Exercise 1 Page 32

To solve the equation, we need to isolate z using inverse operations and the Properties of the Equality.

z-12=30

AddEqn

LHS+12=RHS+12

z-12+12=30+12

AddTerms

Add terms

z=42

To check our solution, we will substitute 42 for z in our equation.

z-12=30

Substitute

z= 42

42-12? =30

SubTerm

Subtract term

30=30

Since the left-hand side and right-hand side are equal, we know the solution is correct.