**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