Simultaneous Equations Questions, Revision and Worksheets

Example: Find the solution to the following simultaneous equations.

3x+y=10,\,\,\,\,x+y=4.

To do this, we’ll use a process called elimination – we’re going to eliminate one of the variables by subtracting one equation from the other. We will write one equation on top of the other and draw a line underneath, as with normal subtraction (see: right).

So, we eliminated the y term and now have 2x=6. Dividing both sides of this equation by 2, we get x=3. It’s worth noting, if one equation had positive y and the other had negative y, we would’ve had to add the equations.

Now that we’ve found x, finding y is easier – we use substitution, which means that we pick one of the equations, we’ll pick x+y=4, and replace x with 3, as that’s what we now know it to be. The equation x+y=4 becomes

3+y=4

Then, subtracting 3 from both sides of the equation, we get y=4-3=1. Thus, the solution to these simultaneous equations is: x=3, y=1.

In this next example, we’ll see how to set up our own simultaneous equations.

Example: A store sells milkshakes and ice creams. If you buy 4 milkshakes and 3 ice creams, it will cost £12. If you buy 2 milkshakes and 2 ice creams, it will cost £7. Work out the cost of an individual milkshake and an individual ice cream.

In these sentences, there is a hidden pair of simultaneous equations.

Let’s say that the price of a milkshake is a, and the price of an ice cream is b. Then, the sentence “4 milkshakes and 3 ice creams will cost £12” is the same as saying “4 lots of a and 3 lots of b will make 12”. This admits the equation

4a+3b=12

If we do the same thing with “2 milkshakes and 2 ice creams cost £7”, we get

2a+2b=7

Thus, we have our simultaneous equations. The only problem is, the elimination process won’t work just yet, because if you subtract one of the equations from the other, no variables will disappear. However, if we will multiply both sides of the second equation by 2, we get a new equation:

4a+4b=14

This equation has a 4a just like the first one, so if we subtract one from the other, the a term will disappear. So, subtracting the first equation from the third, we get

We have immediately that the price of an ice cream, b, is £2. Then, substituting this back into the first equation, we get

4a+6=12

Subtract 6 from both sides to get 4a= , and then divide both sides by 4 to get a=1.5, therefore the price of a milkshake is £1.50.

On the higher course, simultaneous equations may include a quadratic. Let’s see what we must do.

Example: Solve the following simultaneous equations.

x^2+2y=9,\,\,\,\,y=x+3

Because one of these equations is quadratic, we can’t use elimination like before. Instead, we have to use substitution – this time, with algebra. The 2nd equation tells us y is equal to x+3, so we will replace the y in the first equation with x+3:

x^2+2(x+3)=9

If we expand the bracket and then subtract 9 from both sides, this becomes

x^2+2x-3=0

Observing that -1\times3=-3 and -1+3=2, we can factorise this quadratic:

(x-1)(x+3)=0

The solutions for this are clearly x=1 and x=-3. Then, to find the corresponding y solutions, we must put these values back into the 2nd equation from the question: y=x+3. Doing so, we get

\text{When }x=1, y=1+3=4,\,\,\,\text{ and when }x=-3, y=-3+3=0.

Thus, we have two solution pairs: x=1, y=4, and x=-3, y=0.