**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.

## 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

## 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.

**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

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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]**

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]**

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]**

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]**

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]**

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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