+q−4Subtract terms-4=qRearrange equationq=-4 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 0=4+qq=-40=?4+(-4)a+(-b)=a−b0=?4−4Subtract term0=0 Since substituting q=-4 resulted in an identity, our solution is correct.
Exercises 13 To solve for r, we first need to rewrite the addition of the negative term as subtraction. Then we will add 8 to both sides of the equation to isolate the variable. We want to add 8 because addition and subtraction are inverse operations — they cancel each other out. This is allowed by the Addition Property of Equality. r+(-8)=10a+(-b)=a−br−8=10LHS+8=RHS+8r−8+8=10+8Add termsr=18 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. r+(-8)=10r=1818+(-8)=?10a+(-b)=a−b18−8=?10Subtract terms10=10 Since substituting r=18 resulted in an identity, our solution is correct.
Exercises 14 To solve for t, we first need to rewrite the subtraction of the negative term as addition. Then we will subtract 5 from both sides of the equation to isolate the variable. 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. t−(-5)=9a−(-b)=a+bt+5=9LHS−5=RHS−5t+5−5=9−5Subtract termst=4 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. t−(-5)=9t=44−(-5)=?9a−(-b)=a+b4+5=?9Add terms9=9 Since substituting t=4 resulted in an identity, our solution is correct.
Exercises 15 To find the original price of the amusement park ticket, we first need to write an equation. …=…​ On one side, we will write an expression for the price of a discounted ticket. We are told that the discounted ticket is "$12.95 less than the original price p." The phrase less than indicates that we should subtract $12.95 from p. p−12.95=…​ Observe the image of the ticket. Since the left-hand side of the equation is an expression for the price of a discounted ticket, it will be equal to the given price of a discounted ticket, which is $44. p−12.95=44​ With this we can use inverse operations to isolate the original price of the ticket p. p−12.95=44LHS+12.95=RHS+12.95p=56.95 The original price of the ticket is $56.95.
Exercises 16 To find your final score, we first need to write an equation. …=…​ We are told that our score x is 12 less than our friend's score. Since x comes before "is" in the verbal expression, it will be written on the left-hand side of the equation. x=…​ Observing the score card, we can see that our friend scored 195 points. "12 less than your friend's score" indicates that we should subtract 12 from 195. x=195−12​ To calculate our score x we need to subtract terms on the right-hand side. x=195−12Subtract termx=183 Our score was 183 points.
Exercises 17 It is given that the sum of all interior angles in the given quadrilateral is 360∘. We can thus write an equation that adds all of the angles together and is equal to 360∘. x+100+120+100=360 Solving this equation for x will give us the measure of the unknown angle. x+100+120+100=360Add termsx+320=360LHS−320=RHS−320x+320−320=360−320Subtract termx=40 The unknown angle measures 40∘.
Exercises 18 It is given that the sum of all interior angles in the given quadrilateral is 360∘. We can thus write an equation that adds all of the angles together and is equal to 360∘. x+48+150+77=360 Solving this equation for x will give us the measure of the unknown angle. x+48+150+77=360Add termsx+275=360LHS−275=RHS−275x+275−275=360−275Subtract termx=85 The unknown angle measures 85∘.
Exercises 19 It is given that the sum of all interior angles in the given quadrilateral is 360∘. We can thus write an equation that adds all of the angles together and is equal to 360∘. x+122+92+76=360 Solving this equation for x will give us the measure of the unknown angle. x+122+92+76=360Add termsx+290=360LHS−290=RHS−290x+290−290=360−290Subtract termx=70 The unknown angle measures 70∘.
Exercises 20 It is given that the sum of all interior angles in the given quadrilateral is 360∘. We can thus write an equation that adds all of the angles together and is equal to 360∘. x+60+115+85=360 Solving this equation for x will give us the measure of the unknown angle. x+60+115+85=360Add termsx+260=360LHS−260=RHS−260x+260−260=360−260Subtract termx=100 The unknown angle measures 100∘.
Exercises 21 We can solve for g by dividing both sides of the equation by 5. We want to divide because multiplication and division are inverse operations. This is allowed by the Division Property of Equality. 5g=20LHS/5=RHS/555g​=520​Calculate quotientg=4 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 5g=20g=45(4)=?20Multiply20=20 Since substituting g=4 resulted in an identity, our solution is correct.
Exercises 22 We can solve for q by dividing both sides of the equation by 4. We want to divide because multiplication and division are inverse operations. This is allowed by the Division Property of Equality. 4q=52LHS/4=RHS/444q​=452​Calculate quotientq=13 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 4q=52q=134(13)=?52Multiply52=52 Since substituting q=13 resulted in an identity, our solution is correct.
