We can use the Properties of Equality in solving systems of linear equations algebraically to eliminate a variable. There are several steps to follow in order to eliminate a variable in a system of equations. We will present these steps and the possible Properties of Equality that can be used in these steps in a table.

	What is done?	What properties can be used?
Step I	Gather all like terms on the same sides of the equations.	Addition or Subtraction Property of Equality
Step II	Multiply or divide the equations by a constant so that one of the variable terms has the same or opposite coefficient.	Multiplication or Division Property of Equality
Step III	Add or Subtract equations in the system.	Addition or Subtraction Property of Equality
Step IV	Solve the obtained equation with one variable.	Properties of Equality used to solve an equation
Step V	Solve the equation for the other variable by substituting the obtained variable into one of the equations in the system.	Substitution Property of Equality and other Properties of Equality used to solve an equation

Let's give an example.

Example

Consider the following system of linear equations.

{4 x + 2 y = 63 3 x = 3.5 + y (I) (II)

Step I

In Equation (I), the variable terms are on the same side of the equation. However, in Equation (II) the variable terms are on both sides of the equations. We can apply the Subtraction Property of Equality to gather the variables on the left-hand side of the equation.

3 x - y = 3.5 + y - y ⇕ 3 x - y = 3.5

Step II

Using the Multiplication Property of Equality, we will multiply both sides of Equation (II) by

2

to have opposite coefficients for the

y -

variable.

2 (3 x - y) = 2 (3.5) ⇕ 6 x - 2 y = 7

Step III

We will add the obtained equation to Equation (I) by using the Addition Property of Equation.

4 x + 2 y + (6 x - 2 y) = 63 + 7

▼

Simplify

RemovePar

Remove parentheses

4 x + 2 y + 6 x - 2 y = 63 + 7

AddSubTerms

Add and subtract terms

10 x = 70

Step IV

Note that we have an equation with only one variable. We can solve this equation for

x

by using the Division Property of Equality.

10 x = 70

DivEqn

$LHS / 10 = RHS / 10$

\frac{1 0 x}{1 0} = \frac{7 0}{1 0}

CalcQuot

Calculate quotient

\frac{1 0 x}{1 0} = 7

SimpQuot

Simplify quotient

x = 7

Step V

We will now substitute

x = 7

into one of the equations by using the Substitution Property of Equality. Let's use Equation (I). We will then apply the Subtraction and Division Properties of Equality to solve the equation for

y .

4 x + 2 y = 63

Substitute

$x = 7$

4 (7) + 2 y = 63

▼

Solve for

y

Multiply

28 + 2 y = 63

SubEqn

$LHS - 28 = RHS - 28$

2 y = 35

DivEqn

$LHS / 2 = RHS / 2$

y = 17.5

Exercise

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