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 )