Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
3. Solving Systems Using Elimination
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Exercise 6 Page 381

What are the steps you have to take to solve a system of linear equations by elimination?

See solution.

Practice makes perfect
There are several steps to follow in order to solve a system of linear equations using the Elimination Method. We will present these steps in a table.
What is done?
Step I Gather all like terms on the same sides of the equations.
Step II If the coefficients of one of the variables in both equations are not additive inverses of each other or the exact same number, multiply or divide the equations by a constant so that one of the variable terms has the same or opposite coefficients.
Step III Add or subtract equations in the system. This will result in one equation with a single variable.
Step IV Solve the obtained equation with one variable.
Step V Substitute the result into one of the equations to find the other variable.

Extra

Example

Consider the following system of linear equations. 4x+2y=5 & (I) 3x=2.5+y & (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 will first gather the variables on the left-hand side of the equation by Subtraction Property of Equality. 3x - y=2.5+y - y ⇕ 3x-y=2.5

Step II

We will multiply both sides of Equation (II) by 2 to have opposite coefficients for the y-variable by using the Multiplication Property of Equality. 2 (3x-y)= 2(2.5) ⇕ 6x-2y= 5

Step III

We will add the obtained equation to Equation (I) by using the Addition Property of Equality.
4x+2y+( 6x-2y)=5+ 5
â–Ľ
Simplify
4x+2y+6x-2y=5+5
10x=10

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.
10x=10
10x/10 = 10/10
x = 1

Step V

Finally, we will substitute x = 1 into one of the equations and then solve the equation for y. Let's use Equation (I).
4x+2y=5
4( 1)+2y=5
â–Ľ
Solve for y
4+2y=5
2y= 1
y=0.5