Mathematics » Equations and Inequalities » Solving Simultaneous Equations

# Solving by Elimination

## Example

### Question

Solve the following system of equations:

\begin{align*} 3x + y & = 2 \qquad \ldots (1) \\ 6x – y & = 25 \quad \ldots (2) \end{align*}

### Make the coefficients of one of the variables the same in both equations

The coefficients of $$y$$ in the given equations are $$\text{1}$$ and $$-\text{1}$$. Eliminate the variable $$y$$ by adding equation $$(1)$$ and equation $$(2)$$ together:

$\begin{array}{cccc} & 3x + y & = & 2 \\ + & 6x – y & = & 25 \\ \hline & 9x + 0 & = & 27 \end{array}$

### Simplify and solve for $$x$$

\begin{align*} 9x & = 27 \\ \therefore x & = 3 \end{align*}

### Substitute $$x$$ back into either original equation and solve for $$y$$

\begin{align*} 3(3) + y & = 2 \\ y & = 2 – 9 \\ \therefore y & = -7 \end{align*}

### Check that the solution $$x=3$$ and $$y=-7$$ satisfies both original equations

\begin{align*} x & =3 \\ y & =-7 \end{align*}

## Example

### Question

Solve the following system of equations:

\begin{align*} 2a – 3b & = 5 \qquad \ldots (1) \\ 3a – 2b & = 20 \qquad \ldots(2) \end{align*}

### Make the coefficients of one of the variables the same in both equations

By multiplying equation $$(1)$$ by $$\text{3}$$ and equation $$(2)$$ by $$\text{2}$$, both coefficients of $$a$$ will be $$\text{6}$$.

$\begin{array}{cccc} & 6a – 9b & = & 15 \\ – & (6a – 4b & = & 40) \\ \hline & 0 – 5b & = & -25 \end{array}$

(When subtracting two equations, be careful of the signs.)

### Simplify and solve for $$b$$

\begin{align*} b & = \cfrac{-25}{-5} \\ \therefore b & = 5 \end{align*}

### Substitute value of $$b$$ back into either original equation and solve for $$a$$

\begin{align*} 2a – 3(5) & = 5 \\ 2a – 15 & = 5 \\ 2a & = 20 \\ \therefore a & = 10 \end{align*}