To solve the above system of equation, let's use substitution method. Here are the steps:
1. Equate second equation to y = .
[tex]\begin{gathered} 10x+5y=-50 \\ 5y=-10x-50 \\ \text{Divide both sides by 5.} \\ \frac{5y}{5}=-\frac{10x}{5}-\frac{50}{5} \\ y=-2x-10 \end{gathered}[/tex]2. Use this value of "y" and plug it in to the first equation.
[tex]\begin{gathered} 4x+2y=-20 \\ 4x+2(-2x-10)=-20 \end{gathered}[/tex]Then, solve for x.
[tex]\begin{gathered} 4x-4x-20=-20 \\ \text{Add 20 on both sides.} \\ 4x-4x-20+20=-20+20 \\ 0=0 \end{gathered}[/tex]It seems like both equations are just equal to each other. With this, the system has infinite number of solutions.