Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Solving Equations With Variables on Both Sides
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Exercise 20 Page 106

Write expressions to represent the two scenarios and then equate them.

40

Practice makes perfect

We have been asked to find the number of haircuts needed to make the profits at both locations the same given the associated cost of a haircut and monthly rent. Let x represent the number of haircuts. It is important to note that rent is an expense.

Location A

We need to write an expression to represent the situation. The hairdresser earns $45 per haircut at location A, but she loses $1200 per month. This gives us the following expression to represent profit. 45x-1200

Location B

At location B, the hairdresser makes $60 per haircut and loses $1800 to rent per month. Let's make an expression to represent the profit in this scenario.

60x-1800

Making an Equation

Since each expression represents profit and we want to have the same profit in both locations, we can set the expressions equal to each other to form an equation. 45x-1200=60x-1800 To find the number of haircuts that will make the profit at both locations equal, we solve for x.

Solving the Equation

Let's solve our equation from above.
60x-1800=45x-1200
60x-1800-45x=45x-1200-45x
15x-1800=- 1200
15x-1800+1800=- 1200+1800
15x=600
15x/15=600/15
x=40

Checking the Solution

To be sure that we solved the equation correctly, we substitute the value that we found back into the equation. If we get a statement that is always true, we have solved the equation correctly.
60x-1800=45x-1200
60* 40-1800 ? = 45* 40-1200
2400-1800 ? = 1800-1200
600=600
This means that we have solved the equation correctly and that to make the same profit in both locations, the hairdresser will need to give 40 haircuts.