Question

algebra 2 help please!!!

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3.


Solve for x: 2 over quantity x plus 2 plus 1 over 5 equals 6 over quantity x plus 5.



Classify the solutions of 1 over x plus 4, plus one half, equals 1 over x plus 4 as extraneous or non-extraneous.



Solve for x: 3 over 3 x plus 1 over quantity x plus 4 equals 10 over 7x .

Answer 1

@Abhisar

Answer 2

@zepdrix

Answer 3

@Luigi0210

Answer 4

@Leader

Answer 5

\[\Large\rm \frac{4}{x}+\frac{4}{x^2-9}=\frac{3}{x-3}\]This is our problem to solve?

Answer 6

i should have updated that one i got that answer

Answer 7

@zepdrix

Answer 8

Solve for x: 2 over quantity x plus 2 plus 1 over 5 equals 6 over quantity x plus 5.
\[\Large\rm \frac{2}{(x+2)}+\frac{1}{5}=\frac{6}{(x+5)}\]Mmmm ok something like this?

Answer 9

We would like to multiply both sides by the `Least Common Multiple` of the denominators.
It's pretty much the same process as finding a common denominator, but we're multiplying through by that denominator so it cleans things up.

Answer 10

ok and yea thats the right
one

Answer 11

Our denominators have nothing in common, so our LCM is going to be \(\Large\rm (x+2)\cdot5\cdot(x+5)\)

Answer 12

ok

Answer 13

\[\rm 5(x+2)(x+5)\left(\frac{2}{(x+2)}+\frac{1}{5}\right)=\left(\frac{6}{(x+5)}\right)5(x+2)(x+5)\]We'll multiply both sides by these denominators. It will get rid of all of the fractions for us.

Answer 14

So what do we get for our first term?
\[\Large\rm 5(x+2)(x+5)\cdot \frac{2}{x+2}=?\]
Can you figure it out? :)

Answer 15

im looking for the LCD? cause i got x+2

Answer 16

@zepdrix

Answer 17

Ah sorry I ran off :c busy morning.

Least Common Multiple?
The first fraction has (x+2), the second one has a 5, the last one has an (x+5).
So our LCM is the product of all three of those, since they have nothing in common.

Answer 18

\[\rm 5(x+2)(x+5)\left(\frac{2}{(x+2)}+\frac{1}{5}\right)=\left(\frac{6}{(x+5)}\right)5(x+2)(x+5)\]So I'm multiplying both sides by the LCM.
Distributing it to each fraction gives us:\[\rm \frac{2\cdot5(x+2)(x+5)}{(x+2)}+\frac{5(x+2)(x+5)}{5}=\frac{6\cdot5(x+2)(x+5)}{(x+5)}\]

Answer 19

From there, we have some nice cancellations\[\rm \frac{2\cdot5\cancel{(x+2)}(x+5)}{\cancel{(x+2)}}+\frac{\cancel{5}(x+2)(x+5)}{\cancel{5}}=\frac{6\cdot5(x+2)\cancel{(x+5)}}{\cancel{(x+5)}}\]

Answer 20

Which leaves you with:\[\Large\rm 10(x+5)+(x+2)(x+5)=30(x+2)\]No more fractions, yay!

Answer 21

its all good and what do i do from here?

Answer 22

Multiply out the brackets,
get everything on the same side,
combine like-terms.

Answer 23

Boy these are longs problems :O
flvs so mean hehe

Answer 24

hence my name lol

Answer 25

im not good at multiplying the like terms especially when there are so many, my dyslexia kicks in can you help out?

Answer 26

Distribute the 10, what do you get for the first set of terms?

Answer 27

like distribute to the 30?

Answer 28

or among the brackets?

Answer 29

\[\Large\rm \color{orangered}{10(x+5)}+(x+2)(x+5)=30(x+2)\]Just this orange part to start.
What two terms do you get out of that when you give the 10 to the x and 5?

Answer 30

oh, i just did the whole thing and got 0 and 13

Answer 31

0 and 13?
Yayyyy good job \c:/

Answer 32

SWEET!!!!!

Answer 33

can you help me with the next one?

Answer 34

Sorry gotta get ready for work :c