Exercises 23 To solve for p, we need to isolate it by multiplying both sides of the equation by 5. We should multiply because division and multiplication are inverse operations. We are allowed to do this by the Multiplication Property of Equality. p÷5=3Write as a fraction5p​=3LHS⋅5=RHS⋅55p​⋅5=3⋅55a​⋅5=ap=3⋅5Multiplyp=15 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. p÷5=3p=1515÷5=?3Write as a fraction515​=?3Calculate quotient3=3 Since substituting p=15 resulted in an identity, our solution is correct.
Exercises 24 To solve for y, we need to isolate it by multiplying both sides of the equation by 7. We should multiply because division and multiplication are inverse operations. We are allowed to do this by the Multiplication Property of Equality. y÷7=1Write as a fraction7y​=1LHS⋅7=RHS⋅77y​⋅7=1⋅77a​⋅7=ay=1⋅7Multiplyy=7 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. y÷7=1y=77÷7=?1Write as a fraction77​=?1Calculate quotient1=1 Since substituting y=7 resulted in an identity, our solution is correct.
Exercises 25 To solve the equation, we have to isolate the variable. The given equation is -8r=64. Notice that -8 is multiplying the variable term. To undo this multiplication we will use the Division Property of Equality to divide by -8 on both sides. -8r=64LHS/(-8)=RHS/(-8)-8-8r​=-864​Calculate quotientr=-8 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. -8r=64r=-8-8(-8)=?64-a(-b)=a⋅b8⋅8=?64Multiply64=64 Since substituting r=-8 resulted in an identity, our solution is correct.
Exercises 26 To solve the equation, we will isolate the variable. To undo the division in the given equation, we will use the Multiplication Property of Equality to multiply both sides of the equation by -2. x÷(-2)=8Write as a fraction-2x​=8LHS⋅(-2)=RHS⋅(-2)-2x​⋅(-2)=8(-2)(-2)a​⋅(-2)=ax=8(-2)Multiplyx=-16 We will check our solution by substituting this result back into the original equation and simplifying. x÷(-2)=8x=-16-16÷(-2)=?8Write as a fraction-2-16​=?8-b-a​=ba​216​=?8Calculate quotient8=8 Substituting x=-16 resulted in true statement, so our solution is correct.
Exercises 27 To solve for x, we need to isolate it by multiplying both sides of the equation by 6. We should multiply because division and multiplication are inverse operations. We are allowed to do this by the Multiplication Property of Equality. 6x​=8LHS⋅6=RHS⋅66x​⋅6=8⋅66a​⋅6=ax=8⋅6Multiplyx=48 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 6x​=8x=48648​=?8Calculate quotient8=8 Since substituting x=48 resulted in an identity, our solution is correct.
Exercises 28 To solve for w, we need to isolate it by multiplying both sides of the equation by -3. We should multiply because division and multiplication are inverse operations. We are allowed to do this by the Multiplication Property of Equality. -3w​=6LHS⋅(-3)=RHS⋅(-3)-3w​⋅(-3)=6(-3)-3a​⋅-3=aw=6(-3)a(-b)=-a⋅bw=-18 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. -3w​=6w=-18-3-18​=?6-b-a​=ba​318​=?6Calculate quotient6=6 Since substituting w=-18 resulted in an identity, our solution is correct.
Exercises 29 We can solve for s by dividing both sides of the equation by 9. We want to divide because multiplication and division are inverse operations. This is allowed by the Division Property of Equality. -54=9sLHS/9=RHS/99-54​=99s​Calculate quotient-6=sRearrange equations=-6 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. -54=9ss=-6-54=?9(-6)a(-b)=-a⋅b-54=-54 Since substituting s=-6 resulted in an identity, our solution is correct.
Exercises 30 To solve for t, we need to isolate it by multiplying both sides of the equation by 7. We should multiply because division and multiplication are inverse operations. We are allowed to do this by the Multiplication Property of Equality. -7=7t​LHS⋅7=RHS⋅7-7⋅7=7t​⋅77a​⋅7=a-7⋅7=tMultiply-49=tRearrange equationt=-49 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. -7=7t​t=-49-7=?7-49​Put minus sign in front of fraction-7=?-749​Calculate quotient-7=-7 Since substituting t=-49 resulted in an identity, our solution is correct.
