Elimination method

The elimination method is also known as the method of equal coefficients and consists in making the coefficients in front of one unknown equal in the two equations and then adding or subtracting the equations so that the unknown disappears.

We are then left with an equation with only one unknown, which we can solve.

 

Unlike the substitution method, we do not isolate an unknown directly, but change both equations so that one unknown can be eliminated.

 

Coefficients are the numbers in front of the unknowns. For example, in this equation \(\large 8y-4x=4\). Here 8 and 4 are the coefficients.

We take the two equations from earlier:

 

$$ \large 8y-4x=4 $$

$$ \large 2y+4x=20 $$

 

We need equal coefficients for one of the unknowns, e.g. \(\large y\). We can obtain this by multiplying the lower equation by 4, so that we get \(\large 8y\), which we already have in the upper equation.

 

$$ \large \textcolor{red}{4\cdot}2y+\textcolor{red}{4\cdot}4x=\textcolor{red}{4\cdot}20 \Leftrightarrow $$

$$ \large 8y+16x=80 $$

 

Now we have two equations with two equal coefficients, namely \(\large 8y\).

 

Subtracting the equations

Now the two equations with equal coefficients must be subtracted. We do it like this:

 

$$ \large 8y-4x\textcolor{red}{-(8y+16x)}=4\textcolor{red}{-80} $$

 

Remember to use parentheses so that no sign errors occur.

We remove the minus-parenthesis, so the signs must change:

 

$$ \large 8y-4x-(8y+16x)=4-80 \Leftrightarrow $$

$$ \large 8y-4x-8y-16x=4-80 $$

 

We continue by isolating \(\large x\).

\(\large 8y\) disappears because \(\large 8y-8y=0\).

 

$$ \large \begin{aligned} -4x-16x &= -76 \Leftrightarrow \\[12pt] -20x &= -76 \Leftrightarrow \\[12pt] \frac{-20x}{20} &= \frac{-76}{20} \Leftrightarrow \\[12pt] -x &= -3.8 \Leftrightarrow \\[12pt] x &= 3.8 \end{aligned} $$

 

Find the second unknown

Now that we have found \(\large x\), we must find \(\large y\).

This is no different from the previous examples. We insert our \(\large x\) into one of the equations and find \(\large y\). It does not matter which of the two equations you use:

 

$$ \large \begin{aligned}2y+4x&=20 \Leftrightarrow \\[12pt] 2y+4\cdot 3.8&=20 \Leftrightarrow \\[12pt] 2y+15.2&=20 \Leftrightarrow \\[12pt] 2y&=20-15.2 \Leftrightarrow \\[12pt] \frac{2y}{2}&=\frac{4.8}{2} \Leftrightarrow \\[12pt] y&=2.4 \end{aligned} $$

 

Check

We insert the solution into both equations to check:

 

$$ \large 8\cdot 2.4 - 4\cdot 3.8 = 4 $$

$$ \large 2\cdot 2.4 + 4\cdot 3.8 = 20 $$

 

Both are true, so the solution is correct.

 

Note: The elimination method always works, but sometimes it shows that the system has no solution or infinitely many:

 

• If you end up with something impossible, e.g. \( \large 0 = 5 \), it means the system has no solution.
• If you end up with something trivial, e.g. \( \large 0 = 0 \), it means the system has infinitely many solutions.