Mathematics » Equations and Inequalities » Simultaneous Equations Continued

# Solving by Elimination II

## 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

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

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.