As given by the question
There are given that the system of the equation
[tex]\begin{gathered} 3x-2y=10\ldots(1) \\ 5x+10y=-10\ldots(2) \end{gathered}[/tex]Now,
From equation (1), find the value of x
[tex]\begin{gathered} 3x-2y=10 \\ 3x=10+2y \\ x=\frac{10+2y}{3}\ldots(3) \end{gathered}[/tex]Then,
Put the value of x into the equation (2)
So,
[tex]\begin{gathered} 5x+10y=-10 \\ 5(\frac{10+2y}{3})+10y=-10 \\ \frac{50+10y}{3}+10y=-10 \\ \frac{50+10y+30y}{3}=-10 \\ 50+10y+30y=-30 \\ 50+40y=-30 \\ 40y=-30-50 \\ 40y=-80 \\ y=-2 \end{gathered}[/tex]Now,
Put the value of y into the equation (3)
So,
From the equation (3)
[tex]\begin{gathered} x=\frac{10+2y}{3} \\ x=\frac{10+2(-2)}{3} \\ x=\frac{10-4}{3} \\ x=\frac{6}{3} \\ x=2 \end{gathered}[/tex]Hence, the value of x is 2, and the value of y is -2.
