The given system of equations is:
[tex]\begin{gathered} 2x+4y=-4\text{ Equation (1)} \\ 3x+5y=-3\text{ Equation (2)} \end{gathered}[/tex]
Multiply equation 1 by 3 on both sides, and equation 2 by 3 on both sides:
[tex]\begin{gathered} 3(2x+4y)=3(-4) \\ 6x+12y=-12\text{ Equation (1)} \\ 2(3x+5y)=2(-3) \\ 6x+10y=-6\text{ Equation (2)} \end{gathered}[/tex]
Subtract equation (2) from equation (1):
[tex]\begin{gathered} +6x+12y=-12 \\ -6x+10y=-6 \\ ----------- \\ 0+2y=-6 \end{gathered}[/tex]
Now, solve for y:
[tex]\begin{gathered} 2y=-6 \\ y=\frac{-6}{2} \\ y=-3 \end{gathered}[/tex]
Replace y in equation (1) and find x:
[tex]\begin{gathered} 2x+4(-3)=-4 \\ 2x-12=-4 \\ 2x=-4+12 \\ 2x=8 \\ x=\frac{8}{2} \\ x=4 \end{gathered}[/tex]
Answer: The solution is (4, -3)
