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Simultaneous Equations

Simultaneous equations are multiple equations that share the same variables and which are all true at the same time.

When an equation has $2$ variables its much harder to solve, 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 equations are called simultaneous for this reason.

There are 2 main types of equation you need to be able to solve.

Make sure you are happy with the following topics before continuing.

Level 4-5

Type 1: Linear Simultaneous equations.

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.

Example: Find the solution to the following simultaneous equations.

$4x + 3y = 14 \,\,\,\,,\ 5x+7y=11$

Step 1: Write one equation above the other.

Both equations need to be in the form $ax+by=c$ , so rearrange if needed.

$\begin{aligned}4x + 3y &= 14 \\ 5x+7y &= 11 \end{aligned}$

Step 2: Get the coefficients to match

The coefficients are the numbers before $x$ and $y$ , make the $x$ coefficients the same by scaling up both equations

$(\times5) \,\,\,\,\,\,\,\,\,4x + 3y = 14\,\,\,$ gives $\,\,\, 20x + 15y = 70$

$\,\, (\times4)\,\,\,\,\,\,\,\,\, 5x+7y = 11 \,\,\,$ gives $\,\,\, 20x+28y=44$

Step 3: Add or subtract the equations to eliminate terms with equal coefficients.

As both equations are $+20x$ we must subtract the equations.

$\begin{aligned}20x + 15y &= 70 \\ (-)\,\,\,\,\,\,\,\,\, 20x+28y&=44 \\ \hline-13y&=26\end{aligned}$

Step 4: Solve the resulting equation

$\begin{aligned}(\div-13)\,\,\,\,\,\,\,\,\,-13y&=26 \\ y &=-2\end{aligned}$

Step 5: Substitute the answer into the simplest of the two equations to find the other variable.

$\begin{aligned}4x + 3y &= 14 \\ 4x +3(-2) &=14 \\ 4x -6 &= 14 \\ 4x &= 20 \\ x &= 5\end{aligned}$

This gives the final answer to be;

$x = 5, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, y = -2$

Level 4-5

Level 6-7

Type 2: Non-linear Simultaneous Equations

Because one of these equations is quadratic (Non-linear), we can’t use elimination like before. Instead, we have to use substitution.

Example: Solve the following simultaneous equations.

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

Step 1: Rearrange the linear equation to get one of the unknowns on its own and on one side of the equals sign.

$\begin{aligned}(+x)\,\,\,\,\,\,\,\,\,y-x&=3 \\ y &= x+3\end{aligned}$

Step 2: Substitute the linear equation into the non-linear.

We know $y=x+3$ so we can replace the $y$ in the first equation with $x+3$ :

$\begin{aligned} x^2+2y&=9 \\ x^2+2(x+3)&=9\end{aligned}$

Step 3: Expand and solve the new quadratic formed.

$\begin{aligned}x^2+2(x+3)&=9 \\ x^2 + 2x+6 &= 9 \\ x^2+2x-3&=0 \\ (x-1)(x+3)&=0\end{aligned}$

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

Step 4: Substitute both values back into the simplest equation to find both versions of the other variable.

$\text{When }x=1, \,\,\,\,\,\,\,\,\, y=1+3=4,$

$\text{When }x=-3, \,\,\,\,\,\,\,\,\, y=-3+3=0.$

Thus, we have two solution pairs:

$x=1, y=4,$ and $x=-3, y=0$ .

Level 8-9

Level 4-5

Example: Applied Simultaneous Equation Question

A store sells milkshakes and ice creams.

$2$ milkshakes and $2$ ice creams, costs $£7$

$4$ milkshakes and $3$ ice creams, costs $£12$

Work out the cost of an individual milkshake and an individual ice cream.

[4 marks]

Step 1: we need to do is form the two simultaneous equations.

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

This creates the following two equations.

$2a+2b=7$

$4a+3b=12$

Step 2: Now we must get the coefficients to match, in this case we can multiply the first equation by $2$

$(\times 2) \,\,\,\,\,\,\,\,\, 2a+2b=7 \,\,\, \text{Gives} \,\,\, 4a + 4b = 14$

Step 3: Subtract the equations to eliminate terms with equal coefficient.

$\begin{aligned}4a + 4b &= 14 \\ (-)\,\,\,\,\,\,\,\,\, 4a+3b&=12 \\ \hline b&=2\end{aligned}$

Step 4: Substitute the answer into the simplest of the two equations to find the other variable.

$\begin{aligned}2a+2b&=7 \\ 2a + 2(2) &=7 \\ 2a+4&=7 \\ 2a &=3 \\ a& = 1.5\end{aligned}$

This gives the final answer to be

Milkshake ( $a$ ) = $£1.50$

Ice Cream ( $b$ ) = $£2.00$

Level 4-5

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Example Questions

Question 1: Solve the following simultaneous equations,

$\begin{aligned}y&=2x-6\\y&=\dfrac{1}{2}x+6\end{aligned}$

[4 marks]

Level 4-5

Straight 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 rearrange 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,

$x=8, y=10$

Question 2: Solve the following simultaneous equations,

$\begin{aligned}2x-3y&=16\\x+2y&=-6\end{aligned}$

[4 marks]

Level 4-5

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 rearrange 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,

$x=2, y=-4$

Question 3: Solve the following simultaneous equations,

$\begin{aligned}x+2y+10&=0\\3x-5y-14&=0\end{aligned}$

[4 marks]

Level 4-5

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,

$\begin{aligned}3x+6y+30&=0 \\ 3x-5y-14&=0 \end{aligned}$

$(3x-3x)+(6y-(-5y))+(30-(-14)) = 0$

$\begin{aligned} 11y &= -44 \\ y &= - 4\end{aligned}$

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,

$x=-2, y=-4$

Question 4: There are two types of ticket on a train: an adult ticket and a child ticket. The cost of buying $2$ adult tickets and $3$ child tickets would be $£20$ . The cost of buying one of each type would be $£8.50$ .

Work out the cost of a single adult ticket and a single child ticket.

[3 marks]

Level 4-5

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 $£5.50$ .

Question 5: Solve the following simultaneous equations,

$\begin{aligned}x^2-y&=14\\2y-4&=12x\end{aligned}$

[5 marks]

Level 8-9

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 $x=-2,y=-10$