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Real-life situations modeled by an equation include a coefficient of the variable which can be a rational number. Both sides of the equation are multiplied or divided by a number to isolate the variable. This lesson begins its focus on solving equations involving multiplication and division. Then, the focus shifts to how to write these equations when modeling real-life problems.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Challenge

Diego is going through some old stuff in storage. He discovers an entire rack of vinyls! There are $200$ in total and they belong to his dad. Diego asks his father how long this collection took to gather.

a Write an equation in terms of $m$ that represents the situation.

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b Solve the equation to find the time it took Diego's father to gather his collection of records.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"months","answer":{"text":["20"]}}

Discussion

Multiplication and division are inverse operations. They can be used to solve equations by the following properties of equality.

Rule

Given an equation, multiplying each side of the equation by the same number yields an equivalent equation. Let $a,$ $b,$ and $c$ be real numbers.

If $a=b,$ then $a×c=b×c.$

$x÷4x÷4×4x =2=2×4=8 $

Here, by multiplying both sides of the equation by $4,$ the variable $x$ was isolated and the solution of the equation was found.Discussion

Dividing each side of an equation by the same nonzero number yields an equivalent equation. Let $a,$ $b,$ and $c$ be real numbers.

If $a=b$ and $c =0,$ then $a÷c=b÷c.$

$5x5x÷5x =10=10÷5=2 $

As can be observed, by dividing both sides of the equation by $5,$ the variable $x$ was isolated and the solution of the equation was found.Example

a Diego's father stored his vinyl collection in five boxes. Each box contains the same number of records and, in total, there are $200$ vinyl records in the collection. This situation can be represented by an equation.

$5b=200 $

Here, $b$ represents the number of records in one box. Solve this equation to find the number $b$ of records in each box. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.69444em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">b<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["40"]}}

b Diego's grandfather also found some records.
Diego wants to put an equal number of those records grandpa found into each of the five boxes. He notices that he would have to put seven records into each box to do that. The following equation represents the situation.

$5r =7 $

In this equation, $r$ represents the number of records found by Diego's grandfather. Solve this equation to find the number $r$ of records Diego's grandfather found. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.02778em;\">r<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["35"]}}

a Use the Division Property of Equality.

b Use the Multiplication Property of Equality.

a When solving equations in one variable, the variable is isolated on one side of the equation. This can be done by using inverse operations because inverse operations undo each other. Consider the given equation.

$5b=200 $

Here, the variable $b$ is multiplied by $5.$ The inverse operation of multiplication is division, so each side of the equation is divided by $5$ to isolate $b.$ The reason why this operation can be done is the Division Property of Equality, which ensures that both sides of the equation remain equal.
$5b=200$

DivEqn

$LHS/5=RHS/5$

$55b =5200 $

CrossCommonFac

Cross out common factors

$5 5 b =5200 $

SimpQuot

Simplify quotient

$b=5200 $

CalcQuot

Calculate quotient

$b=40$

b Recall that the variable is isolated on one side when solving equations in one variable. Inverse operations play a role in isolating the variable because they

undoeach other. Consider the given equation.

$5r =7 $

In this equation, the variable $r$ is divided by $5.$ The inverse operation of division is multiplication, so the Multiplication Property of Equality is used to multiply each side of the equation by $5.$
$5r =7$

MultEqn

$LHS⋅5=RHS⋅5$

$5r ⋅5=7⋅5$

FracMultDenomToNumber

$5a ⋅5=a$

$r=7⋅5$

Multiply

Multiply

$r=35$

Discussion

The Multiplication Property of Equality can be used instead of the Division Property of Equality anytime when solving an equation with a coefficient that is a rational number. Consider an equation in the form $ba x=c.$

$ba x=c $

The coefficient $ba $ is a fraction. For that reason, the equation can be solved by multiplying both sides by the reciprocal of $ba .$ As an example, consider the following equation.
$53 x=6 $

The coefficient next to the variable is $53 .$ Interchange the numerator and denominator to find its reciprocal. The reciprocal of $53 $ is $35 ,$ so the equation can be solved by multiplying both sides by $35 .$ $53 x=6$

MultEqn

$LHS⋅35 =RHS⋅35 $

$53 x⋅35 =6⋅35 $

▼

Simplify

CommutativePropMult

Commutative Property of Multiplication

$x⋅53 ⋅35 =6⋅35 $

MultFracByInverse

$ba ⋅ab =1$

$x⋅1=6⋅35 $

MultByOne

$a⋅1=a$

$x=6⋅35 $

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$x=36⋅5 $

Multiply

Multiply

$x=330 $

CalcQuot

Calculate quotient

$x=10$

Example

Diego finds himself wondering about the speed at which a record rotates. His father tells Diego that the record rotates $190$ times while playing a song on the album. Diego later reads online that a record rotates $95 $ of a full rotation every second it plays.

