# Solving Simultaneous Equations with Graphs *KS3 Revision and Worksheets*

## What you need to know

We say that two equations are simultaneous if they are both true at the same time and for the same values of x and y (or whatever other variables are used in the question). Simultaneous equations can be solved using algebra, but here we’re going to go through the process of solving them using graphs. This process has two steps:

1. Plot each of the equations as individual straight-line graphs.

2. Read-off the coordinates of the point where they cross – this is the solution.

Example: By plotting their graphs, for values of x between -1 and 4, on the same axes, find the solution to the two simultaneous equations below.

\begin{aligned}y&=2x-5 \\ y&=-x+4\end{aligned}

To plot these straight lines, we’re going to find 3 points on each line and then join these points up. We do this by substituting some x values into each equation to find their matching y coordinates. For the first equation, our three y values will be

x=-1 \text{ gives } y=2\times (-1)-5=-7

x=1 \text{ gives } y=2\times (1)-5=-3

x=4 \text{ gives } y=2\times (4)-5=3

So, our 3 coordinates to plot are

(-1, -7),\,\,(1, -3),\,\,(4, 3)

Then, for the second equation, our three y values will be

x=-1 \text{ gives } y=-(-1)+4=5

x=1 \text{ gives } y=-(1)+4=3

x=4 \text{ gives } y=-(4)+4=0

So, our 3 coordinates to plot are

(-1, 5),\,\,(1, 3),\,\,(4, 0)

Plotting these coordinates and drawing the lines, we get the graph shown below.

As we can see, these straight lines cross at 1 point – circled in red. The coordinates of this point, (3, 1), provide the solution to this problem. Therefore, the solution is

x=3,\,\,\,y=1Example: By plotting their graphs for values of x between -4 and 2 on the same axes, find the solution to the following pair of simultaneous equations.

\begin{aligned}y&=x+2 \\ 3y&+x=-4\end{aligned}The process for this will be exactly the same as the last one with one extra step right here at the beginning. The second equation is written it an unusual form, and in order to be able to substitute x values into it, we need to have it written in the form ”y=...”.

So, we must rearrange! Firstly, subtract x from both sides to get

3y=-x-4Then, divide both sides by 3 to get

y=-\dfrac{1}{3}x-\dfrac{4}{3}Yikes, this is pretty unpleasant. That’s okay, we have a calculator. Remember, we only need 3 points, so try substituting in different values of x with your calculator until you get some nice whole number answers (just to make plotting easier). Doing this, we get

x=-4 \text{ gives } y=-\dfrac{1}{3}\times (-4)-\dfrac{4}{3}=0

x=-1 \text{ gives } y=-\dfrac{1}{3}\times (-1)-\dfrac{4}{3}=-1

x=2 \text{ gives } y=-\dfrac{1}{3}\times (2)-\dfrac{4}{3}=-2

So, our 3 coordinates to plot are

(-4, 0),\,\,(-1, -1),\,\,(2, -2)Then, for the second equation, our three y values will be

x=-4 \text{ gives } y=-4+2=-2

x=0 \text{ gives } y=0+2=2

x=2 \text{ gives } y=2+2=4

So, our 3 coordinates to plot are

(-4, -2),\,\,(0, 2),\,\,(2, 4)Plotting these points and drawing the lines, we get the graph shown below.

Looking at where the graphs cross, we can see that – whilst the coordinates of it are not whole numbers – the point is halfway between -2 and -3 on the x axis and halfway between 0 and -1 on the y axis.

So, it’s coordinates are: (-2.5, -0.5), and therefore the solution is

x=-2.5,\,\,\,y=-0.5NOTE: this will only be clear if the graphs are drawn as accurately as possible. Rough sketches of lines will not give you accurate solutions.

## Example Questions

1) By plotting their graphs for values of x between -2 and 3 on the same axes, find the solution to the following pair of simultaneous equations.

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

So, we want 3 points in order to plot each line. Subbing in some values of x into the first equation, we get

x=-2 \text{ gives } y=-2\times (-2)+1=5

x=0 \text{ gives } y=-2\times (0)+1=-1

x=3 \text{ gives } y=-2\times (3)+1=-5

So, our 3 coordinates to plot are

(-2, 5),\,\,(0, -1),\,\,(3, -5)

Then, doing the same for the second equation, we get

x=-2 \text{ gives } y=-(-2)-1=1

x=0 \text{ gives } y=-(0)-1=-1

x=3 \text{ gives } y=-(3)-1=-4

So, our 3 coordinates to plot are

(-2, 1),\,\,(0, -1),\,\,(3, -4)

Plotting these points and drawing the lines, we get the graph shown below.

As we can see, the coordinates where the lines cross are (2, -3). Therefore, the solution is

x=2,\,\,\,\,\,y=-3

2) By plotting their graphs for values of x between -3 and 2 on the same axes, find the solution to the following pair of simultaneous equations.

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

Before we can plot any points, we need to rearrange the second equation to be in a form we can use. So, subtracting 3x from both sides, we get

2y=-3x-3

Then, dividing both sides by 2 we get the desired form:

y=-1.5x-1.5

Now we’ve sorted that out, we want 3 points in order to plot each line. Subbing in some values of x into the first equation, we get

x=-3 \text{ gives } y=2\times (-3)+2=-4

x=0 \text{ gives } y=2\times (0)+2=2

x=2 \text{ gives } y=2\times (2)+2=6

So, our 3 coordinates to plot are

(-3, -4),\,\,(0, 2),\,\,(2, 6)

Then, doing the same for the (rearranged) second equation, we get

x=-3 \text{ gives } y=-1.5 \times (-3)-1.5=3

x=0 \text{ gives } y=-1.5 \times (0)-1.5=-1.5

x=1 \text{ gives } y=-1.5 \times (1)-1.5=-3

So, our 3 coordinates to plot are

(-3, 3),\,\,(0, -1.5),\,\,(1, -3)

Plotting these points and drawing the lines, we get the graph shown below.

As we can see, the coordinates where the lines cross are (-1, 0). Therefore, the solution is

x=-1,\,\,\,\,\,y=0