Help with 4 part question =)

Question

zarkam21 · Answer

attachment

zarkam21 · Answer

attachment

563blackghost · Answer

Still need help with this?

zarkam21 · Answer

Yes

563blackghost · Answer

If so, we would first need to make `4x-8y=0` into an equation of x.

First we would need to get `-8y` to the other side, so we would do the opposite process done in the equation. So we would `add 8y` to both sides...

$\LARGE\bf{4x-8y+8y=8y+0 \rightarrow 4x = 8y+0}$

Now we need to isolate x so we would divide by 4 from each side...

$\huge\bf{\frac{4x}{4} = \frac{8y}{4}}$

What does that equal?

zarkam21 · Answer

That equals 2

563blackghost · Answer

Well 2 is partially correct, when doing expressions you must include the variable so it would be...

$\huge\bf{x=2y}$

So that is for Step 1.

zarkam21 · Answer

Oh okay, so I would include the Y?

563blackghost · Answer

Yes.

zarkam21 · Answer

Okay , thanks now for the next part

zarkam21 · Answer

I would just write 2y in the equation?

563blackghost · Answer

We would input the expression in place of x in the second equation..

$\huge\bf{2y+y^{2}=25}$

Yes :)

zarkam21 · Answer

Okay and then it says to solve for y

zarkam21 · Answer

Ya there?

563blackghost · Answer

Yes sorry the next part is a bit confusing for me one second :)

zarkam21 · Answer

no, it's alright, take your time =)

563blackghost · Answer

The next process is tricky so i'll type it out :)

zarkam21 · Answer

Okay =)

563blackghost · Answer

Aww man I forgot to include the power of 2 for x, so sorry let me write that again.

zarkam21 · Answer

Okay =)

563blackghost · Answer

$\huge\bf{(2y)^{2}+y^{2}=25}$

We simplify the parenthesis.

$\huge\bf{4y+y^{2}=25}$

Combine.

$\huge\bf{5y^{2}=25}$

Divide by 5.

$\huge\bf{y^{2}=5}$

Find the square root.

$\huge\bf{y= \pm \sqrt{5}}$

So the answer for y is $\large\bf{-\sqrt{5}, \sqrt{5}}$.

563blackghost · Answer

Thats a bit simpler. I was wondering why it was so complicated ^.^

zarkam21 · Answer

is that for step 2 or 3

563blackghost · Answer

So we then input into the first equation.

$\huge\bf{4x-8(\sqrt{5})=0}$

$\huge\bf{4x-8(-\sqrt{5})=0}$
~~~~~~~~~~~~~~

So we would simplify the parathesis.

$\huge\bf{4x-8\sqrt{5}=0}$

Add `-8 sqrt5`.

$\huge\bf{4x = 8\sqrt{5}}$

Divide by 4.

$\huge\bf{x= \pm 2\sqrt{5}}$

So our solutions is....

$\huge\bf{(-2\sqrt{5},-\sqrt{5})}$

and

$\huge\bf{(2\sqrt{5},\sqrt{5})}$

563blackghost · Answer

The solution to finding y is for step 2.

zarkam21 · Answer

and this second thing that you typed out is for step 3, just to make it clear?

563blackghost · Answer

Yup :) And the solutions are for step 4.

zarkam21 · Answer

Okay thank you so much. I really appreciate guidance like this where I can work and excel in my experience as a student. Thank you again =)

563blackghost · Answer

Your welcome :)