Exercises 31 To solve the equation, we have to isolate t on one of the sides. By subtracting 23​ from both sides, we can eliminate 23​ from the left side of the equation. 23​+t=21​LHS−23​=RHS−23​23​+t−23​=21​−23​ Subtract terms Subtract termt=21​−23​Subtract fractionst=21−3​Subtract termt=2-2​Put minus sign in front of fraction t=-22​Calculate quotientt=-1 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 23​+t=21​t=-123​+(-1)=?21​a+(-b)=a−b23​−1=?21​ Write -1 as a fraction a=22⋅a​23​−22⋅1​=?21​Multiply 23​−22​=?21​Subtract fractions23−2​=?21​Subtract term21​=21​ Since substituting t=-1 resulted in an identity, our solution is correct.
Exercises 32 To solve the equation, we have to isolate b on one of the sides. By adding 163​ to both sides, we can eliminate 163​ from the left side of the equation. b−163​=165​LHS+163​=RHS+163​b−163​+163​=165​+163​Add termsb=165​+163​Add fractionsb=165+3​Add termsb=168​ba​=b/8a/8​b=21​ We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. b−163​=165​b=21​21​−163​=?165​ba​=b⋅8a⋅8​168​−163​=?165​Subtract fractions168−3​=165​Subtract terms165​=165​ Since substituting b=21​ resulted in an identity, our solution is correct.
Exercises 33 To solve for m, we need to eliminate the coefficient. We can do this by multiplying both sides of the equation by the fraction's inverse. 73​m=6LHS⋅37​=RHS⋅37​73​m⋅37​=6⋅37​ba​⋅ab​=1m=6⋅37​Write as a fractionm=16​⋅37​Multiply fractionsm=1⋅36⋅7​Multiplym=342​Calculate quotientm=14 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 73​m=6m=1473​(14)=?6ca​⋅b=ca⋅b​73⋅14​=?6Multiply742​=?6Calculate quotient6=6 Since substituting m=14 resulted in an identity, our solution is correct.
Exercises 34 To solve for y, we need to eliminate the coefficient. First, let's rewrite the product on the left-hand side. -52​y⇔5-2y​ This makes the steps we need to take to isolate y a bit simpler. By multiplying both sides by 5, we have only -2y on the left-hand side. Then we can divide by -2 to fully isolate y. 5-2y​=4LHS⋅5=RHS⋅55-2y​⋅5=4⋅55a​⋅5=a-2y=4⋅5Multiply-2y=20LHS/(-2)=RHS/(-2)-2-2y​=-220​Calculate quotienty=-10 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 5-2y​=4y=-105-2(-10)​=?4-a(-b)=a⋅b52⋅10​=?4Multiply520​=?4Calculate quotient4=4 Since substituting y=-10 resulted in an identity, our solution is correct.
Exercises 35 By adding 0.4 to both sides, we can eliminate the -0.4 on the right-hand side and isolate a. 5.2=a−0.4LHS+0.4=RHS+0.45.2+0.4=a−0.4+0.4Add terms5.6=aRearrange equationa=5.6 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. 5.2=a−0.4a=5.65.2=?5.6−0.4Subtract terms5.2=5.2 Since substituting a=5.6 resulted in an identity, our solution is correct.
Exercises 36 Remember that π is just a constant, not another variable. When we keep that in mind, we can solve for f by subtracting 3π from both sides of the equation. We want to subtract 3π because addition and subtraction are inverse operations - they cancel each other out. This is allowed by the Subtraction Property of Equality. f+3π=7πLHS−3π=RHS−3πf+3π−3π=7π−3πSubtract termsf=4π We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. f+3π=7πf=4π4π+3π=?7πAdd terms7π=7π Since substituting f=4π resulted in an identity, our solution is correct.
Exercises 37 Remember that π is just a constant, not another variable. When we keep that in mind, we can solve for j by dividing both sides of the equation by 6π. We want to divide because multiplication and division are inverse operations. This is allowed by the Division Property of Equality. -108π=6πjLHS/6π=RHS/6π6π-108π​=6π6πj​Calculate quotient-18=jRearrange equationj=-18 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. -108π=6πjj=-18-108π=?6π(-18)Multiply-108π=-108π Since substituting j=-18 resulted in an identity, our solution is correct.
