Elimination Method

Given a system of two equations in two variables, replacing one equation with the sum of that equation and a multiple of the other equation produces an equivalent system. This fact is used to solve systems of equations by the Elimination Method. Consider an example system of linear equations. 3x+2y=6 & (I) y=2x-11 & (II) To solve the system by using the Elimination Method, there are five steps to follow.

Write the Equations in the Same Form

First, all like terms must be gathered on the same sides of the equations. In Equation (I), the variable terms are on the same side of the equation. However, the variable terms are on both sides of the equations in Equation (II). Like terms can be gathered on the same sides of the equations by applying the Properties of Equality.

3x+2y=6 y=2x-11

SubEqn

(II): LHS-2x=RHS-2x

3x+2y=6 - 2x+y=- 11

Multiply an Equation

Multiply one of the equations by a constant so that one of the variable terms of the resulting equation is equal to or is the opposite of the corresponding variable term in the other equation. In this case, multiplying Equation (II) by - 2 will produce opposite coefficients for the y-variable. - 2 (- 2x+y)= - 2(- 11) ⇕ 4x-2y= 22 Both the original and the resulting equations have the same solutions because they are equivalent equations.

Add the New Equation and the Other Equation

After the rewrites, the system of equations looks the following way. 3x + 2y=6 4x - 2y=22 Add these two equations by adding the right-hand sides together and the left-hand sides together. This way one variable will be eliminated.

3x+2y+( 4x-2y)=6+ 22

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Simplify

RemovePar

Remove parentheses

3x+2y+4x-2y=6+22

CommutativePropAdd

Commutative Property of Addition

3x+4x+2y-2y=6+22

AddSubTerms

Add and subtract terms

7x= 28

Note that this step results in an equation in only one variable. This equation can be solved by dividing both sides by 7.

7x = 28

DivEqn

.LHS /7.=.RHS /7.

7x/7 = 28/7

CalcQuot

Calculate quotient

x = 4

Write an Equivalent System

Substitute the value for the solved equation in one variable for any of the equations of the system. This produces an equivalent system of equations. In this case, Equation (I) will be replaced. & 3x+2y=6 & (I) - 2x+y=- 11 & (II) & ⇕ & x= 4 - 2x+y=- 11 Note that the first equation is the solution value of x.

Solve the Equivalent System

To solve the new system, substitute the found value into the other equation. In this case, 4 will be substituted into Equation (II) for x to find the value of y.

x=4 - 2x+y=- 11

Substitute

(II): x= 4

x=4 - 2( 4)+y=- 11

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(II): Solve for y

MultNegPos

(II): (- a)b = - ab

x=4 - 8+y=- 11

AddEqn

(II): LHS+8=RHS+8

x=4 y=- 3

In this system, the value of y is - 3. Therefore, the solution to the system of equations, which is the point of intersection of the lines, is (4,- 3), or x=4, y=- 3.

If at any step of the method a true statement is found, then the lines represented by the equations of the system are coincidental. The system would be a dependent system and have infinitely many solutions. Conversely, if at any step a false statement is found, then the lines are parallel. In this case, the system would be inconsistent and have no solution.

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