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Solve the system of equations? 10x+y=-20 y=2x^2-4x-16

User Fathi
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You have one of those equations already solved for y, so part of the hard stuff is done, at least. Let's sub that y value in for y in the other equation and simplify, like this:
10x + (2 x^(2) -4x-16)=-20. Combining like terms and setting this equal to 0 so we can solve for x gives us this:
2 x^(2) +6x+4=0. When you factor that, you will get 2 values for x. They are x = -1 and x = -2. Let's see which one works. When x = -1:
10(-1)+y=-20 and -10 + y = -20. Therefore, y = -10. Let's see if -10 works in the other equation too when x is -1:
-10=2(-1) ^(2) -4(-1)-16. That simpifies down to -10 = -10, so the solution of (-1, -10) works. Let's see what happens if we try x = -2. Solving for y, we have 10(-2) + y = -20 and we find that y = 0. If we replace 0 in the other equation for y we get:
0=2(-2) ^(2) -4(-2)-16 and that simplifies to 0 = 0. Therefore, both of these solutions work. Let me tell you why. We have a line intersecting a parabola. By the fact that both of the solutions work, that means that the line intersects the parabola in 2 places, which is common in this type of system. 2 lines can only intersect in 1 place, but this is a line and a quadratic.
User Chollier
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