Identifying and verifying a solution to a system | Grade 8 (TX TEKS) | Khan Academy

We're told the system of linear equations below is graphed on the coordinate grid. So we can see the graph of ( y = -2X - 2 ) in blue here, and then ( Y = -\frac{1}{4}x + 5 ) in brown here.

What I want you to first do before I do it with you is see if you can visually think about what the solution is to this system. That is, an ( X ) and ( Y ) pair that satisfy both of these equations. Then I want you to verify that it is indeed the solution.

All right, now when I visually inspect it, it looks like this point right over here is on both lines. If I eyeball it, that looks like the point ( x = -4 ) and ( Y = 6 ), so ((-4, 6)).

But let's verify that that indeed is a point on both of these lines. To do that, let's see what ( Y ) is equal to in each of these when ( X = -4 ).

So in that first one, and maybe I'll do it in that same color just to make it a close color. If I say ( Y = -2 \cdot -4 - 2 ), that's equal to positive ( 8 - 2 ), which is indeed equal to ( 6 ).

So for this blue line, when ( X = -4 ), ( Y ) is indeed equal to ( 6 ). Now let's also do it for this brownish-looking line. There, ( Y = -\frac{1}{4} \cdot -4 + 5 ).

So here we have ( -\frac{1}{4} \cdot -4 ) is ( 1 + 5 ), which is indeed equal to ( 6 ). So that point ((-4, 6)) is indeed on both lines.