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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

Chapter Review

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Exercise 24 Page 154

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side.

-90

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other using the Properties of Equality. In this case, we start by subtracting 4 from both sides of the equation.

2/3x+4=3/5x-2

LHS-4=RHS-4

2/3x+4-4=3/5x-2-4

Subtract term

2/3x=3/5x-6

MoveRightFacToNum

a/c* b = a* b/c

2x/3=3x/5-6

LHS-3x/5=RHS-3x/5

2x/3-3x/5=3x/5-6-3x/5

Subtract term

2x/3-3x/5=- 6

Now, we can continue to solve using the Properties of Equality.

2x/3-3x/5=- 6

a/b=a * 5/b * 5

10x/15-3x/5=- 6

a/b=a * 3/b * 3

10x/15-9x/15=- 6

Subtract fractions

10x-9x/15=- 6

Subtract term

x/15=- 6

LHS * 15=RHS* 15

x/15* 15=- 6* 15

Multiply

x/15* 15=-90

MoveRightFacToNum

a/c* b = a* b/c

x* 15/15=-90

CancelCommonFac

Cancel out common factors

x=-90

The solution to the equation is x=-90. We can check our solution by substituting it into the original equation.

2/3x+4=3/5x-2

x= -90

2/3( -90)+4 ? = 3/5( -90)-2

▼

Simplify LHS and RHS

MoveRightFacToNum

a/c* b = a* b/c

2(-90)/3+4 ? = 3(-90)/5-2

a(- b)=- a * b

-180/3+4 ? = -270/5-2

Calculate quotient

-60+4 ? = -54-2

Add and subtract terms

-56=-56

Since the left-hand side is equal to the right-hand side, our solution is correct.

Subchapter links

Exercises p.152-156