Diego combines these two pieces of information to write the following equation.

$95 t=190 $

Here, $t$ represents the length of the song in seconds. Solve the equation to find the length of the song. {"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.61508em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">t<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["342"]}}

Identify the coefficient of the variable. Multiply both sides of the equation by the reciprocal of the coefficient.

Consider the given equation.
The solution to the equation is $t=342,$ which means that the song lasts $342$ seconds.

$95 t=190 $

Here, the variable is $t$ and its coefficient is the fraction $95 .$ Since it is a fraction, isolate the variable by multiplying both sides of the equation by the reciprocal of that fraction. Recall that the reciprocal of a fraction is found by interchanging the numerator and denominator. The reciprocal of $95 $ is $59 .$ Now, multiply the equation by $59 .$
$95 t=190$

MultEqn

$LHS⋅59 =RHS⋅59 $

$95 t⋅59 =190⋅59 $

▼

Simplify

CommutativePropMult

Commutative Property of Multiplication

$t⋅95 ⋅59 =190⋅59 $

MultFracByInverse

$ba ⋅ab =1$

$t⋅1=190⋅59 $

MultByOne

$a⋅1=a$

$t=190⋅59 $

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$t=5190⋅9 $

Multiply

Multiply

$t=51710 $

CalcQuot

Calculate quotient

$t=342$

Pop Quiz

Solve the equations using the Multiplication Property of Equality or the Division Property of Equality. If necessary, round answers to two decimal places.

Discussion

Many real-life situations can be algebraically modeled by equations. These equations can involve a variable that represents an unknown quantity. Consider modeling the following situation.

Diego categorizes his father's record collection by genre. He finds out that there are $8$ different genres, each with the same number of records. The collection contains $200$ records. How many records are there in each genre? |

$The product of the number of genresandthe number of records in each genreis equal to the total number of records. $

Here, the unknown quantity is the number of records in each genre. Let $x$ be the variable representing this unknown quantity. Then, the verbal sentence can be translated into an algebraic equation. $The product ofthe number of genresandthe number of records in each genreis equal tothe total number of records.⇓8⋅x=200 $

The equation can now be solved to find the number $x$ of records in each genre. Use the Division Property of Equality.
$8x=200$

DivEqn

$LHS/8=RHS/8$

$88x =8200 $

CrossCommonFac

Cross out common factors

$8 8 x =8200 $

SimpQuot

Simplify quotient

$x=8200 $

CalcQuot

Calculate quotient

$x=25$

Example

Diego's father's old turntable is broken. Diego is so eager to listen to the records that he decides to make some money to buy the replacement parts. The parts that Diego needs to buy costs $$96.$ Diego's neighbor offers him $$8$ per walk to walk her dog.

a Write an equation for the number $x$ of times Diego has to walk his neighbor's dog to earn enough money to buy the parts he needs.

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b Solve the equation.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.43056em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["12"]}}

c Which of the following is a true statement?

{"type":"choice","form":{"alts":["Diego needs to walk the dog exactly <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span> times.","Diego needs to walk the dog more than <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span> times.","Diego needs to walk the dog less than <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.64444em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><\/span><\/span><\/span> times."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}

a Assign the variable to the unknown quantity.

b Use the Properties of Equality to solve the equation written in Part A.

c What does the variable represent?

a The number of times Diego has to walk the dog is represented by the variable $x.$ Diego gets $$8$ per walk, so the total amount he earns is equal to $8x.$ He wants to save $$96,$ so the amount earned should be equal to $$96.$ By combining all this information, an equation can be written as follows.

$8x=96 $

This equation models the given situation.
b Solving an equation in one variable means isolating the variable on one side.

$8x=96 $

In this case, the variable $x$ is multiplied by $8.$ Since the inverse operation of multiplication is division, the Division Property of Equality is used to isolate $x.$ Now divide both sides of the equation by $8.$
$8x=96$

DivEqn

$LHS/8=RHS/8$

$88x =896 $

CrossCommonFac

Cross out common factors

$8 8 x =896 $

SimpQuot

Simplify quotient

$x=896 $

CalcQuot

Calculate quotient

$x=12$

c The variable $x$ in the equation from Part A represents the number of times Diego needs to walk his neighbor's dog to earn enough money for the parts he needs to repair the broken turntable. The calculations show that $x=12.$ This means that Diego needs to walk the dog exactly $12$ times to earn enough.

