Exercise 115 Page 267

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 the properties of equality. In this case, we need to start by using the Distributive Property to simplify the left- and right-hand sides of the equation.

3(2+x)=4-(x-2)

Distr

Distribute 3

3(2)+3x=4-(x-2)

Distr

Distribute -1

3(2)+3x=4-x+(-1)(- 2)

Multiply

6+3x=4-x+2

AddTerms

Add terms

6+3x=6-x

Now we can continue to solve using the properties of equality.

6+3x=6-x

AddEqn

LHS+x=RHS+x

6+3x+x=6-x+x

SubEqn

LHS-6=RHS-6

6-6+3x+x=6-6-x+x

AddSubTerms

Add and subtract terms

4x=0

DivEqn

.LHS /4.=.RHS /4.

4x/4=0/4

CalcQuot

Calculate quotient

x=0

The solution to the equation is x=0.

We are asked to solve the following equation. x/2+x/3-1=x/6+3To do so, we will try to group all the variable terms on the left-hand side and the constant terms on the right-hand side of the equation. Here we will start by using the Multiplication Property of Equality to get rid of the fractions.

x/2+x/3-1=x/6+3

MultEqn

LHS * 6=RHS* 6

6(x/2+x/3-1)=6(x/6+3)

Distr

Distribute 6

6(x/2)+6(x/3)-6(1)=6(x/6)+6(3)

MoveLeftFacToNum

a*b/c= a* b/c

6x/2+6x/3-6(1)= 6x/6 +6(3)

CalcQuot

Calculate quotient