Question

What are the solutions to the following system of equations y=x^2 + 3x-7
3x-y= -2

(3,11) and (-3,-7)
(11,3) and (-3,-7)
(3,11) and (-7,-3)
No real solutions.

Answer 1

$y = {x}^{2} + 3x - 7 \\ and \\ 3x - y = 2$
From the second equation:
$3x + 2 = y$
Substitute the value of y in the first equation:
$3x + 2 = {x}^{2} + 3x - 7$
Rearrange the equation:
$0 = {x}^{2} + 3x - 3x - 7 - 2 \\ \\ {x}^{2} - 9 = 0$
Using the "difference of two squares" approach:
$(x + 3)(x - 3) = 0$
Therefore the values of x can be
$x + 3 = 0 \: or \: x - 3 = 0 \\ \\ x = - 3 \: or \: x = + 3$
Substitute either values of x in the second equation:
$3( - 3) - y = 2 \: or \: 3(3) - y = 2 \\ \\ - 9 - y = 2 \: or \: 9 - y = 2$
y =-2 - 9 or y = 9 - 2
y = - 11 or y = 7

The answer is (-3, - 11) and (3, 7). The answer is D

Answer 2

Answer:

its D

Step-by-step explanation: