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Pearson Algebra 1 Common Core, 2011 View details

2. Solving Systems Using Substitution

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Exercise 48 Page 377

To solve for x, first isolate 2x.

5.5

Practice makes perfect

To solve equation, we will use the Properties of Equality to isolate the variable. For this exercise, we will start by using the Addition Property of Equality to isolate the variable term 2x.

2x-3=8

▼

LHS+3=RHS+3

Evaluate

2x-3+3=8+3

AddTerms

Add terms

2x=11

Next, we need to isolate x on the left-hand side. Using the Division Property of Equality, we will divide both sides of the equation by 2 to eliminate the coefficient of x.

2x=11

▼

.LHS /2.=.RHS /2.

Evaluate

2x/2=11/2

CalcQuot

Calculate quotient

x=11/2

WriteDec

Write as a decimal

x=5.5

Therefore, x=5.5 is the solution to the equation.