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5.04 Systems of equations using elimination method

Elimination method

We've already had a look at two ways of solving a system of equations:  graphically  and by the  substitution method  .

The last method of solving a system of equations that we'll be learning is the elimination method. It works by adding or subtracting equations from one another in order to eliminate one variable, leaving us with only the other variable to solve for.

We should use addition when the coefficients of the variable we want to eliminate are equal and opposite in sign, and subtraction when they're equal and the same sign. But what happens when they don't have the same coefficients at all?

When we don't have the same value coefficients for like terms we want to eliminate, we can multiply or divide the whole equation by a constant until it gets to the coefficient that we want.

Let's look at an example where we solve the following system of equations:2x - y = 1 \\ 5x + y = 2

Because the variable y has the same coefficient and opposite signs in the two equatons, we can add our two equations together to eliminate y and sovle for x:\begin{array}{c} & &2x &- &y &= &1 \\ &+ &5x &+ &y &= &2 \\ \hline \\ & &7x & & &= &3 \\ & &x & & &= &\dfrac{3}{7} \end{array}

Now, substitute the x-value into one of the equations from the system:

\displaystyle 5x + y\displaystyle =\displaystyle 2Seconds equation
\displaystyle 5 \times \dfrac{3}{7} + y\displaystyle =\displaystyle 2Substitute the value of x
\displaystyle \dfrac{15}{7} + y\displaystyle =\displaystyle 2Evaluate the multiplication
\displaystyle \dfrac{15}{7}+y-\dfrac{15}{7}\displaystyle =\displaystyle 2- \dfrac{15}{7}Subtract \dfrac{15}{7} from both sides
\displaystyle y\displaystyle =\displaystyle -\dfrac{1}{7}Evaluate

The solution of the system is \left(\dfrac{3}{7}, -\dfrac{1}{7}\right).

When solving systems of equations with eliminaton, it can be helpful to do the following:

  • Being organized is key. Name your equations by writing (1) & (2) next to them, and whenever you create new equations out of one or both of them, you can name them (3), \, (4) , etc. When adding two equations you can write it in shorthand as (3) + (1), and similarly for all the other operations. This helps to keep your ideas in order and not get confused.
  • When dealing with equations with big numbers, see if you can simplify them before beginning to solve them. eg. 2x - 4 = 6y can be simplified to x - 2 = 3y without changing the values of x & y.
  • If you have time, remember to check your answers by substituting your x and y values back into the original two equations.

Examples

Example 1

How would we eliminate a variable in the following system of equations?

2x + 3y = 4 \\ 5x - y = 3

A
Eliminate y by multiplying the second equation by -3 and then adding the equations.
B
Eliminate y by multiplying the second equation by 3 and then adding the equations.
C
Eliminate x by multiplying the second equation by 2 and then subtracting the equations.
D
Eliminate x by multiplying the first equation by -5 and then adding the equations.
Worked Solution
Create a strategy

Decide which variable you will eliminate. Make the coefficients of one variable opposites by multiplying one or both equations so that the coefficients of that variable are opposites.

Apply the idea

By looking on the x terms, we have a coefficient of 2 and a coefficient of 5. There is no number that we could multiply 2 by to get 5 or multiply 5 by to get 2. There are ways to elimate the x variable but it is not the easiest way.

Let's look at the y variable. In the first equation, the coefficient of y is 3 and in the second equation the coefficient of y is -1. We can multiply the second equation by 3 to get the opposite of the coefficient of the first equation.

\displaystyle 5x-y\displaystyle =\displaystyle 3Write the second equation
\displaystyle 15x-3y\displaystyle =\displaystyle 9Multiply the equation by 3

Now we can add the new equation to the first equation to eliminate y:\begin{array}{c} & &2x &+ &3y &= &4 \\ &+ &15x &- &3y &= &9 \\ \hline \\ & &17x &+ &0y &= &13 \end{array}

The y is eliminated since it has 0 coefficient.

So to eliminate the y variable of this system, we first multiply the second equation by 3 and add the first equation to the resulting equation.

So option B is the correct answer.

Example 2

Use the elimination method by subtraction to solve for x and y.

  • Equation 1: 3x - 2y = -3

  • Equation 2: 3x + 5y = 39

a

First solve for y.

Worked Solution
Create a strategy

Arrange the equations in vertical alignment and subtract to eliminate x.

Apply the idea

Looking at the two equations, the x terms have the same coefficients.

So equation 2 can be subracted from equation 1 to eliminate the x term.

\begin{array}{c} & &3x &- &2y &= &-3 \\ &- &3x &+ &5y &= &39 \\ \hline \\ & & & &-7y &= &-42 \\ & & & &y &= &6 \end{array}

b

Now solve for x.

Worked Solution
Create a strategy

Substitute the y-value into one of the original equations to find the value of x.

Apply the idea

Substitute the y-value into equation 1 from the system:

\displaystyle 3x - 2y\displaystyle =\displaystyle -3Write the equation 1
\displaystyle 3x - 2 \times 6\displaystyle =\displaystyle -3Substitute the value of y
\displaystyle 3x - 12\displaystyle =\displaystyle -3Simplify
\displaystyle 3x - 12+12\displaystyle =\displaystyle -3+12Add 12 to both sides
\displaystyle \dfrac{3x}{3}\displaystyle =\displaystyle \dfrac{9}{3}Evaluate and divide both sides by 3
\displaystyle x\displaystyle =\displaystyle 3Simplify

Example 3

Use the elimination method to solve for x and y.

  • Equation 1: -6x - 2y = 46

  • Equation 2: -30x - 6y = 246

a

First solve for x.

Worked Solution
Create a strategy

Multiply equation 1 by a number to get the same coefficient of x or y.

Apply the idea

Multiply both sides of Equation 1 by 3 to get the same coefficient of y.

\displaystyle -6x-2y\displaystyle =\displaystyle 46Write equation 1
\displaystyle -18x-6y\displaystyle =\displaystyle 138Multiply by 3

Now that both coefficients of y are the same we can subtract equation 2 from the new equation.

\begin{array}{c} & &-18x &- &6y &= &138 \\ &- &-30x &- &6y &= &246 \\ \hline \\ & &12x & & &= &-108 \\ & &x & & &= &-9 \end{array}

b

Now solve for y.

Worked Solution
Create a strategy

Substitute the x-value into one of the original equations to find the value of y.

Apply the idea

Substitute x=-9 into equation 1:

\displaystyle -6x - 2y\displaystyle =\displaystyle 46Write equation 1
\displaystyle -6 \times -9 - 2y\displaystyle =\displaystyle 46Substitute the value of x
\displaystyle 54 - 2y\displaystyle =\displaystyle 46Simplify
\displaystyle -2y\displaystyle =\displaystyle 46 - 54Subtract 54 from both sides
\displaystyle \dfrac{-2x}{-2}\displaystyle =\displaystyle \dfrac{-8}{-2}Evaluate and divide both sides by -2
\displaystyle y\displaystyle =\displaystyle 4Evaluate
Idea summary

Elimination method works by adding or subtracting equations from one another to eliminate one variable, leaving us with the other variable to solve on its own.

We should use addition when the coefficients of the variable we want to eliminate are equal and opposite in sign, and subtraction when they're equal and the same sign.

When we don't have the same value coefficients for like terms we want to eliminate, we can multiply or divide the whole equation by a constant to get the coefficient that we want.

Outcomes

8.EE.C.8

Analyze and solve pairs of simultaneous linear equations.

8.EE.C.8.B

Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection.

8.EE.C.8.C

Solve real-world and mathematical problems leading to two linear equations in two variables.

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