Exercise 7 Page 94

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 side using inverse operations. In this case, we need to start by adding 2.6b to both sides of the equation to isolate the b-terms on the right-hand side. Now, we will add 17 to both sides of the equation to isolate the constant terms on the left-hand side.

4=3.5b-17

AddEqn

LHS+ 17=RHS+ 17

4+ 17=3.5b-17+ 17

AddTerms

Add terms

21=3.5b

Finally, to solve the equation we should divide both sides by the number standing next to the variable. To avoid division by a decimal, we will first multiply both sides of the equation by 2.

21=3.5b

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Solve for b

MultEqn

LHS * 2=RHS* 2

21* 2=3.5b* 2

CommutativePropMult

Commutative Property of Multiplication

21* 2=3.5* 2* b

Multiply

Multiply

42=7b

DivEqn

.LHS / 7.=.RHS / 7.

42 ÷ 7 = 7b ÷ 7

CalcQuot

Calculate quotient

6=b

RearrangeEqn

Rearrange equation

b=6

The solution to the equation is b=6.