We have to solve the system of equations:
[tex]\begin{gathered} 2x+3y=-10 \\ 5x+2y=8 \end{gathered}[/tex]
We can solve it by substitution.
We find the value of x in function of y from the first equation:
[tex]\begin{gathered} 2x+3y=-10 \\ 2x=-10-3y \\ x=\frac{-10-3y}{2} \end{gathered}[/tex]
Then, we substitute the value of x in the second equation and solve for y:
[tex]\begin{gathered} 5x+2y=8 \\ 5(\frac{-10-3y}{2})+2y=8 \\ -\frac{5\cdot10}{2}-\frac{5\cdot3y}{2}+2y=8 \\ -25-\frac{15}{2}y+2y=8 \\ (2-\frac{15}{2})y=8+25 \\ \frac{4-15}{2}y=33 \\ \frac{-11}{2}y=33 \\ y=\frac{33\cdot2}{-11} \\ y=-\frac{66}{11} \\ y=-6 \end{gathered}[/tex]
With the value of y=-6, we can solve for x:
[tex]x=\frac{-10-3y}{2}=\frac{-10-3\cdot(-6)}{2}=\frac{-10+18}{2}=\frac{8}{2}=4[/tex]
Answer: x=4 and y=-6
