## Solving by Elimination

Make one of the variables the subject of both equations.

Equate the two equations; 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 either original equation, to find the corresponding value of the other unknown variable.

## Example

### Question

Solve for \(x\) and \(y\): \begin{align*} y &= x^2 – 6x \qquad \ldots (1) \\ y + \cfrac{1}{2}x – 3 &= 0 \qquad \ldots (2) \end{align*}

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

\begin{align*} y + \cfrac{1}{2}x – 3 &= 0 \\ y &= -\cfrac{1}{2}x + 3 \end{align*}

### Equate the two equations and solve for \(x\)

\begin{align*} x^2 – 6x &= -\cfrac{1}{2}x + 3 \\ x^2 – 6x + \cfrac{1}{2}x – 3 &= 0\\ 2x^2 – 12x + x – 6 &= 0\\ 2x^2 -11x – 6 &= 0\\ (2x +1)(x-6) &= 0\\ \text{Therefore } x = -\cfrac{1}{2}&\text{ or } x = 6 \end{align*}

### Substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = -\dfrac{1}{2}\): \begin{align*} y &= -\cfrac{1}{2} ( -\cfrac{1}{2} ) + 3 \\ \therefore y &= 3\cfrac{1}{4} \end{align*} This gives the point \((-\dfrac{1}{2};3\dfrac{1}{4})\).

If \(x = 6\): \begin{align*} y &= -\cfrac{1}{2}(6) + 3 \\ &= -3 + 3 \\ \therefore y &= 0 \end{align*} This gives the point \((6;0)\).

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

### Write the final answer

The solution is \(x = -\dfrac{1}{2} \text{ and } y = 3\dfrac{1}{4}\) or \(x = 6 \text{ and } y = 0\). These are the coordinate pairs for the points of intersection as shown below.

## Example

### Question

Solve for \(x\) and \(y\): \begin{align*} y &= \dfrac{5}{x – 2} \qquad \ldots (1)\\ y + 1 &= 2x \qquad \ldots (2) \end{align*}

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

\begin{align*} y + 1 &= 2x \\ y &= 2x – 1 \end{align*}

### Equate the two equations and solve for \(x\)

\begin{align*} 2x – 1 &= \cfrac{5}{x – 2} \\ (2x – 1)(x – 2) &= 5 \\ 2x^2 – 5x + 2 &= 5 \\ 2x^2 – 5x – 3 &= 0 \\ (2x + 1)(x – 3) &= 0 \\ \text{Therefore } x = -\cfrac{1}{2}&\text{ or } x = 3 \end{align*}

### Substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x -\dfrac{1}{2}\): \begin{align*} y &= 2(-\cfrac{1}{2}) – 1 \\ \therefore y &= -2 \end{align*} This gives the point \((-\dfrac{1}{2};-2)\).

If \(x = 3\): \begin{align*} y &= 2(3) -1 \\ &= 5 \end{align*} This gives the point \((3;5)\).

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

### Write the final answer

The solution is \(x = -\dfrac{1}{2} \text{ and } y = -2\) or \(x = 3 \text{ and } y = 5\). These are the coordinate pairs for the points of intersection as shown below.