Question

Help w/ Quadratic Equations 1?

Help w/ Quadratic Equations 1?

Hi Can you please solve the following, please be visual and show your work!
Just #18

Answer 1

The fraction is `=2/5` or `=3/2`

Explanation:

Let the denominator be `=x`

Then,

the first fraction is `y=(x-3)/x`

The second fraction is `2y=(x-3+6)/(x+5)=(x+3)/(x+5)`

Therefore,

`(x+3)/(x+5)=(2(x-3))/x`

`x(x+3)=2(x-3)(x+5)`

`x^2+3x=2(x^2+2x-15)`

`x^2+3x=2x^2+4x-30`

`x^2+x-30=0`

`(x-5)(x+6)=0`

So,

`x=5` or `x=-6`

The original fraction is

`y=(x-3)/x=(5-3)/5=2/5`

or `y=(x-3)/x=(-6-3)/(-6)=9/6=3/2`

Answer 2

`2/5`

Explanation:

Let the original fraction be `p/q`

We are told that:

`p=q-3` [A]

`(p+6)/(q+5)=(2p)/q` [B]

[A] in [B] `-> (q-3+6)/(q-5)=(2(q-3))/q`

`(q+3)/(q-5)=(2q-6)/q`

`q^2+3q=2q^2-6q+10q=30`

`q^2+q-30=0`

`(q+6)(q-5)=0`

`q=-6` or `q=5`

Since `p=q-3 ->p=-9` or `p=2`

However, we can reject the solution `(-9)/-6` as this reduces to `3/2` and does not satisfy [A] above.

Hence `p=2, q=5` is the only solution.

The original fraction was `2/5`