Exercise 38 Page 23

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

g=-5

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side, using the Properties of Equality. In this case, we need to start by combining like terms.

5g+18-7g+4g=8

SubTerm

Subtract term

-2g+18+4g=8

AddTerms

Add terms

2g+18=8

Now we can isolate g using Subtraction and Division Properties of Equality.

2g+18=8

▼

LHS-18=RHS-18

SubEqn

LHS-18=RHS-18

2g+18-18=8-18

SubTerms

Subtract terms

2g=-10

▼

.LHS /2.=.RHS /2.

DivEqn

.LHS /2.=.RHS /2.

2g/2=-10/2

MoveNegNumToFrac

Put minus sign in front of fraction

2g/2=-10/2

CalcQuot

Calculate quotient

g=-5

The solution to the equation is g=-5. We can check our solution by substituting it into the original equation.

5g+18-7g+4g=8

Substitute

g= -5

5( -5)+18-7( -5)+4( -5) ? = 8

Multiply

-25+18+35-20 ? = 8

AddSubTerms

Add and subtract terms

8=8

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

Subchapter links

Check Your Understanding p.22

Practice and Problem Solving p.22-24

H.O.T. Problems p.24

Standardized Test Practice p.25

Spiral Review p.25

Skills Review p.25

Exercise 38 Page 23

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