Question

solve the equation for B:

((5x+34)/(x^2+x-12))=(7/(x-3))-(B/(x+4))

Answer 1

\[ \frac{5x+34}{x^2+x-12}=\frac{7}{x-3}-\frac{B}{x+4}\\ ~ \\\frac{B}{x-4}=\frac{7}{x-3}-\frac{5x+34}{x^2+x-12}\]

I just added B/(x+4) on both sides. And then subtracted (5x+34)(x^2+x-12) on both sides. Do you see how you can solve for B now?

Answer 2

oops That should stay as B/(x+4) on the left side of the = on the second line.. typo :(

Answer 3

All I can honestly see is to cross multiply? @kirbykirby

Answer 4

well to give rid of the divided by (x-4), multiply both sides by (x-4) :)

Answer 5

to get rid*

Answer 6

I really dont even know what multiplying by (x+4) does.. lol @kirbykirby

Answer 7

\[\frac{B}{x+4}(x+4)=\frac{B\cancel{(x+4)}}{\cancel{x+4}}=B\] (sorry I made the same mistake by writing x-4 instead of x+4, but I think you caught that in your post :P

Answer 8

Ah I saw that it seperates and puts B by itself but what does it do to (5x+34)/(x^2+x-12) and do you subtract (7/(x-3)) after multiplying (x+4) by (5x+34)/(x^2+x-12)?

Answer 9

I'm not sure if I fully understood what you meant, but on the right side you have \[ \left(\frac{7}{x-3}-\frac{5x+34}{x^2+x-12}\right)(x+4)\]
At first glance I would have just left it like that, but you could multiply in the (x+4):
\[\frac{7(x+4)}{x-3}-\frac{(5x+34)(x+4)}{x^2+x-12} \] and you could simplify this further by factoring the denominator of the second fraction.
\[\frac{7(x+4)}{x-3}-\frac{(5x+34)\cancel{(x+4)}}{\cancel{(x+4)}(x-3)}=\frac{7x+28+5x+34}{x-3}=\frac{12x+62}{x-3} \]

Answer 10

When multiplying the (x+4) to the right side why exactly do you just apply it to the numerator and not the denominator?

and also after you multiply the (x+4) times the 7 in the numerator of (7/(x-3)) how come the (5x+34) doesnt stay negative in the last step? because originally before you cancel out those (x+4)'s the (5x+34)/(x-3) is being subtracted.

@kirbykirby

Answer 11

Since the goal is to just eliminate the denominator (x+4) under B, you must multiply both sides of the equality by (x+4). Since you only multiply the left side in the numerator, you must do exactly the same operation on the right side.

Yes you are right about the negative! I completely ignored it. It should be: \[\frac{7(x+4)}{x-3}-\frac{(5x+34)\cancel{(x+4)}}{\cancel{(x+4)}(x-3)}=\frac{7x+28-5x-34}{x-3}=\frac{2x-6}{x-3}=\frac{2(x-3)}{x-3}=2\] (I should have been more suspicious that there was no cancellation of the (x-3) ;) I apologize for that. I hope I didn't confuse you

Answer 12

Thanks a lot @kirbykirby I completely understand now!

Answer 13

yay :)