# How do I solve a simple simultaneous equation?

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Simultaneous equations pop up all the time in maths, science, and engineering, so being able to solve them is really useful. Although they can look a bit daunting at first, the same general rules apply to solving all of them.

Let's look at the simplest type of simultaneous equation - two equations, with two 'unknowns' - we'll call them 'x' and 'y'.

Example

2x + y = 7 (equation 1)

3x - y = 8 (equation 2)

Now, we want to find out what the values of x and y are, just like in a normal equation. This is tricky though, because here there are two things we don't know.

We solve the problem by saying that the x and y in equation one are the same as the x and y in equation two. That way, if we can find one, we will find the other.

In the example above, the solution is straightforward. Looking at the two equations, we can notice that the y terms are opposite to each other - one positive, one negative.

What we do now is take the two equations and add them together, remembering to keep the bits on the correct side of the equals sign. We get:

2x + 3x + y - y = 7 + 8

which becomes

5x = 15

so we've got rid of the y bits! This means we can find out what x is really easily. We know that 5x = 15, so:

x = 15/5 = 3

Now that we now what x is, we can find out what y is just by looking back at one of our first equations (it doesn't matter which one). Looking at equation 1:

2x + y = 7

Because we know that x = 3, we can say that

6 + y = 7

and so

y = 7 - 6

y = 1

so we've found our two unknowns, x = 3 and y = 1! We can check these by putting them back into the equations and making sure we get the right answer.

It's not always as simple as this though. We were lucky because in our example, the equations 'matched' - one had a '+ y' bit, the other one had a '- y' bit, so we could just add them to get rid of the y parts.

Often, the equations don't fit so well, and so we have to do things to them before we can use them - we can multiply, divide and rearrange equations to make them fit.

But, once they fit, we can use the same method we just used in the example to solve them!

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