The idea behind solving equations by elimination is similar to

*solving using substitution*.

We again begin by looking to find an initial value for one of the variables, by adding or subtracting one equation from the other, or where appropriate a multiple of one equation.

## Solving Equations by Elimination Approach

If we consider solving the simultaneous equations,5x + y = 7

x + y = 3

Subtracting the 2nd equation from the 1st eliminates the y variable.

\begin{array}{r} \space{5x + y = 7}\space\space\\ {\text{--}}\space\space\space\space{x + y = 3}\space\space\\ \hline {4x \space\space{\color{white}+ y} = 4}\space\space \end{array}

We can see from the subtraction that x = 1.

Now using the 2nd equation. 1 + y = 3

So y = 2.

Checking in the 1st equation.

5(1) + 2 = 7 => 5 + 2 = 7

**( CORRECT )**

### Solving Equations by Elimination

Example

*(1.1)*Solve the simultaneous equations:

4x + y = 8

x \space {\text{--}} \space 2y = 11

by elimination.

*Solution*Adding or subtracting the equations as they sit, won’t eliminate a variable in one of the equations.

But multiplying the 1st equation by 2, then adding the equations together will achieve a variable elimination.

4x + y = 8 ( × 2 ) = 8x + 2y = 16

\begin{array}{r} \space\space{8x + 2y = 16}\space\space\\ +\space\space\space{x \space {\text{--}} \space 2y = 11}\space\space\\ \hline {9x \space\space{\color{white}+ 2y} = 27}\space\space \end{array}

The result tells us that x = 3.

Which we can input into the 2nd equation to obtain y.

3 \space {\text{--}} \space 2y = 11

3 \space {\text{--}} \space 3 \space {\text{--}} \space 2y = 11 \space {\text{--}} \space 3 => {\text{-}}2y = 8 => y = {\text{-}}4

Checking with 1st equation:

4(3) + ({\text{-}}4) = 8 => 12 \space {\text{--}} \space 4 = 8

**( CORRECT )**

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