So the question is Solve for the system of equations:
y =3x - 10 (-4,-2)
4x-3y=-19
2x+y=13
You have copied (and high-lighted) the last line of problem #10 as if it were part of problem #11. Did you realize that? Problem #11 asks you to solve the two equations 4x- 3y= -19 and 2x+ y= 13. Multiply the second equation by 3, 6x+ 3y= 39, and add that to the first equation.
That technique is trying to eliminate one of the variables. So, the first equation has -3y. The second equation has y. If you multiply the second equation by 3, you have a -3y in the first equation and a +3y in the modified second equation, so now when you add them, you have 0y (this gives an equation only in x).
Another method: In the second equation, solve for y in terms of x (I choose the second equation because the coefficient of y is 1):
$y = 13-2x$
Now, plug that into the first equation:
$4x-3(13-2x)=-19$
After multiplying out, you can solve for $x$ since it is an equation in one variable. Once you know x, plug it into the formula you just created for y.