Question

24. 2c/c-4 - 2/1=4/c+5
I know the answer is supposed to be c=-14 but I'm not really sure what the LCD is

Answer 1

\[\frac{ 2c }{ c-4 }-2=\frac{ 4 }{ c+5 }\]

Answer 2

LCD = least common denominator = (c-4)*(c+5)

Answer 3

I don't understand how that is the LCD, can you please explain?

Answer 4

@Vocaloid are you still online? I really need help, sorry

Answer 5

the denominator of the first term is (c-4), correct?
the denominator of 2 is 1
the denominator of the third term is (c+5)

multiply these together to find the LCD

(c-4)*(c+5)*1

Answer 6

Okay, so then I have to multiply the other terms by that?
Like:
\[(c-4*c+5) \frac{ 2c }{c-4 }-(c-4*c+5)2=(c-4*c+5)\frac{ 4 }{ c+5 }\]
Is this correct?

Answer 7

you need to multiply each term by the quantity that will make the denominator equal to (c-4)*(c+5)*1

for example, the first term is 2c/(c-4). since we already have (c-4) in the denominator, we have to multiply by (c+5)/(c+5) to make the denominator equal to (c+5)*(c-4)

Answer 8

since all the denominators are the same now, we can cancel out the denominators and look at the numerators only now

(c+5)*2c - 2(c-4)(c+5) = 4*(c-4)
and now we can solve for c

Answer 9

Oh, okay, I see what you're saying sort of
So multiplied that would all be \[2c^{2}+10c-2c+8-2c-10c=4c-16\]
so i combine like terms and then I just solve for x?

Answer 10

yes

Answer 11

Double check your work. There is an error on the left of the equal sign. The c^2 term should of disappeared, plus there may be other errors.

Answer 12

@Vocaloid thank you so much for helping!!
@radar was I supposed to cancel out the top c in 2c/c-4 along with cancelling the c-4?

Answer 13

After the combining in the numerator your fractions should appear as: