## Solving by Substitution

Contents

- Solving by Substitution
- Example
- Question
- Make \(y\) the subject of the first equation
- Substitute into the second equation and simplify
- Factorise the equation
- Substitute the values of \(x\) back into the first equation to determine the corresponding \(y\)-values
- Check that the two points satisfy both original equations
- Write the final answer

Use the simplest of the two given equations to express one of the variables in terms of the other.

Substitute into the second equation. By doing this we reduce the number of equations and the number of variables by one.

We now have one equation with one unknown variable which can be solved.

Use the solution to substitute back into the first equation to find the value of the other unknown variable.

## Example

### Question

Solve for \(x\) and \(y\): \begin{align*} y-2x &= -4 \qquad \ldots (1) \\ x^2 + y &= 4 \qquad \ldots (2) \end{align*}

### Make \(y\) the subject of the first equation

\[y = 2x – 4\]

### Substitute into the second equation and simplify

\begin{align*} x^2 + (2x-4) &= 4 \\ x^2 + 2x – 8 &= 0 \end{align*}

### Factorise the equation

\begin{align*} (x+4)(x-2) &= 0 \\ \therefore x = -4 &\text{ or } x = 2 \end{align*}

### Substitute the values of \(x\) back into the first equation to determine the corresponding \(y\)-values

If \(x = -4\): \begin{align*} y &= 2(-4)-4 \\ &= -12 \end{align*}

If \(x = 2\): \begin{align*} y &= 2(2) – 4 \\ &= 0 \end{align*}

### Check that the two points satisfy both original equations

### Write the final answer

The solution is \(x = -4 \text{ and } y = -12\) or \(x = 2 \text{ and } y = 0\). These are the coordinate pairs for the points of intersection as shown below.