to solve this,
Step 1
the purpose of this method is eliminate a variable by adding the two equations, to do this, you need to be sure that the add will make that variable disappear.
Let's see y
[tex]\begin{gathered} \frac{y}{3} \\ \text{and} \\ -\frac{y}{2} \end{gathered}[/tex]to eliminate y make
[tex]\begin{gathered} multiply\text{ the first equation }by\text{ }\frac{1}{2} \\ \\ \frac{5x}{6}+\frac{y}{3}=\frac{4}{3}\text{ by }\frac{1}{2} \\ \frac{1}{2}\cdot\frac{5x}{6}+\frac{1}{2}\cdot\frac{y}{3}=\frac{1}{2}\cdot\frac{4}{3} \\ \frac{5x}{12}+\frac{y}{6}=\frac{2}{3}\text{ equation (3)} \\ \end{gathered}[/tex]Now, multiply the second equation by 1/3
[tex]\begin{gathered} \frac{2x}{3}-\frac{y}{2}=\frac{11}{6}\text{ by }\frac{1}{3} \\ \frac{1}{3}\cdot\frac{2x}{3}-\frac{1}{3}\cdot\frac{y}{2}=\frac{1}{3}\cdot\frac{11}{6} \\ \frac{2x}{9}-\frac{y}{6}=\frac{11}{18}\text{ equation (4)} \end{gathered}[/tex]Now, add equations 3 and 4
[tex]\begin{gathered} \frac{5x}{12}+\frac{y}{6}=\frac{2}{3} \\ \frac{2x}{9}-\frac{y}{6}=\frac{11}{18} \\ x(\frac{5}{12}+\frac{2}{9})=\frac{2}{3}+\frac{11}{18} \\ x(\frac{23}{36})=\frac{23}{18} \\ x=\frac{23\cdot18}{36\cdot23} \\ x=\frac{414}{828} \\ \\ x=\frac{1}{2} \\ \end{gathered}[/tex]Now, with this value of x, find y, replacing in the equation 1 or 2
[tex]\begin{gathered} \frac{2x}{3}-\frac{y}{2}=\frac{11}{6} \\ \frac{2(\frac{1}{2})}{3}-\frac{y}{2}=\frac{11}{6} \\ \frac{1}{3}-\frac{y}{2}=\frac{11}{6} \\ -\frac{y}{2}=\frac{11}{6}-\frac{1}{3} \\ -\frac{y}{2}=\frac{3}{2} \\ y=\frac{3\cdot-2}{2} \\ y=-3 \end{gathered}[/tex]so, the solution is x=0.5 and y =-3
I really hope this helps