# Simultaneous Equations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Simultaneous Equations

Simultaneous equations are equations that share the same variables. There will be a solution, or solutions, that work for all equations. In A level maths, you will only see simultaneous equations in two variables, e.g. $x$ and $y$.

There are 2 main methods used to solve simultaneous equations.

A Level

## Method 1: Elimination

The method of elimination is used when there are two linear simultaneous equations. We eliminate one variable by subtracting one equation from the other.

Example: Solve the following equations:

\begin{aligned} 3x + 7y -44 &= 0 \\ 2x + 3y &= 21 \end{aligned}

Step 1: Write both equations in the form $ax+by=c$, if necessary:

\begin{aligned} 3x + 7y &= 44 \\ 2x + 3y &= 21 \end{aligned}

Step 2: Manipulate the equations so that the coefficients match – multiply the equations to make either the $x$‘s or $y$‘s equal in size (ignoring the signs). Always multiply to get the LCM of the coefficients.

$\,(\times2) \,\,\,\,\,\,\,\,\,3x + 7y = 44\,\,\, \Rightarrow \,\,\, 6x + 14y = 88$

$(\times3)\,\,\,\,\,\,\,\,\, 2x+3y = 21 \,\,\, \Rightarrow \,\,\, 6x+9y=63$

Step 3: Add or subtract the equations to eliminate variable that has terms with equal coefficients, so that you can find the other variable.

In this case, both equations have $+6x$, so we need to subtract:

\begin{aligned}6x + 14y &= 88 \\ (-)\,\,\,\,\,\,\,\,\, 6x+9y&=63 \\ \hline 5y&=25\end{aligned}

Step 4: Solve the resulting equation.

In this case, we need to solve the equation to find $y$

\begin{aligned}(\div 5)\,\,\,\,\,\,\,\,\,5y&=25 \\ \textcolor{red}{y} &\textcolor{red}{= 5} \end{aligned}

Step 5: Find the variable that you eliminated.

In this case, replace (substitute) $y = 5$ into one of the equations and solve to find $x$:

\begin{aligned} 2x + 3y &= 21 \\ 2x + 3(5) &= 21 \\ 2x + 15 &= 21 \\ 2x &= 6 \\ \textcolor{blue}{x} &\textcolor{blue}{= 3} \end{aligned}

Hence,

$\textcolor{blue}{x = 3}$  and  $\textcolor{red}{y = 5}$

A Level

## Method 2: Substitution

We use the method of substitution when one of the simultaneous equations is quadratic (non-linear), since we can’t use the method of elimination.

Example: Solve

\begin{aligned} x^2 + 2y &= 31 \\ x+y &= 14 \end{aligned}

Step 1: Rearrange the linear equation so that one of the variables is on its own (in this case it will be easier to get $y$ on its own so that we don’t have to do any squaring of brackets)

\begin{aligned} (-x) \,\,\,\,\,\, x+y &= 14 \\ y &= -x + 14 \end{aligned}

Step 2: Substitute this variable into the quadratic equation, so that there is an equation in only one variable.

$y = -x + 14$ so replace $y$ with $-x + 14$:

$x^2 + 2(-x+14) = 31$

Step 3: Expand and solve to find the values for one variable.

\begin{aligned} x^2 + 2(-x+14) &= 31 \\ x^2 - 2x + 28 &= 31 \\ x^2 - 2x - 3 &= 0 \\ (x-3)(x+1) &= 0 \end{aligned}

Hence,

$\textcolor{red}{x = 3}$  and  $\textcolor{red}{x = -1}$

Step 4: Substitute these values into the linear equation (since this will be easier) and solve to find the corresponding values of the other variable.

