Question

System Of Equations~

Answer 1

@zepdrix

Answer 2

\[\large \color{orangered}{x^2-2x-35=y}\]So I guess what we can do is ummm.. substitute this in for our y in the other equation.

\[\large x=\color{orangered}{y}-5 \qquad \rightarrow \qquad x=\color{orangered}{x^2-2x-35}-5\]

And from here, we can solve for x.
Following along so far?

Answer 3

yeh :)

Answer 4

40

Answer 5

True, we can move the -40 to the other side, but we don't want to do that.

We have a quadratic involving x.
Let's get everything to one side.

Answer 6

Subtracting x from each side gives us,\[\large 0=x^2-3x-40\]

Answer 7

This quadratic will factor quite nicely, do you see the factors? :)

Answer 8

the 3x and the 40?

Answer 9

We can break this down into binomials.

We wants `factors of -40` that `add to give us -3`.

So let's try ummmm -8 and 5.

\(\large -8 \cdot 5=-40\)
\(\large -8+5=-3\)

So we've found the values we want for our binomial.

\[\large 0=x^2-3x-40\]So our quadratic will factor into,\[\large 0=(x-8)(x+5)\]

Answer 10

Ok

Answer 11

Using the `Zero Factor Property`, we can set each factor equal to zero and solve for x.

\[\large 0=(x-8)(x+5)\]
\[\large 0=x-8 \qquad \qquad \qquad 0=x+5\]
Solving for x gives us,
\[\large x=8 \qquad\qquad\qquad x=-5\]

Let's substitute these x values back into our \(\large x=y-5\) and see what values we get for y.

Answer 12

whops pluged in the wrong onw, hold up

Answer 13

y=13

Answer 14

Which x value gave you that?

Answer 15

8=y-5

Answer 16

The x=8 gave you y=13 ?

Ok good, we can write that as a coordinate pair.

(8,13).

If we were to plug x=-5 into the equation, we could get another solution.
But since that other solution isn't listed as one of our options, let's not worry about it.

Answer 17

The answer shall be D

Answer 18

Yah, sounds right. Good job!

Answer 19

tnx u too :D