Exercises 38 To solve the equation, we will isolate the variable. To undo the division in the given equation, we will use the Multiplication Property of Equality to multiply both sides of the equation by -2. x÷(-2)=1.4Write as a fraction-2x​=1.4LHS⋅(-2)=RHS⋅(-2)-2x​(-2)=1.4(-2)-2a​⋅-2=ax=1.4(-2)a(-b)=-a⋅bx=-2.8 We will check our solution by substituting this result back into the original equation and simplifying. If doing so results in an identity, our solution is correct. An identity is when both sides of an equation are exactly when same when fully simplified. x÷(-2)=1.4x=-2.8-2.8÷(-2)=?1.4 Calculate quotient Write as a fraction-2-2.8​=?1.4-b-a​=ba​22.8​=?1.4Calculate quotient 1.4=1.4 Since substituting x=-2.8 resulted in an identity, our solution is correct.
Exercises 39 Looking at the erroneous solution, we can see that the mistake made was related to whether -0.8 should be added or subtracted from both sides. The critical mistake was that -0.8 was added to both sides of the equation, rather than subtracted. Incorrect: r=12.6 + (-0.8)Correct: r=12.6 − (-0.8)​ One way to avoid this mistake would be to apply the Commutative Property of Addition to rearrange the terms on the left-hand side of the equation. See our correct solution below for an example. -0.8+r=12.6Commutative Property of Additionr+(-0.8)=12.6a+(-b)=a−br−0.8=12.6LHS+0.8=RHS+0.8r−0.8+0.8=12.6+0.8Add termsr=13.4
Exercises 40 Looking at the erroneous solution, we can see that the mistake was made between Line 2 and Line 3. While multiplying both sides of the equation by 3, according to the Multiplication Property of Equality, the negative sign in front of -3m​ was lost. Incorrect Line 3: Correct Line 3: ​-m=-12-m=-12​ To correct the error, we need to either continue the given solution by dividing both sides of the equation by -1 or, in Line 2, multiply the equation by -3 rather than 3. It may also help to begin solving by moving the negative to the denominator. -3m​=-4Put minus sign in denominator-3m​=-4LHS⋅-3=RHS⋅-3-3(-3m​)=-3(-4)-3a​⋅-3=am=-3(-4)-a(-b)=a⋅bm=3⋅4Multiplym=12
Exercises 41 We are told that the baker ordered 162 eggs in x cartons, each containing 18 eggs. In other words, after the eggs are delivered to the bakery, the baker will have to open cartons of 18 eggs x times to crack all 162 eggs for the recipes.VerbalAlgebraic eggs per carton 18 times × the number of cartons x equals = total number of eggs 162 18×x=162⇔18x=162 This equation corresponds with option C.
Exercises 42 Let x represent the change in temperature. Since the beginning temperature is 20∘, subtracting x from 20 gives the ending temperature of -5∘. 20−x=-5 Solving this equation for x gives the change in temperature. 20−x=-5LHS−20=RHS−20-x=-25LHS/-1=RHS/-1x=25 The solution x=25 means that the temperature decreases 25∘.
Exercises 43 Let w represent the width of the flag. It is given that the length of the flag is 1.9 times as long as the width and that the length is 9.5 feet. 9.5=1.9w Let's solve the equation by dividing by 1.9 on both sides. 9.5=1.9wLHS/1.9=RHS/1.95=wRearrange equationw=5 The solution w=5 means that the width of the flag is 5 feet.
Exercises 44 Let y represent the balance in the account 4 years ago. It is given that the balance has increased $308 over the last four years and that the current balance is $4708 dollars. Adding 308 to y results in a balance of 4708. y+308=4708 Solving this equation for y will give us the balance of the account 4 years ago. y+308=4708LHS−308=RHS−308y=4400 Four years ago, the balance in the account was $4400.
Exercises 45 Let's think about the x-term on the left-hand side in both equations. In Equation 1, we have x. In Equation 2, we have 4x. This means that we can multiply the x-term in Equation 1 by 4 to make it equal the x-term in Equation 2. x⋅4=4x Let's check to see if multiplying all of Equation 1 by 4 makes it equivalent to Equation 2. x−21​=4x​+3LHS⋅4=RHS⋅4(x−21​)4=(4x​+3)4Distribute 44x−24​=44x​+12Calculate quotient4x−2=x+12 Multiplying Equation 1 by 4 did, in fact, make it equivalent to Equation 2. Thus, we used the Multiplication Property of Equality.
