Here, we want to get the value of x + y
We proceed as follows;
From the second equation, we have that;
[tex]\frac{2}{y-3}\text{ = }\frac{4}{x+1}-\frac{2}{3}[/tex]We can rewrite the first equation as;
[tex]\frac{2}{x+1}+4(\frac{2}{y-3})\text{ = }\frac{10}{3}[/tex]We substitute the first equation above in the second
Thus, we have that;
[tex]\begin{gathered} \frac{2}{x+1}\text{ + 4(}\frac{4}{x+1}-\frac{2}{3})\text{ = }\frac{10}{3} \\ \\ \frac{2}{x+1}+\frac{16}{x+1}-\frac{8}{3}\text{ = }\frac{10}{3} \\ \\ \frac{18}{x+1}\text{ = }\frac{10}{3}+\frac{8}{3} \\ \\ \frac{18}{x+1}\text{ = }\frac{18}{3} \\ \\ x+1\text{ = 3} \\ x\text{ = 3-1} \\ x\text{ = 2} \end{gathered}[/tex]We then proceed to get the value of y by substituting the obtained value of x
We have this as;
[tex]\begin{gathered} \frac{2}{y-3}\text{ =}\frac{4}{x+1}-\frac{2}{3} \\ \\ \frac{2}{y-3}\text{ = }\frac{4}{3}-\frac{2}{3} \\ \\ \frac{2}{y-3}\text{ = }\frac{2}{3} \\ \\ y-3\text{ = 3} \\ y\text{ = 3+ 3 = 6} \end{gathered}[/tex]Thus, we have the value of x + y as;
[tex]x\text{ + y = 2 + 6 = 8}[/tex]