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

SubEqn

LHS-4=RHS-4

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

SubTerm

Subtract term

2/3x=3/5x-6

MoveRightFacToNum

a/c* b = a* b/c

2x/3=3x/5-6

SubEqn

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

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

SubTerm

Subtract term

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

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

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

ExpandFrac

a/b=a * 5/b * 5

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

ExpandFrac

a/b=a * 3/b * 3

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

SubFrac

Subtract fractions

10x-9x/15=- 6

SubTerm

Subtract term

x/15=- 6

MultEqn

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

Substitute

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

MultPosNeg

a(- b)=- a * b

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

CalcQuot

Calculate quotient

-60+4 ? = -54-2

AddSubTerms

Add and subtract terms

-56=-56

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

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