Solution by elimination -

Here is an example of a system of equations in two unknowns:

$\begin{cases}9a+4b=68\\ 5a+b=39 \end{cases}$

We need to find the value of $a$ and of $b$ that works in both equations. The solution to this system is $a=8$ and $b=-1$ . You can check this by substituting the $a$ and $b$ into the equations:

$\begin{align*}9a+4b&=9(8)+4(-1)=72-4=68\checkmark\\[10pt] 5a+b&=5(8)+(-1)=40-1=39\checkmark \end{align}$

See the details of this solution in Example 1 below.

Khan Academy video:

Worked examples

The goal is to force a ‘match’ between the coefficients of one or the other uknown. A ‘match’ is when the coefficients are equal or have opposite sign of the same number.

Example 1

$\begin{cases}9a+4b=68 \\ 5a+b=39 \end{cases}$

Let’s force a match with the coefficient of $b$ . Multiply the second equation by 4.

$\begin{cases}9a+4b=68\\ 4(5a+b)=4(39) \end{cases}$

leads to:

$\begin{cases}9a+4b=68\\ 20a+4b=156 \end{cases}$

Now subtract the first equation from the second equation:

$\begin{align*}(20a+4b)-(9a+4b)&=156-68\\[6pt] 11a&=88\end{align}$

We have eliminated the $b$ term! We can now solve for $a$ .

$\begin{align*}11a&=88\\[6pt]a&=8\end{align}$

Now solve for $b$ . Use either one of the original equations. Substitute $a=8$

$\begin{align*}5a+b&=39\\[6pt]5(8)+b&=39\\[6pt]40+b&=39\\[6pt]b&=-1\end{align}$

Finally, substitute our $a$ and $b$ values into the other equation to check the solution:

$\begin{align*}9a+4b&=68\\[6pt]9(8)+4(-1)&=68\\[6pt]72-4&=68\checkmark\end{align}$

Our solutions are $\fbox{\text{a=8, \quad b=-1}}$

Note: when we ‘subtract equation 1 from equation 2’ we can do this more efficient writing:

When you solve a system by elimination, you have to make two decisions:

Decision 1: Is there already a ‘match’ or do you need to force one by multiplying?

Decision 2: Should you add or subtract the equations?

Example 2

$\begin{cases} m+4n=0\\-m+2n=-6 \end{cases}$

Decision 1: Is there already a ‘match’ or do you need to force one by multiplying?

Answer: There is already a match. The coefficients of $m$ are $1$ and $-1$ in each equation respectively.

Decision 2: Should you add or subtract the equations?

Answer: Since we have opposite signs (plus, minus) we should add the equations.

$6n=-6$

We have eliminated the $m$ term! We can now solve for $n$ .

$\begin{align*}6n&=-6\\[6pt]n&=-1\end{align}$

Now solve for $m$ . Use either one of the original equations and substitute $n=-1$ :

$\begin{align*}m+4n&=0\\[6pt]m+4(-1)&=0\\[6pt]m-4&=0\\[6pt]m&=4\end{align}$

Finally, check your solutions by substituting them both into the other original equation:

$\begin{align*}-m+2n&=-6\\[6pt]-(4)+2(-1)&=-6\\[6pt]-4-2&=-6\checkmark\end{align}$

Our solutions are $\fbox{m=4, \quad n=-1}$

Example 3

$\begin{cases}2x-5y=-9 \\3x+4y=44 \end{cases}$

Decision 1: Is there already a ‘match’ or do you need to force one by multiplying?

Answer: The isn’t a match. We need to multiply both equations to force a match.

To get a match, choose either the coefficients of the $x$ or of the $y$ . Let’s choose $x$ . The Lowest Common Multiple of 2 and 3 is 6. This is how we get $6x$ in both equations:

$\begin{cases}3(2x-5y)=3(-9) \\2(3x+4y)=2(44) \end{cases}$

leads to:

$\begin{cases}6x-15y=-27 \\6x+8y=88 \end{cases}$

Now we have a match with the coefficient of $x$ .

Decision 2: Should you add or subtract the equations?

Answer: Since we have the same sign (plus $6x$ , plus $6x$ ) we should subtract the equations.

Take care with the ‘double negatives’:

$\begin{align*}(6x-6x)+(8y--15y)&=88--27\\[6pt]23y&=115\\[6pt]y&=5\end{align}$

Now substitute $y=5$ into one of the original equations:

$\begin{align*}2x-5y&=-9\\[6pt]2x-5(5)&=-9\\[6pt]2x-25&=-9\\[6pt]2x&=16\\[6pt]x&=8\end{align}$

Finally substitute $y=5$ and $x=8$ into the other original equation to check the solutions:

$\begin{align*}3x+4y&=44\\[6pt]3(8)+4(5)&=44\\[6pt]24+20&=44\checkmark\end{align}$

Our solutions are $\fbox{x=8, \quad y= 5}$

PRACTICE 1

Example problems (with context problems) to print:

Exercise solutions:

Practice 2

Solve the following systems using elimination. Do your working on paper. Type your solutions in to see the corresponding graphs and the point of intersection.

further Learning

Understand the concepts behind the methods.

Concepts for manipulating one equation in one unknown:

Concepts for manipulating two equations in two unknowns: