Thread: part of a fractional equation w/ binomial denominator

1. part of a fractional equation w/ binomial denominator

$8(5-y)(6/5-y)$

I don't get how to solve this part.. the result should be 48.. would anyone explain, thanks

2. Is the question $8(5-y)(\frac{6}{5}-y)=0$ or something?

(clearly it isn't this since y=48 is not a solution).

3. no, the whole equation is

3/8=6/5-y

LCD=8(5-y)

(3/8)8(5-y)=8(5-y)(6/5-y)

3(5-y)=48

y=-11

to clarify, the fraction is 6/5-y, where 5-y is the binomial denominator and it's not as this shows it: $6/5-y$
I dont get how to show it in the proper way

4. Sorry but I can't really see the problem.

You've just worked it out yourself to get y=-11, isn't that the question done?

Anyway, if you have $
8(5-y)(6/5-y)
$
it must be equal to something in order for it to have solutions.

Are you trying to solve:

$\frac{3}{8}=\frac{6}{5}-y$?

You can solve this by:

$y=\frac{6}{5}-\frac{3}{8}$

$y=\frac{48-15}{40}=\frac{33}{40}$

5. Originally Posted by strunz
no, the whole equation is

3/8=6/5-y

LCD=8(5-y)
From your last line above, I will guess that the equation is actually "3/8 = 6/(5 - y)", rather than "3/8 = 6/5 - y".

A good start for solving this proportion is to "cross-multiply" to get:

. . . . .3(5 - y) = 6(8)

Solve the resulting linear equation by first multiplying things out:

. . . . .15 - 3y = 48

Rearrange to get:

. . . . .15 - 48 = 3y

. . . . .-33 = 3y

Then divide through. To confirm your solution, plug it back in to the original equation:

. . . . .left-hand side: 3/8

. . . . .right-hand side: 6/(5 - [-11]) = 6/(5 + 11) = 6/16 = 3/8

And you're done!

6. Yeah that's perfect, thanks.