$x=12⇓Diego needs to walk the dog12times. $

Example

The dog Diego walked loved him so much — just look at the two of them! His neighbor, the dog owner, was so impressed that affter $12$ walks, she decided to pay Diego handsomely. Diego received $$10$ for each walk.

a Write an equation for the amount $t$ of money Diego got in total.

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b Solve the equation.

{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.61508em;vertical-align:0em;\"><\/span><span class=\"mord mathdefault\">t<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["120"]}}

c Which of the following is a true statement?

{"type":"choice","form":{"alts":["Diego got less than <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mord\">$<\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span>","Diego got more than <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mord\">$<\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span>","Diego got exactly <span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.80556em;vertical-align:-0.05556em;\"><\/span><span class=\"mord\">$<\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><span class=\"mord\">0<\/span><span class=\"mord\">.<\/span><\/span><\/span><\/span>"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":2}

a Assign the variable to the unknown quantity.

b Use the Properties of Equality to solve the equation written in Part A.

c What does the variable represent?

a The total amount of money Diego got for walking the dog is represented by the variable $t.$ Diego walked the dog $12$ times, so the amount he earned for one walk is $12t .$ The neighbor paid Diego $$10$ per walk, so the amount earned for one walk should be equal to $10.$ Now, an equation can be written.

$12t =10 $

Note that different equations, equivalent to this one, can also be used to model this situation. An example of such an equation is $t=12⋅10.$
b Isolate the variable $t$ on one side to solve the equation.

$12t =10 $

Notice that $t$ is divided by $12.$ This means that the Multiplication Property of Equality can be used to isolate $t$ because division and multiplication are inverse operations.
$12t =10$

MultEqn

$LHS⋅12=RHS⋅12$

$12t ⋅12=10⋅12$

FracMultDenomToNumber

$12a ⋅12=a$

$t=10⋅12$

Multiply

Multiply

$t=120$

c The variable $t$ in the equation from Part A represents the total amount of money Diego earned walking his neighbor's dog. The calculations show that $t=120.$ This means that Diego earned exactly $$120.$

$t=120⇒Diego earned$120. $

It is safe to say that Diego earned enough to fix the turntable. Play that jam! Closure

The challenge presented at the beginning of the lesson can be solved by writing an equation that models the situation and then solving the equation. It stated that Diego's father collected $200$ vinyl records and bought $10$ records every month.

a Write an equation in terms of $m$ that represents the situation.

b Solve the equation to find the time it took Diego's father to gather his collection of records.

a Assign a variable to the number of months.

b Use the Division Property of Equality.

a When modeling a real-life situation with an equation, the first step is to assign a variable to the unknown quantity. In this case, the unknown quantity is the number of months it took Diego's father to gather his collection, so let it be $m.$

$Number of Months:m $

Diego's father bought $10$ records every month. Then, the number of records Diego's father collected in $m$ months is $10m.$ There are $200$ records in the collection, so $10m$ must be equal to $200.$
$10m=200 $

Diego figured out the equation!
b The $m-$variable must be isolated to solve the equation. In this case, $m$ is multiplied by $10.$ Divide both sides of the equation by $10$ to isolate $m.$ Recall that dividing both sides by the same number does not change the solution to the equation because of the Division Property of Equality.

$10m=200$

DivEqn

$LHS/10=RHS/10$

$1010m =10200 $

FracMultDenomToNumber

$10a ⋅10=a$

$m=10200 $

CalcQuot

Calculate quotient

$m=20$

On a final note, consider an equation where the coefficient is a decimal.
Similarly, when solving equations where the variable is divided by a decimal, or a decimal is added or subtracted from the variable, we use the same methods we would when solving an equation with integers or fractions.

$1.6x=12 $

Equations where the coefficient is a decimal can be solved the same way as equations where the coefficient is an integer or a fraction. In this case, this means dividing both sides of the equation by the coefficient $1.6.$
$1.6x=12$

DivEqn

$LHS/1.6=RHS/1.6$

$1.61.6x =1.612 $

FracMultDenomToNumber

$1.6a ⋅1.6=a$

$x=1.612 $

CalcQuot

Calculate quotient

$x=7.5$