## How do you solve a pair of simultaneous equations?

Simultaneous equations are easy if you follow these simple steps. Rearrange the first equation to get an expression for x, substitute that into the second equation and solve it for the value of y. Then once you have a number for y you can find the value of x by substituting into either equation.

## Explanation

An equation like y=mx + c defines a set of points for which a relationship between the x value and y value are true. For example, if we have the line y=x

- Point (1,1) lies on the line because the equation is true for that point (y=x=1).
- Point (2,1) doesn’t lie on the line because y=1 which is not equal to the x value 2.

In this way * all points on a line are solutions of the line’s equation*.

A set of simultaneous equations defines a set of points for which *multiple* statements are true. For example if both y=x *and* y=2x+3 then only a single point in the plane can be true for both equations. The solution is now a single point and in this article we’ll see how to calculate that point. We’ll also see how to calculate the solutions for more complex curves and lines.

## Two Lines

Let’s start simple and work our way up to more complex examples.

\(\begin{align} y &= x \\\\ y &= 2x + 3 \end{align}\)If both statements are true then we can solve this problem in two ways, but both use the same idea. We are trying to get rid of one variable, either x or y, so that we can solve for the other variable. First, let’s do it by substitution. We know that \(y=x\) so we can substitute that into \(y=2x+3\) and solve. Let’s try it.

### Substitution

- Label your equations (1) and (2) so you know what you’ve got
- Choose one equation, say (1), and rearrange it to get just y on the left hand side
- Substitute for y in equation (2), now you have an equation with just y’s and numbers
- Rearrange this equation to solve for the value of y
- Substitute the numerical value of y into equation (1) and rearrange to get the value of x

All we did was substitute \(x=y\) in \(y=2x+3\) then move all the y’s onto one side of the equals sign and all the numbers onto the other side. Now we can find x easily because we know \(y=x\) so the solution is (-3,-3). We can easily check this is a solution for both equations because \(-3 = -3\) (from the equation \(y=x\)) and \(-3=2 \times -3 + 3=-6 + 3\).

### Subtract Equations

- Label your equations (1) and (2) so you know what you’ve got
- Choose to get rid of x or y. Let’s say you choose to get rid of y.
- Multiply (1) and/or (2) by some number so you have the same number of y’s
- Subtract the multiplied up version of (1) from (2)
- Rearrange this equation to solve for the value of x
- Substitute the numerical value of x into equation (1) and rearrange to get the value of y

The other method is to subtract one equation from the other to get rid of x or y. Then we solve for x or y just as we did by substitution. We have a y on its own in both equations so by subtracting the bottom equation from the top

\(\begin{align}y – y &= (2x+3) – x \\\\ 0 &= x + 3 \end{align}\)We know \(x+3=0\) so \(x=-3\) and as before we know \(y=x\) so the solution is (-3,-3).

## Line and Quadratic

Two lines can only ever cross at one point (or no point if they’re parallel). A quadratic and a line are more complicated because they can cross at two points, one point or no point. Two solutions must be reminding you of something… yes to solve these problems you will have to solve a quadratic equation!

Let’s visualise what’s going on here. A quadratic equation is a parabola and the solutions are the points where the line cuts the parabola. The equations of the line and the parabola are

\(\begin{align} y &= x-2 \\\\ y &= x^2 + x – 6 \end{align}\)

To solve this we just substitute to get rid of one of the variables, like this.

\(\begin{align} y &= x^2 + x – 6 \\\\ y &= (y+2)^2 + (y-2) – 6 \\\\ y &= (y+2)(y+2) + y – 4 \\\\ y &= y^2 + 4y + 4 + y – 4 \\\\ 0 &= y^2 + 4y \\\\ 0 &= y(y+4) \end{align}\)This must have two solutions: either y+4=0, in which case y=-4 or y=0. That’s because we’re multiplying two numbers together and they’re producing zero so either one or the other must be zero. By substituting we can then work out the values of x for the these two solutions. We know that x=y+2, so the solutions are (-2,-4) and (2,0). You can also see these are the solutions by looking at the graph above.

## Line and Circle

The equation of a circle with radius r is

\(x^2+y^2=r^2\)If we have a pair of simultaneous equations that consist of a circle and a line then, just as with a parabola, there are three possibilities:

- Two solutions because the line meets the circumference of the circle in two places
- One solution where the line is a tangent to the circle at a single point
- No solutions if the line does not cut the circle

This means the equation we end up solving is a quadratic because this has two, one or zero solutions. Here’s an example.

The equations of the line and the circle are:

\(\begin{align} y &= -2x + 5 \\\\ x^2 + y^2 & = 10 \end{align}\)Let’s solve this by substitution. We can replace the y in the circle equation with -2x + 5 from the line equation then it becomes a quadratic like so:

\(\begin{align} x^2 + (-2x + 5)^2 &= 10 \\\\ x^2 + 4x^2 – 10x – 10x + 25 &= 10 \\\\ 5x^2 – 20x + 15 &= 0 \\\\ x^2 – 4x + 3 &= 0 \\\\ (x-1)(x-3) &= 0 \end{align}\)So the solutions are x=1 and x=3 and substituting into the line equation we can work out the two points where the line intersects the circle are (1,3) and (3,-1).

## Some More Examples

Here is a worksheet to give you a bit more practice: