Question

Solve the system of equations. 10x+y=−20 y=2x2−4x−16 ( , ) and ( , )

Answer 1

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 = $-1 \frac{1}{3}$

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

10($-1 \frac{1}{3}$) + y = -20
$-13 \frac{1}{3}$ + y = -20 (subtract $-13 \frac{1}{3}$)
y = $-6 \frac{2}{3}$

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

Answer 2

Answer:

$(-2,0)$ and $(-1,-10)$.

Step-by-step explanation:

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

$10x+y=-20...(1)$

$y=2x^2-4x-16...(1)$

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

$y=-20-10x$

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

$-20-10x=2x^2-4x-16$

$-20+20-10x=2x^2-4x-16+20$

$-10x=2x^2-4x+4$

$-10x+10x=2x^2-4x+10x+4$

$0=2x^2+6x+4$

Now we will use quadratic formula to solve for x.

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

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

$x=\frac{-6\pm \sqrt{36-32}}{4}$

$x=\frac{-6\pm \sqrt{4}}{4}$

$x=\frac{-6}{4}\pm \frac{\sqrt{4}}{4}$

$x=-1.5\pm \frac{2}{4}$

$x=-1.5\pm 0.5$

$x=-1.5-0.5\text{ or }-1.5+0.5$

$x=-2\text{ or }-1$

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

$10(-2)+y=-20$

$-20+y=-20$

$-20+20+y=-20+20$

$y=0$

Now, we will substitute $x=-1$ in equation (1) as.

$10(-1)+y=-20$

$-10+y=-20$

$-10+10+y=-20+10$

$y=-10$

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