Exercises 46 aLet x be the area of one rectangular mat. It is given that there are four rectangle mats and one square mat. We also know that the area of the square mat is half the area of one rectangular mat. Adding these two areas, we have an expression for the total area. 4x+21​x It is also given that the total area is 81 feet squared. We can set the above expression equal to 81 and solve for x. This will give us the area of one rectangular mat. 4x+21​x=81LHS⋅2=RHS⋅28x+x=162Add terms9x=162LHS/9=RHS/9x=18 The area of one rectangular mat is 18 square feet.bBefore using Guess, Check, and Revise, we will write an equation using the given information. Let ℓ be the length of a rectangular mat and w be the width. It is given that the length is twice the width. ℓ=2w The area of a rectangle can be found by multiplying the length and the width.Let's substitute x=18 and try to solve for w. x=2w⋅wx=1818=2w⋅wMultiply18=2w2LHS/2=RHS/29=w2Rearrange equationw2=9 Now, we can arbitrarily choose values for w ("Guess") and substitute them into the equation. Then, we will evaluate the equation for each ("Check"). We "Revise" by trying different values for w until we have w2=9.ww2= 1122 2228 3329Our work in the table shows that the width of the rectangle is 3 feet. Substituting w=3 into our equation for ℓ will give us the length of one rectangular mat. l=2ww=3l=2(3)Multiplyl=6 The dimensions of a rectangular mat are 3 feet by 6 feet.
Exercises 47 aLet x represent the discounted price of each CD. It is given that we buy 4 CDs and that the total we spend is $30.40 4x=30.40 Solving this equation for x will give us the discounted price of each CD. 4x=30.40LHS/4=RHS/4x=7.60 The solution x=7.60 means that the discounted price of one CD is $7.60.bIt is given in Part B that the CDs are no longer on sale. In order to determine if $25 is enough to buy 3 CDs, we must find the full price of one CD.What is the full price of one CD? In Part A, it was given that the price of CDs were discounted. In fact, we only paid 80% of the original cost. We need to determine the full price of one CD. We know that $7.60 is 80% of the original cost. If we let c represent the original cost, we can write the following equation. 7.60=0.80c Solving the equation for c will give us the full price of one CD. 7.60=0.80cLHS/0.80=RHS/0.809.50=cRearrange equationc=9.50 The solution c=9.50 means that the full cost of one CD is $9.50.Is $25 enough for 3 CDs? Now that we know the cost of one CD, we can determine the cost of three CDs by multiplying 9.50 by 3. 9.50⋅3=28.50 At full price, three CDs will cost $28.50. Thus, $25 is not enough.
Exercises 48 Let's solve each equation for x and then substitute increasing values of c if it's not obvious how x will behave.Equation 1 Let's solve for x. x−c=0LHS+c=RHS+cx=c In this equation, x is equal to c so as c increases, x also increases.Equation 2 Let's solve for x. cx=1LHS/c=RHS/cx=c1​ By substituting c with increasing numbers, we can determine what happens to the value of x.cc1​= 111​1 221​0.5 331​0.33333… From the table, we see that as c increases, x decreases.Equation 3 Let's solve for x. cx=cLHS/c=RHS/cx=cc​aa​=1x=1 Since x equals 1, it doesn't depend on the value of c. In other words, it stays the same.Equation 4 Let's solve for x. cx​=1LHS⋅c=RHS⋅cx=c Again, x equals c which means that as c increases, x increases too.
Exercises 49 aFirst, let's isolate x on the left-hand side. ax=b−5LHS/a=RHS/ax=ab−5​ Now we can substitute each of the integers for b and check the value of the numerator.bab−5​Simplify -2a-2−5​a-7​ 5a5−5​a0​ 9a9−5​a4​ 10a10−5​a5​ Right away we can disregard -2 and 5 as values of b. Inputting these values resulted in fractions that won't give a positive quotient when dividing by any of the remaining integers. We only have two options now. b=9 or b=10​ While 4 is divisible by -2, this would result in a negative quotient so we can disregard b=9. Dividing 5 by 5 equals 1, a positive integer! When a=.10.​ and b=.5.​,x is a positive integer.​bFrom Part A, we know that substituting the various values for b gives us the following fractions. a-7​, a0​, a4​, and a5​ Yet again, we can disregard the two first fractions because neither of these will give us a negative integer when substituting any of the remaining values for a. For the remaining two fractions, we are going to need a negative denominator to get a negative quotient. Therefore, the denominator must be a=-2. The only numerator that is divisible by -2 is 4! When a=-2.​ and b=.9.​,x is a positive integer.​
Exercises 50 aThe entire circle is represented by 100%. Whenever we have the entirety of something, we have 100% of it. Think about when you get 100% on a test, that means you got every question correct. Or, when you have completed every quest and collected every coin in a video game, you have completed 100% of the game.bThe left-hand side of the equation represents the sum of the percentages of all the animals that are in the pet store. We are equating it to 100 because the sum of the percentages should equal 100%. Since x represents the percent of cats sold, to find it we should solve the equation for x.
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