Question

Find the solutions to the following linear-quadratic systems algebraically. Select the ordered pair that is one of the correct solutions from among the choices below.
y = x² +3x + 8
y = 2x + 10

(-2, 6)
(0, 8)
(0,10)
(2, 14)

Answer 1

Answer:

(-2, 6)

Step-by-step explanation:

we have

$y=x^{2} +3x+8$ ----> equation A

$y=2x+10$ ----> equation B

equate equation A and equation B

$x^{2} +3x+8=2x+10$

$x^{2} +3x+8-2x-10=0$

$x^{2} +x-2=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +x-2=0$

so

$a=1\\b=1\\c=-2$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$x=\frac{-1\pm\sqrt{9}} {2}$

$x=\frac{-1\pm3} {2}$

$x=\frac{-1+3} {2}=1$

$x=\frac{-1-3} {2}=-2$

Find the values of y

For x=1

$y=2(1)+10=12$

For x=-2

$y=2(-2)+10=6$

therefore

The solutions are (1,12) and (-2,6)