Given the system of equations:
[tex]\begin{cases}3x+2y=14 \\ 3x-2y=10\end{cases}[/tex]We'll multiply the second equation by -1 and then add both equations up:
[tex]\begin{gathered} \begin{cases}3x+2y=14 \\ 3x-2y=10\end{cases}\rightarrow\begin{cases}3x+2y=14 \\ -3x+2y=-10\end{cases} \\ \\ \rightarrow4y=4 \end{gathered}[/tex]Solving the resulting equation for y ,
[tex]\begin{gathered} 4y=4\rightarrow y=\frac{4}{4} \\ \\ \Rightarrow y=1 \end{gathered}[/tex]We'll plug in this y-value in the first equation and solve for x ,
[tex]\begin{gathered} 3x+2y=14 \\ \rightarrow3x+2(1)=14 \\ \rightarrow3x+2=14\rightarrow3x=12\rightarrow x=\frac{12}{3} \\ \\ \Rightarrow x=4 \end{gathered}[/tex]Therefore, we can conclude that the solution to our system is:
[tex]\begin{gathered} x=4 \\ y=1 \end{gathered}[/tex]