When $\textcolor{red}{x = 3}$,   $\textcolor{blue}{y} = -3+14 = \textcolor{blue}{11}$

When $\textcolor{red}{x = -1}$,   $\textcolor{blue}{y} = -(-1) + 14 = \textcolor{blue}{15}$

So, there are two pairs of solutions

$\textcolor{red}{x=3}$, $\textcolor{blue}{y = 11}$  and  $\textcolor{red}{x = -1}$, $\textcolor{blue}{y = 15}$

Note: You may need to use the quadratic formula if you get a quadratic equation that is too difficult to solve by factorising.

A Level

## Interpreting Simultaneous Equations Geometrically

To interpret simultaneous equations geometrically, we need to draw a sketch of the two functions and describe what we see.

The number of solutions is equal to the number of intersections between the graphs:

Two solutions – the graphs intersect twice

One solution – the graphs meet at a single point – the graph is a tangent to the curve at this point

No solutions – the graphs do not intersect

A Level
A Level

## Example: Interpreting Simultaneous Equations Geometrically

Interpret the following geometrically:

$y = x^2 - 3x$  and  $y = x - 4$

[4 marks]

Substitute $y=x-4$ into $y = x^2 - 3x$, and then solve for $x$:

\begin{aligned} x - 4 &= x^2 - 3x \\ x^2 - 4x + 4 &= 0 \\ (x-2)^2 &= 0 \\ \textcolor{red}{x} &\textcolor{red}{= 2} \end{aligned}

Then, substitute $\textcolor{red}{x = 2}$ into $y = x - 4$ and solve for $y$:

\begin{aligned} y &= x - 4 \\ y &= 2 - 4 \\ \textcolor{blue}{y} &\textcolor{blue}{= -2} \end{aligned}

Hence, there is only one solution $\textcolor{red}{x = 2}$, $\textcolor{blue}{y = -2}$

Therefore, the graphs will meet at a single point: $(\textcolor{red}{2}, \textcolor{blue}{-2})$

So, the straight line is a tangent to the curve at the point $(\textcolor{red}{2},\textcolor{blue}{-2})$

A Level

## Example Questions

Multiply the first equation by $2$ so that they coefficients of $y$ match:

\begin{aligned} 8x + 2y &= 36 \\ 3x + 2y &= 21 \end{aligned}

Subtract the second equation from the first, to eliminate the $y$ variable:

$5x = 15$

And then solve:

$x = 3$

Then, substitute $x =3$ into either equation and solve:

\begin{aligned} 4(3) + y &= 18 \\ 12 + y &= 18 \\ y &= 6 \end{aligned}

Hence,

$x = 3$  and  $y = 6$

Rearrange $x-y=2$ so that $x$ is on its own:

$x = y+2$

Then, substitute this into $x^2 + y^2 = 100$ and expand and solve for $y$:

\begin{aligned} (y+2)^2 + y^2 &= 100 \\ y^2 + 4y + 4 + y^2 &= 100 \\ 2y^2 + 4y - 96 &= 0 \\ (2y-12)(y+8) &= 0 \end{aligned}

Hence, $2y - 12 = 0$  and  $y + 8 = 0$ $\Rightarrow y = 6$  and  $y = -8$

Substitute these values into the non-linear equation, to find the values of $x$:

$y = 6 \Rightarrow x = 6+2 = 8$

$y = -8 \Rightarrow x = -8 + 2 = -6$

Hence, the points of intersection are $(8, 6)$ and $(-6, -8)$

Substitute $y = x$ into $y = x^2 + 5x + 5$, and solve:

\begin{aligned} x^2 + 5x + 5 &= x \\ x^2 + 4x + 5 &= 0 \\ (x+2)^2 - 4 + 5 &= 0 \\ (x+2)^2 + 1 &= 0 \\ (x+2)^2 &= -1 \end{aligned}

You cannot get a real number from square rooting a negative number. Therefore there are no real solutions.

You could have found the discriminant of $x^2 +4x + 5$:

$b^2 - 4ac = 4^2 - 4 \times 1 \times 5 = 16 - 20 = -4$

The discriminant is $<0$, so there are no real roots, and therefore no real solutions.

A Level

A Level

## You May Also Like...

### A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99

### A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99

### Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99