$\displaystyle 8(5-y)(6/5-y)$
I don't get how to solve this part.. the result should be 48.. would anyone explain, thanks
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: $\displaystyle 6/5-y$
I dont get how to show it in the proper way
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 $\displaystyle
8(5-y)(6/5-y)
$ it must be equal to something in order for it to have solutions.
Are you trying to solve:
$\displaystyle \frac{3}{8}=\frac{6}{5}-y$?
You can solve this by:
$\displaystyle y=\frac{6}{5}-\frac{3}{8}$
$\displaystyle y=\frac{48-15}{40}=\frac{33}{40}$
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!