Question

What is the solution to the rational equation

Answer 1

And the equation is?

Answer 2

\(\frac{x}{x^2 - 9} + \frac{1}{x+3} = \frac{1}{4x-12}\)

x^2 - 9 = x^2 - 3^2 = (x+3)(x-3) (Perfect Square identity)
4x - 12 = 4(x - 3)

So now we have \(\frac{x}{(x+3)(x-3)} + \frac{1}{x+3} - \frac{1}{3(x-3)}= 0\). The LCM of the denominators is 3(x+3)(x-3). The first term is missing 3, so we multiply 3 to become 3x. The second term is missing 3(x-3), so it becomes 3(x-3). The third term is missing (x+3), so it becomes (x+3). Now the equation becomes \(\frac{3x + 3(x-3) - (x+3)}{3(x+3)(x-3)} = 0\). Simplify it and you get \(\frac{5x-12}{27-3x^2} = 0\). Multiply both sides by 27-3x^2 to get rid of the denominator, and you get 5x-12. This should now be solvable.

Answer 3

what do i do?

Answer 4

@andreadesirepen Do you understand the explanation?

Answer 5

sort of

Answer 6

he doesn't

Answer 7

choices are: -9/7, 15, 15/7 and 9

Answer 8

It looks like I have some calculation error in between in my example... my apologies. The idea is to find the LCM of the denominators, making 1 big term with a common denominator and the other side zero. Multiply both sides by the denominator and continue solving.

Answer 9

what denominator?

Answer 10

@andreadesirepen The denominators... A denominator is the lower part of a fraction.

Answer 11

(In contrast, the numerator is the upper part of it)

Answer 12

yes, but which one of the fraction?

Answer 13

@andreadesirepen All of them. You're trying to find the LCM of these denominators and make this LCM the new denominator of all of them.

Answer 14

This is diffucult

Answer 15

@andreadesirepen Where does this question appear on?

Answer 16

a hw sheet

Answer 17

@andreadesirepen Of which level?

Answer 18

10th grade

Answer 19

@andreadesirepen You might want to review your material again... I suppose this would be on the harder end of the worksheet

Answer 20

would it be -9/7 though?

Answer 21

No.

Answer 22

then def. 15/7

Answer 23

Correct. If it's just multiple-choice, then you just need to plug in the numbers and compute.