Reduction or elimination method

Example

$$\begin{array}{c}x+y=2\\ -x+y=-4\end{array}\}$$ If we add both equations together, $x$ disappears. We can express this like follows: $$\begin{array}{rcl}& & -x+y=-4\\ & +& \underset{―}{x+y=2}\\ & & 0+2y=2\end{array}$$ The resulting equation is equivalent to the second one, so we can use it instead of the one we had in the initial system: $$\begin{array}{c}x+y=2\\ 2y=-2\end{array}\}$$ From this second equation we can obtain a value for $y$ straight away: $$2y=-2\Rightarrow y=-{\displaystyle \frac{2}{2}}=-1$$ With this value we can use the first equation to obtain the value for $x$: $$x+y=2\Rightarrow x-1=2\Rightarrow x=2+1=3$$ So that the solution to the system is $x=3,y=-1$.

Example

$$\begin{array}{c}{\displaystyle \frac{x}{2}}+4y={\displaystyle \frac{3}{2}}\\ -x-y=-4\end{array}\}$$ We can multiply the second equation by $4$ and add it to the first one, so we would eliminate $y$, or alternatively we can divide the first equation by $2$ and then add it to the second one. We better use this second option since this way denominators disappear:

$$[{\displaystyle \frac{x}{2}}+4y={\displaystyle \frac{3}{2}}]\cdot 2\Rightarrow x+8y=3$$

This equation is equivalent to the first one, so we obtain a new and equivalent system: $$\begin{array}{c}x+8y=3\\ -x-y=-4\end{array}\}$$ The sum of both equations allows us to know the value of $y$ straight away: $$\begin{array}{rcl}& & x+8y=3\\ & +& \underset{―}{-x-y=-4}\\ & & 0+7y=-1\end{array}\Rightarrow y=-{\displaystyle \frac{1}{7}}$$ Now it is possible to replace this value in the first equation to find $x$: $$x=3-8y\Rightarrow x=3-8\cdot (-{\displaystyle \frac{1}{7}})\Rightarrow x={\displaystyle \frac{21+8}{7}}={\displaystyle \frac{29}{7}}$$ Then, the solution to the system is $x={\displaystyle \frac{29}{7}},y=-{\displaystyle \frac{1}{7}}$.