Question

Please help me solve this and check for extraneous solutions. 10 over x+4=15 over 4x+4

Answer 1

@oaktree

Answer 2

Right here.

Answer 3

Is it this:\[\frac{ 10 }{ x+4 } = \frac{ 15 }{ 4x+4 }\]

Answer 4

Yay, yes

Answer 5

@oaktree ?

Answer 6

"cross multiply" and get
\[\frac{ 10 }{ x+4 } = \frac{ 15 }{ 4x+4 }\iff 10(4x+4)=15(x+4)\]

Answer 7

you good from there?

Answer 8

Okay 40x+40=15x+60

Answer 9

yeah three more steps

Answer 10

Yes I believe so. Thank you but what would I do with a problem like....x over x-2 plus 1over x-4=2 over x squared-6x+8

Answer 11

It has 3 instead of two @satellite73

Answer 12

i would try to write it so i can understand it
is it
\[\frac{x}{x-2}+\frac{1}{x-4}=\frac{2}{x^2-6x+8}\]

Answer 13

\[\frac{x}{x-2}+\frac{1}{x-4}=\frac{3}{x^2-6x+8}\]

Answer 14

that right?

Answer 15

It is the first one

Answer 16

add on the left and get
\[\frac{x}{x-2}+\frac{1}{x-4}=\frac{2}{x^2-6x+8}\]
\[\frac{x(x-4)+(x-2)}{(x-2)(x-4)}=\frac{2}{(x-2)(x-4)}\]

Answer 17

cooked up so that the denominator on the left is the same as the denominator on the right after you factor

Answer 18

since the denominators are the same, you can equate
\[x(x-4)+x-2=2\]and solve the quadratic

Answer 19

ooooh you have to be real real careful here

Answer 20

when you solve the quadratic, one solution will be \(x=4\) but that is not a solution to the original equation, because the original expressions are undefined if \(x=4\)
so omit that solution

Answer 21

What does that mean?

Answer 22

(Omit)

Answer 23

what does what mean?

Answer 24

Yes

Answer 25

oh, it means don't put that as a solution, because it is not one

Answer 26

The answer that I got is (x+1)(x+2)

Answer 27

Oh, thankyou

Answer 28

Is that wrong?

Answer 29

\[x(x-4)+x-2=2\]
\[x^2-4x+x-2=2\]
\[x^2-3x-4=0\]
\[(x-4)(x+1)=0\]

Answer 30

solutions to this are \(x=4\) or \(x=-1\) but as i said, \(4\) is not a solution to the original equation as it would make the denominator zero

Answer 31

So then would I put both down answers?

Answer 32

no
you put down \(x=-1\) only

Answer 33

Can you help me with one more that is different from that one?

Answer 34

@satellite73