Simultaneous equations: elimination and substitution

In a nutshell

Simultaneous equations involves solving two (or more) equations, where the answers work in all equations given. The idea is that there is a set of solutions that work in both the equations. For two unknowns, there needs to be two equations to solve. Simultaneous equations can be solved using the elimination method or the substitution method.

Elimination method

The elimination method involves writing the equations in the correct form and in columns so that the equations can be added or subtracted to eliminate a variable.

PROCEDURE

1. Check if the equations need to be rearranged or multiplied up for them to be written in the correct format. Write out the equations, and number them.

2. Decide whether to add or subtract equations, and solve for one of the variables. Use the rule:

Same sign	Subtract
Different sign	Add

3. Substitute the answer obtained into one of the equations, and solve for the other variable.

4. Check both answers in the other equation.

Example 1

$\begin {aligned}7c+2d &=2 &\textcircled{1} \\-7c-5d&= 16 &\textcircled{2} \\\underline{\qquad \quad}& \underline{\qquad \qquad \quad} \\-3d&=18 &\textcircled{2} + \textcircled{1}\\d&=-6 \\\\7c +2\times-6 &=2 \\7c &=14 \\c&=2 \\\\-7\times2 -5\times -6 &=16 \\\underline {c=2, d=-6}\end {aligned}$

Substitution method

The substitution method involves rearranging one of the equations and substituting into the other equation in order to find the solutions.

PROCEDURE

$1.$	Rearrange one of the equations for either $x$ or $y$ . Call this equation $1$ and call the other equation $2$ .
$2.$	Substitute equation $1$ into equation $2$ and solve for the unknown variable.
$3.$	Substitute the answer from step $2$ into equation $1$ and solve.
$4.$	Check both answers in equation $2$ .

Example 2

Solve

$\begin{aligned}x-y&=3 \\2x+y&=3 \\\end{aligned}$

Rearrange the first equation for $x$ and label the equations.

$\begin{aligned}x& = y+3 \qquad &\textcircled{1} \\2x+y &=3 \qquad &\textcircled{2} \\ \\\end {aligned}$

Substitute equation $1$ into equation $2$ and solve.

$\begin{aligned}2(y+3)+y &=3 \\2y+6+y &= 3 \\3y + 6 &= 3 \\3y &=-3 \\y &= -1\end {aligned}$

Substitute the $y=-1$ into equation $1$ .

$\begin{aligned}x &=y+3 \\x &=-1+3 \\x &= 2\end{aligned}$

Check the answers in equation $2$ .

$\begin{aligned}2x+y &= 3 \\2 \times 2 - 1 &= 3 \\\end {aligned}$

$\underline{x = 2, y=-1}$