The given system of equations is:
[tex]\begin{gathered} 5x-4y=-10\text{ Eq.(1)} \\ 3x+2y=16\text{ Eq.(2)} \end{gathered}[/tex]
Start by solving for x in equation 1:
[tex]\begin{gathered} 5x-4y=-10 \\ 5x=-10+4y \\ x=\frac{-10+4y}{5} \\ x=-\frac{10}{5}+\frac{4y}{5} \\ x=-2+\frac{4y}{5}\text{ Eq.(3)} \end{gathered}[/tex]
Now, replace this into equation 2:
[tex]\begin{gathered} 3(-2+\frac{4y}{5})+2y=16 \\ \text{Apply the distributive property} \\ 3\cdot(-2)+3\cdot\frac{4y}{5}+2y=16 \\ -6+\frac{12y}{5}+2y=16 \\ \frac{12y}{5}+2y=16+6 \\ \frac{12y}{5}+2y=22 \\ \frac{12y+5\cdot2y}{5}=22 \\ \frac{12y+10y}{5}=22 \\ \frac{22y}{5}=22 \\ 22y=22\cdot5 \\ y=\frac{22\cdot5}{22} \\ \text{Simplify} \\ y=5 \end{gathered}[/tex]
Now, replace the y-value into Eq.(3) and find x:
[tex]\begin{gathered} x=-2+\frac{4\cdot5}{5} \\ \text{Simplify} \\ x=-2+4 \\ x=2 \end{gathered}[/tex]
Now, let's check these values into the system of equations:
[tex]\begin{gathered} 5\cdot2-4\cdot5=-10\text{ Eq.(1)} \\ 10-20=-10 \\ -10=-10\text{ O.K.} \\ 3\cdot2+2\cdot5=16\text{ Eq.(2)} \\ 6+10=16 \\ 16=16\text{ O.K.} \end{gathered}[/tex]
Answer: x=2 and y=5
