Simultaneous Equations Worksheets | Questions and Revision | MME

# Simultaneous Equations Questions, Revision and Worksheets

Level 4 Level 5

## What you need to know

### Simultaneous Equations

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 already, revise solving equations,) but if there’s an equation with two variables, like

$3x+y=10$,

we can’t really “solve” it in the same way. However, if you have 2 equations both with 2 variables, like

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

then there is a solution for us to find that works for both equations. These are equations are called simultaneous for this reason. In order to be able to solve simultaneous equations you will need to be confident in the following topics.

### How to Solve Simultaneous Equations

There are two ways to solve simultaneous equations:

1. The elimination method
2. The substitution method

For higher tier simultaneous equation questions you will need to be able to solve quadratics by factorising and through use of the quadratic formula.

### Example 1: Solving Simultaneous Equations

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.

So, we eliminated the $y$ term and now have $2x=6$.

Dividing both sides of this equation by 2, we get

$x=3$

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

$x=3, y=1$

### Example 2: Applied Simultaneous Equation Question

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, solving these will require application of knowledge.

Let’s say that the price of a milkshake is $a$, and the price of an ice cream is $b$. This creates 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.

In order to use the elimination method, you need to have the a or b variables the same in both equations. We can multiply both sides of the second equation by 2, to make this work:

$4a+4b=14$

So the a values of the first and third equation are the same. 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

$4a=6$

and then divide both sides by 4 to get

$a=1.5$

Therefore the price of a milkshake is £1.50.

### Example 3: Solving Quadratic Simultaneous Equations

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.

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

$x=1$ and $x=-3$.

Then, to find the corresponding $y$ solutions, we must put these values back into the 2nd equation. 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

Striaght away we can subtract equation 2 from equation 1 so that,

\begin{aligned}y&=2x-6\\ y&=\dfrac{1}{2}x+6 \\ \\ (y-y)&=(2x-\dfrac{1}{2}x)-6-6 \\ 0&= \dfrac{3}{2}x-12 \end{aligned}

If we rearrgne to make $x$ the subject we find,

$x=\dfrac{2\times12}{3}=\dfrac{24}{3}=8$

Substituting $x=8$ back into the original first equation,

\begin{aligned}y&=2(8)-6 \\ y&=10\end{aligned}

Hence, the solution is,

$x=8, y=10$

If we multiply the second equation by 2, we have two equations both with a $2x$ term, hence subtracting our new equation 2 from equation 1 we get,

\begin{aligned}2x-3y&=16\\ 2x+4y&=-12 \\ \\ (2x-2x)+(-3y-4y)&= 16-(-12) \\ 0x -7y &=28\end{aligned}

If we rearrgne to make $y$ the subject we find,

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

Substituting $y=-4$ back into the original second equation,

\begin{aligned}x+2(-4)&=-6 \\ x-8&=-6 \\ x&=2\end{aligned}

Hence, the solution is,

$x=2, y=-4$

If we multiply the first equation by 3, we have two equations both with a $3x$ term, hence subtracting our new equation 2 from equation 1 we get,

\begin{aligned}3x+6y+30&=0 \\ 3x-5y-14&=0 \\ \\ (3x-3x)+(6y-(-5y))+(30-(-14)) &=0 \\ 11y &= -44 \\ y&=-4\end{aligned}

Substituting $y=-4$ back into the original first equation,

\begin{aligned}x+2(-4)+10&=0\\ x-8+10 &=0 \\ x&=-2 \end{aligned}

Hence, the solution is,

$x=-2, y=-4$

Let $A$ be the cost of an adult ticket and let $C$ be the cost of a child ticket, thus we have two simultaneous equations,

\begin{aligned} 2A+3C&=20 \\ A+C&=8.5 \end{aligned}

If we multiply the second equation by 2, we have two equations both with a $2A$ term, hence subtracting our new equation 2 from equation 1 we get,

\begin{aligned} 2A+3C&=20 \\ 2A+2C&=17 \\ \\ (2A-2A)+(3C-2C) &=(20-17) \\ C&=3\end{aligned}

Then, substituting this value back into the original equation 2, we get,

\begin{aligned} A+3&=8.5 \\ A & =5.5\end{aligned}

Therefore, the cost of a child ticket is £3, and the cost of an adult ticket is £5.50

If we multiply the first equation by 2, we have two equations both with a $2y$ term, hence adding our new equation 1 and equation 2 we get,

\begin{aligned} 2x^2-2y&=28 \\ 2y-4&=12x \\ \\ 2x^2+(-4)+(-2y+2y)&=28+12x \\ 2x^2-12x-32&=0\end{aligned}

After rearranging to form a quadratic we can solve for $x$

\begin{aligned} 2x^2-12x-32&=0 \\ 2(x^2-6x-16)&=0 \\ (x-8)(x+2)&=0\end{aligned}

Therefore, the 2 solutions for $x$ are $x=8, x=-2$. To find the two $y$ solutions, we can substitute these values back into the original second equation.

When x=8,

\begin{aligned}2y-4&=96 \\ y&=50 \end{aligned}

When x=-2,

\begin{aligned}2y-4&=-24 \\ y &= -10\end{aligned}

Thus, the two pairs of solutions are,

$x=8,y=50$ and $x=-2,y=-10$

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