Question

What is the solution of the linear-quadratic system of equations? { y = x 2 + 5 x − 3 y − x = 2

Answer 1

The two solutions for the system of equations:

y = x^2+ 5x - 3

y - x = 2

Are:

(-5, -7) and (1, -1)

How to solve the system of equations?

Here we have the following system of equations:

y = x^2+ 5x - 3

y - x = 2

We can replace the first equation into the second one to get:

(x^2+ 5x - 3) - x = 2

x^2 + 4x - 3 = 2

x^2 + 4x - 3 - 2 = 0

x^2 + 4x - 5  = 0

Using the quadratic formula, we will get the solutions:

$x = \frac{-4 \pm \sqrt{4^2 - 4*1*-5} }{2} \\\\x = \frac{-4 \pm 6}{2}$

The two solutions are:

x = (-4 - 6)/2 = -5

x = (-4 + 6)/2 = 1

To find the values of y, we can evaluate the linear equation:

when x = -5

y - (-5) = -2

y + 5 = -2

y = -2 - 5 = -7

So one solution is (-5, -7)

when x = 1

y - 1 = -2

y = -2 + 1 = -1

The other solution is (1, -1)

Learn more about systems of equations:

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