 # Simultaneous Equations Questions, Revision and Worksheets

Level 4 Level 5

## What you need to know

We use the term ‘simultaneous equations’ when we have multiple equations that share a bunch of variables and are all true at the same time. We’ve seen how to solve one equation with one variable (if you haven’t, click here solving equations revision, but if there’s an equation with two variables, like

$3x+y=10$,

then things become different. This equation actually has infinite solutions, so we can’t really “solve” it in the way that we want to. However, if you have 2 equations both with 2 variables, like

$2x+y=10\,\,\,\text{ and }\,\,\,x+y=4$

Then suddenly there is 1 solution for us to find. This is system of simultaneous equations, and we’re going to see how to find the solution to it.

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 $x=-3, y=0$.

### Example Questions

Currently, the equations are not in a state where subtracting one from the other will help eliminate one of the variables. However, if we multiply the second equation by 2, it becomes

$2x+4y=-12$,

Now we have two equations (this one and the first one in the question) with a $2x$ in them. So, we will subtract this equation from the first equation.

Firstly, the $2x$ terms will cancel, then we get $-3y-4y=-7y$ for the $y$ term, and finally we get $16-(-12)=28$ for the constant term. So, this looks like. Then, if we divide both sides of $-7y=28$ by -7, we get

$y=\dfrac{28}{-7}=-4$

Now, we will substitute $y=-4$ back into the equation $x+2y=-6$. Doing so, we get

$x+2(-4)=x-8=-6$

Adding 8 to both sides, we get that $x=-6+8=2$. So, the solution is: $x=2,y=-4$.

#### Is this a topic you struggle with? Get help now.

Let $A$ be the cost of an adult ticket and let $C$ be the cost of a child ticket (it doesn’t matter if you chose different letters). Then, the sentence “the cost of buying 2 adult tickets and 3 child tickets would be £20” becomes the equation

$2A+3C=20$

Furthermore, the sentence “the cost of buying one of each type would be £8.50” (i.e. the cost of one adult and one child) becomes the equation

$A+C=8.5$

To make these equations suitable for the elimination process, we will multiply the second equation by 2 (if you multiplied it by 3 and went down that route, that’s fine too). Doing so, we get

$2A+2C=17$

Then, subtracting this equation from the first equation looks like So, we get immediately that $C=3$. Then, substituting this value back into the equation $A+C=8.5$, we get

$A+3=8.5$

Subtracting 3 from both sides gives us $A=8.5-3=5.5$. Therefore, the cost of a child ticket is £3, and the cost of an adult ticket is £5.50.

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So, we’re going to have to substitute one of these equations directly into the other. Specifically, we’re going to replace the $y$ in the first equation with an expression involving $x$. To do this, we need rearrange the second equation to make $y$ the subject. Firstly, adding 4 to both sides of the second equation, we get

$2y=12x+4$

Then, divide both sides by 2 to get

$y=6x+2$

Now, replacing the $y$ in the first equation with $(6x+2)$, we get

$x^2-(6x+2)=14$

If we expand the bracket and then subtract 14 from both sides, this equation becomes

$x^2-6x-16=0$

Now, if we observe that $-8\times2=-16$ and $-8+2=-6$, then we can factorise this quadratic equation to be

$(x-8)(x+2)=0$

Therefore, the 2 solutions for $x$ are $x=8, x=-2$. To find the two $y$ solutions, we must substitute these values back into one of the equations in the question. We’re going to pick the rearranged version of the second equation: $y=6x+2$. Doing this, we get

$\text{when }x=8, y=6(8)+2=50,\,\,\,\text{ and when }x=-2, y=6(-2)+2=-10$.

Thus, the two pairs of solutions are $x=8,y=50$ and $x=-2,y=-10$.

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