Question

Fractional Equation help? Problem in the comments below.

Answer 1

\[\frac{ 2 }{ a+3 }-\frac{ a^2+3a-10 }{ a^3+5a^2+6a }=\frac{ a-4 }{ a^2+3a }\]

Answer 2

First find the LCD. on the bottom you have \[(a+3)\]\[a(a+3)(a+2)\]\[a(a+3)\]

Answer 3

Ok and then I multiply each fraction by \[\frac{ a(a+3)(a+2) }{ a(a+3)(a+2) }\] Is that right?

Answer 4

Well, let's define our LCD, you agree that it would be \(a(a+3)(a+2)\) correct?

Answer 5

Yes

Answer 6

Then in our first fraction:\[\frac{2}{a+3}\] we look for what part of our LCD is missing in the denominator of this fraction, and multiply both top and bottom by it.

Answer 7

Oh ok so we multiply the first fraction by \[\frac{ a(a+2) }{ a(a+2) }\]

Answer 8

yes :)

Answer 9

\[\frac{ 2a(a+2) }{ a(a+3)(a+2) }\]

Answer 10

You got it.

Answer 11

Now the last fraction: \[\frac{a-4}{a(a+3)}\]

Answer 12

\[\frac{ (a-4)(a+2) }{ a(a+3)(a+2) }\]

Answer 13

Yay! That is correct.

Answer 14

So now we can write our function as: \[\frac{2a(a+2)}{a(a+2)(a+3)} -\frac{a^2+3a-10}{a(a+2)(a+3)}=\frac{(a-4)(a+2)}{a(a+2)(a+3)}\]

Answer 15

Now we can multiply both sides of our equation by \(a(a+2)(a+3)\) to eliminate our denominator, so we can simply solve for our numerator.\[a(a+2)(a+3)\cdot \left[\frac{2a(a+2)}{a(a+2)(a+3)} -\frac{a^2+3a-10}{a(a+2)(a+3)}=\frac{(a-4)(a+2)}{a(a+2)(a+3)}\right]\]

Answer 16

This leaves us with \[2a(a+2) -(a^2+3a-10) = (a-4)(a+2)\]

Answer 17

ok it showed up a little weird but I can see what you mean

Answer 18

What do you mean?

Answer 19

As in question marks?

Answer 20

it looks like part of the equation got cut off but only a little tiny bit

Answer 21

Oh I see.

Answer 22

Do you understand my steps so far though? :)

Answer 23

Yes I do.

Answer 24

Alright! So let's evaluate \[2a(a+2) -(a^2+3a-10) = (a-4)(a+2)\]

Answer 25

Distribute the \(2a\) into \((a+2)\). What do you get?

Answer 26

\[2a^2+4a\]

Answer 27

Good. Now the next part. Distribute the negative into \((a^2+3a-10)\). What do you get?

Answer 28

\[-a^2-3a+10\]

Answer 29

Awesome :) Let's combine both of the functions on the left hand side together now: \[2a^2+4a-a^2-3a+10\]

Answer 30

Rearrange so it's easier to evaluate: \((2a^2-a^2) +(4a-3a) +10\)

Answer 31

OK so you get a^2+a+10
and then distribute a-4 into a+2 and get a^2-2a-8
so you end up with a^2+a+10=a^2-2a-8

Answer 32

yeah. \[a^2 +a+10 = a^2-2a-8\] Good job so far.

Answer 33

the a^2 on each side cancels out so you are left with \[a+10=-2a-8\] so then if you get a by itself you have\[3a=-18\] so a=-6

Answer 34

Awesome! You are good at this :D

Answer 35

You got it.

Answer 36

Thanks for the help.

Answer 37

No problem :)