13Elimination by equating - step by step

You have just seen how the equation procedure was applied in a specific example. Now we will look again in general how the procedure works.

Basic requirement

The two equations in your system of equations must both be solved for the same term! In this term there should be an unknown (often $x$ or $y$), which is then no longer on the other side.

Example 1

$\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I}) & y &= &-3x+5 \\\mathrm{II}) & y &=& 2x + 1\end{array}$

Example 2

$\def\arraystretch{1.25} \begin{array}{rrll}\mathrm{I})& 2x+1&=& 3y \\\mathrm{II}) & 2x+1&=& 6y-5\end{array}$

Step 1

Put together the two sides of the equations that are not equal.

Example 1

$\def\arraystretch{1.25} \begin{array}{rrl}\mathrm{I}&=&\mathrm{II} \\-3x+5&=&2x + 1 \end{array}$

Example 2

$\def\arraystretch{1.25} \begin{array}{rrl}\mathrm{I}&=&\mathrm{II} \\3y&=&6y -5 \end{array}$

Step 2

Solve for the other variable.

Example 1

$\def\arraystretch{1.25} \begin{array}{rlll}\mathrm{I}&=&\mathrm{II}& & \\-3x+5&=&2x + 1& |+3x\\5&=&5x+1& |-1\\4 &=&5x& |:5\\x&=&\frac{4}{5}\end{array}$

Example 2

$\def\arraystretch{1.25} \begin{array}{rlll}\mathrm{I}&=&\mathrm{II}& & \\3y&=&6y-5& |-6y\\-3y&=&-5& |:(-3)\\y&=&\frac{5}{3}\end{array}$

Step 3

Plug the first solution into one of both equations.

Example 1

In $\mathrm{I}$:

$y= 2\cdot \frac{4}{5} +1$

$y=\frac{13}{5}$

$P \; (\frac{4}{5}|\frac{13}{5})$

Example 2

In $\mathrm{II}$:

$\def\arraystretch{1.25} \begin{array}{rrll}2x+1&=&3\cdot \frac{5}{3} \\2x+1&=&5&|-1\\2x&=&4& |:2 \\x&=&2\end{array}$

$P \; (2|\frac{5}{3})$