Question

Solve the system of equations algebraically. Round to the nearest tenth, if necessary. 4x + 6=y, y = 3x^2 + 2x - 10

Answer 1

Since we have both equations set equal to y, we know that:

$4x+6=3x^{2}+2x-10\\ \\ 3x^{2}-2x-16=0 \\ \\ (3x-8)(x+2)=0\\\\x=\frac{8}{3}, -2$

If x=8/3, then y=4(8/3)+6=50/3.

If x=-2, then y=4(-2)+6=-2.

So, the solutions, to the nearest tenth, are:

$x=2, y=-2\\\\x=2.7, y=16.7$