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.

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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: