To solve the given system of linear equations.
For elimination method:
1. Turn the equations to get that one of the terms with variables be opposite. In this case you can multiply one of the equation for -1:
Multiply the first equation for -1:
[tex]\begin{gathered} -1(4x+3y=-28) \\ -4x-3y=28 \end{gathered}[/tex]
2. Add both equations:
3. Solve y in the result of the addition in step 2:
[tex]-12y=48[/tex]
Divide both sides of the eqution into -12
[tex]\begin{gathered} \frac{-12}{-12}y=\frac{48}{-12} \\ \\ y=-4 \end{gathered}[/tex]
4. Use the value of y= -4 to find the value of x:
Substitute in one of the equations the y for -4 and solve for x:
[tex]\begin{gathered} 4x+3y=-28 \\ 4x+3(-4)=-28 \\ 4x-12=-28 \end{gathered}[/tex]
Add 12 in both sides of the equation:
[tex]\begin{gathered} 4x-12+12=-28+12 \\ 4x=-16 \end{gathered}[/tex]
Divide both sides of the equation into 4:
[tex]\begin{gathered} \frac{4}{4}x=-\frac{16}{4} \\ \\ x=-4 \end{gathered}[/tex]Then, the solution for the system is ( -4, -4)
