Respuesta :

akwow0924
I will go about solving this using the elimination method.

First, convert the equations.

10x + y = -20
4x + y = -12

Second, find the easiest variable to get rid of and get rid of it!  (In this case, y)  We will subtract to get rid of y.

6x = -8

Third, you want to solve the equation.

6x = -8 (divide by 6)
x = [tex]-1 \frac{1}{3} [/tex]

Fourth, solve for y by inserting the answer for x into one of the equations.

10([tex]-1 \frac{1}{3} [/tex]) + y = -20
[tex]-13 \frac{1}{3} [/tex] + y = -20 (subtract [tex]-13 \frac{1}{3} [/tex])
y = [tex]-6 \frac{2}{3} [/tex]

The solution for this system of equations is ([tex]-1 \frac{1}{3} [/tex], [tex]-6 \frac{2}{3} [/tex]).

Answer:

[tex](-2,0)[/tex] and [tex](-1,-10)[/tex].

Step-by-step explanation:

We have been given a system of equations. We are asked to solve our given system of equations.

[tex]10x+y=-20...(1)[/tex]

[tex]y=2x^2-4x-16...(1)[/tex]

We will use substitution method to solve our given system. From equation (1) we will get,

[tex]y=-20-10x[/tex]

Substituting this value in equation (1) we will get,

[tex]-20-10x=2x^2-4x-16[/tex]

[tex]-20+20-10x=2x^2-4x-16+20[/tex]

[tex]-10x=2x^2-4x+4[/tex]

[tex]-10x+10x=2x^2-4x+10x+4[/tex]

[tex]0=2x^2+6x+4[/tex]

Now we will use quadratic formula to solve for x.

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]x=\frac{-6\pm \sqrt{6^2-4*2*4}}{2*2}[/tex]

[tex]x=\frac{-6\pm \sqrt{36-32}}{4}[/tex]

[tex]x=\frac{-6\pm \sqrt{4}}{4}[/tex]

[tex]x=\frac{-6}{4}\pm \frac{\sqrt{4}}{4}[/tex]

[tex]x=-1.5\pm \frac{2}{4}[/tex]

[tex]x=-1.5\pm 0.5[/tex]

[tex]x=-1.5-0.5\text{ or }-1.5+0.5[/tex]

[tex]x=-2\text{ or }-1[/tex]

Now to find y values we will substitute [tex]x=-2\text{ and }x=-1[/tex] in equation (1) as.

[tex]10(-2)+y=-20[/tex]

[tex]-20+y=-20[/tex]

[tex]-20+20+y=-20+20[/tex]

[tex]y=0[/tex]

Now, we will substitute [tex]x=-1[/tex] in equation (1) as.

[tex]10(-1)+y=-20[/tex]

[tex]-10+y=-20[/tex]

[tex]-10+10+y=-20+10[/tex]

[tex]y=-10[/tex]

Therefore, there are two solutions for our given system that are [tex](-2,0)[/tex] and [tex](-1,-10)[/tex].

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