We are given the following system of equations
[tex]\begin{gathered} 5x=4y-30\quad eq.1 \\ 2x+3y=-12\quad eq.2 \end{gathered}[/tex]
We are asked to solve the system of equations by using a linear combination.
First of all, re-write eq.1 so that it is in the same form as eq.2
[tex]\begin{gathered} 5x-4y=-30\quad eq.1 \\ 2x+3y=-12\quad eq.2 \end{gathered}[/tex]
As you can see, both equations are in the same form now.
The goal is to add or subtract equations so that any one of the variables gets eliminated.
Let's say we want to eliminate the variable x.
To do that we need to multiply eq.1 by 2 and eq.2 by 5
[tex]\begin{gathered} 2\cdot(5x-4y=-30)\Rightarrow10x-8y=-60\quad eq.1 \\ 5\cdot(2x+3y=-12)\Rightarrow10x+15y=-60\quad eq.2 \end{gathered}[/tex]
Now subtract eq.2 from eq.1
As you can see, x is eliminated, the value of y is
[tex]\begin{gathered} -23y=0 \\ y=\frac{0}{-23} \\ y=0 \end{gathered}[/tex]
To find the value of x, substitute the value of y into any of the equations.
[tex]\begin{gathered} 10x-8y=-60\quad eq.1 \\ 10x-8(0)=-60 \\ 10x=-60 \\ x=\frac{-60}{10} \\ x=-6 \end{gathered}[/tex]
Therefore, the solution of the given system of equations is
[tex](-6,0)[/tex]
