Simultaneous Equations with 3 Variables (AQA GCSE Further Maths)

Revision Note

Test yourself

Written by: Jamie Wood

Reviewed by: Dan Finlay

Simultaneous Equations with 3 Variables

To be able to solve simultaneous equations with 3 unknowns; we will need 3 equations.

We can use similar methods to when we solve with 2 unknowns; elimination or substitution.

How do I solve simultaneous equations with 3 unknowns using elimination?

The general approach is to eliminate one of the variables from two equations; so that we are left with 2 equations in terms of 2 unknowns
Once we have 2 equations with 2 unknowns, we can use the methods we already know to solve them, and then substitute the values back in to find the 3rd unknown
It is very important to label the equations as there are lots to keep track of
Take for example the following simultaneous equations
- $1 2 x + 3 y - z = 17$
- $2 x - 3 y + 2 z = - 12$
- $3 3 x + y + z = 9$
We can eliminate $z$ by adding equations $1$ and $3$
- $1 + 3 : 5 x + 4 y = 26$
- This is equation $4$
We can form another equation containing only $x$ and $y$ , by eliminating $z$ , by taking equation $2$ and adding on 2 times equation $1$
- $2 + 2 \times 1 : 5 x + 3 y = 22$
- This is equation $5$
Equations $4$ and $5$ can then be solved using the previously covered methods to find $x = 2$ and $y = 4$
Substitute these values into any of the original three equations to find $z = - 1$
As a check, we should substitute the found values for $x, y, z$ into each of the original equations to make sure they all work

How do I solve simultaneous equations with 3 unknowns using substitution?

The general approach is the same; eliminate one of the variables from two equations; so that we are left with 2 equations in terms of 2 unknowns
- With elimination we did this by adding or subtracting the equations (or multiples of the equations)
- With substitution we do it by rearranging one of the equations to make $x, y,$ or $z$ the subject, and then substituting this into the other two equations
Once we have 2 equations with 2 unknowns, we can use the methods we already know to solve them, and then substitute the values back in to find the 3rd unknown
It is very important to label the equations as there are lots to keep track of
Take for example the following simultaneous equations
- $1 2 x + 3 y - z = 17$
- $2 x - 3 y + 2 z = - 12$
- $3 3 x + y + z = 9$
We can rearrange equation $1$ to make $z$ the subject
- $z = 2 x + 3 y - 17$
Substitute $z = 2 x + 3 y - 17$ into equation $2$
- $x - 3 y + 2 (2 x + 3 y - 17) = - 12$
- Simplify into the form $a x + b y = c$
- $5 x + 3 y = 22$ , call this equation $4$
Substitute $z = 2 x + 3 y - 17$ into equation $3$
- $3 x + y + (2 x + 3 y - 17) = 9$
- Simplify into the form $a x + b y = c$
- $5 x + 4 y = 26$ , call this equation $5$
Equations $4$ and $5$ can then be solved using the previously covered methods to find $x = 2$ and $y = 4$
Substitute these values into any of the original three equations to find $z = - 1$
As a check, we should substitute the found values for $x, y, z$ into each of the original equations to make sure they all work

Examiner Tips and Tricks

Label each equation you use and write down exactly what you are doing in each step
- e.g. $1 - 2$ or $2 - 3 \times 2$
Always check your final solutions work for all three original equations
- You can leave the exam knowing that you got the answer correct!

Worked Example

Solve the simultaneous equations.

$\begin{array}{rcl} 2 a - 3 b + c & = & 19 \\ 3 a - 4 b - c & = & 15 \\ 5 a + 3 b + 2 c & = & 24 \end{array}$

Method 1: Elimination

Number the equations.

$2 a - 3 b + c = 19 3 a - 4 b - c = 15 5 a + 3 b + 2 c = 24$ $(1) (2) (3)$

Eliminate $c$ by adding equations (1) and (2).

(1) + (2):

$2 a - 3 b + c = 19 + (3 a - 4 b - c = 15) 5 a - 7 b = 34$

$5 a - 7 b = 34$ $(4)$

Find a second equation in $a$ and $b$ by subtracting 2 times equation (1) from equation (3).

(3) - 2(1):

$5 a + 3 b + 2 c = 24 - (4 a - 6 b + 2 c = 38) a + 9 b = - 14$

$a + 9 b = - 14$ $(5)$

Solve equations (4) and (5) using your preferred method of solving a pair of linear simultaneous equations.

$\begin{array}{rcl} 5 a - 7 b & = & 34 \\ a + 9 b & = & - 14 \end{array}$ $\begin{array}{rcl} (4) \end{array} \begin{array}{rcl} (5) \end{array} \begin{array}{rcl} \end{array}$

(4) -5(5):

$5 a - 7 b = 34 - (5 a + 45 b = - 70) - 52 b = 104 b = - 2$

Substitute into equation (4) or (5) and solve to find $a$ .

$\begin{array}{rcl} (4) \end{array}$ $\begin{array}{rcl} 5 a - 7 (- 2) & = & 34 \end{array}$
$\begin{array}{rcl} 5 a + 14 & = & 34 \\ 5 a & = & 20 \\ a & = & 4 \end{array}$

Substitute $a = 4$ and $b = - 2$ back into one of equation (1), (2) or (3) and solve to find $c .$

$\begin{array}{rcl} (1) \end{array}$ $\begin{array}{rcl} 2 (4) - 3 (- 2) + c & = & 19 \end{array}$
$\begin{array}{rcl} 8 + 6 + c & = & 19 \\ 14 + c & = & 19 \\ c & = & 5 \end{array}$

Substitute the three solutions into the other two equations to check that they are correct.

$\begin{array}{rcl} (2) \end{array}$ $\begin{array}{rcl} 3 (4) - 4 (- 2) - 5 & = & 15 \end{array}$
$\begin{array}{rcl} 12 + 8 - 5 & = & 15 \\ 15 & = & 15 \end{array}$

$\begin{array}{rcl} (3) \end{array}$ $\begin{array}{rcl} 5 (4) + 3 (- 2) + 2 (5) & = & 24 \end{array}$
$\begin{array}{rcl} 20 - 6 + 10 & = & 24 \\ 24 & = & 24 \end{array}$

$a = 4, b = - 2, c = 5$

Method 2: Substitution

Number the equations.

$2 a - 3 b + c = 19 3 a - 4 b - c = 15 5 a + 3 b + 2 c = 24$ $(1) (2) (3)$

Rearrange equation (1) to make $c$ the subject.

(1):

$c = 19 - 2 a + 3 b$

Substitute $c = 19 - 2 a + 3 b$ into equation (2) and simplify.

(2):

$3 a - 4 b - (19 - 2 a + 3 b) = 15 3 a - 4 b - 19 + 2 a - 3 b = 15 5 a - 7 b = 34$

$5 a - 7 b = 34$ $(4)$

Substitute $c = 19 - 2 a + 3 b$ into equation (3) and simplify.

(3):

$5 a + 3 b + 2 (19 - 2 a + 3 b) = 24 5 a + 3 b + 38 - 4 a + 6 b = 24 a + 9 b = - 14$

$a + 9 b = - 14$ $(5)$

Solve equations (4) and (5) using your preferred method of solving a pair of linear simultaneous equations.

$\begin{array}{rcl} 5 a - 7 b & = & 34 \\ a + 9 b & = & - 14 \end{array}$ $\begin{array}{rcl} (4) \end{array} \begin{array}{rcl} (5) \end{array}$