1. Solving an equation.

$6/5+3/(w-3)=9/[5(w-3)]$

LCD= 5(w-3)

$(6/5)(5)(w-3)+5(w-3)[3/(w-3)]=9/[5(w-3)](5)(w-3)$

$6w-18+5w-15(3/(w-3)=9/[5(w-3)](5)(w-3)$

5w-15(3/(w-3)=9/[5(w-3)](5)(w-3) that's the part I'm having trouble with..

2. Originally Posted by strunz

$6/5+3/(w-3)=9/[5(w-3)]$
You appear to have:

. . . . . $\frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}$

Multiplying through by 5(w - 3) gives:

. . . . . $6(w\, -\, 3)\, +\, 3(5)\, =\, 9$

I'm not sure how you got your result...?

3. $
\frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}
$

$6(w-3)+3(5)=9$? (multiplying through by $5(w-3)$).

Or is my mental arithmetic terrible?

$6w-18+15=9$

$6w-3=9$

$6w=12$

$w=2$.

4. Originally Posted by Showcase_22
$w=2$.
Check your solution by plugging it back in to the original equation:

. . . . . $\mbox{left-hand side: }\, \frac{6}{5}\, +\, \frac{3}{(2)\, -\, 3}\, =\, \frac{6}{5}\, +\, \frac{3}{-1}\, =\, \frac{6}{5}\, -\, \frac{15}{5}\, =\, -\frac{9}{5}$

. . . . . $\mbox{right-hand side: }\, \frac{9}{5\left[(2)\, -\, 3\right]}\, =\, \frac{9}{5(-1)}\, =\, -\frac{9}{5}$

Looks good to me!

5. Fractional equation w/ binomial denominator

$
\frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}
$

Hi could someone write down in steps how to actually solve this.. thanks

6. Originally Posted by strunz
$
\frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}
$

Hi could someone write down in steps how to actually solve this.. I already have the answer but can't figure out how the steps precede.. thanks
get the same denominator of $5(w-3)$

$\frac{6}{5} \cdot \frac{w-3}{w-3} + \frac{3}{w-3} \cdot \frac{5}{5} - \frac{9}{5(w-3)} = 0$

now you can collect all the terms

$6w-18 + 15 + 9 = 0$

7. $

\frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}
$

My book says the second step precedes like this

(6/5)(5)(w-3)(3/(w-3)=9/[5(w-3)](5)(w-3)

6(w-3)+15=9
w=2

Maybe I'm not at the level you think I am, because my problem is really how to multiplicate those parentheses.. the bold written line.. I don't know the way it's done.

8. Ok, first write that properly:

First line:
$

\frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}
$

Second line:
$

\frac{6 \times 5(w-3)}{5} + \frac{3\times 5(w-3)}{w-3} = \frac{9}{5(w-3)} \times 5(w-3)

$

What has been done between lines 1 & 2:

To get 2, you simply multiply both sides by $5(w-3)$. This is one way of solving an equation with fractions.

From there, you need to simplify the expression. Cross out terms... and get rid of the fractions:

$

6 \times (w-3) + 3 \times 5 = 9

$

That done, multiply the terms and finish to solve!

9. Originally Posted by strunz
$

\frac{6}{5}\, +\, \frac{3}{w\, -\, 3}\, =\, \frac{9}{5(w\, -\, 3)}
$

My book says the second step precedes like this

(6/5)(5)(w-3)(3/(w-3)=9/[5(w-3)](5)(w-3)

6(w-3)+15=9
w=2

Maybe I'm not at the level you think I am, because my problem is really how to multiplicate those parentheses.. the bold written line.. I don't know the way it's done.
You have the same answer I do, I just used the lowest common denominator which is 5(w-3)

10. The simplest way to do this is not to combine fractions but to get rid of the fractions by multiplying on both sides of the equation by the common denominator 5(w-3):

$5(w-3)\left[\frac{6}{5}+ \frac{3}{w-3}\right]= 5(w-3)\left[\frac{9}{5(w- 3)}\right]$
$6(w-3)+ 15= 9$
immediately.

11. Originally Posted by HallsofIvy
The simplest way to do this is not to combine fractions but to get rid of the fractions by multiplying on both sides of the equation by the common denominator 5(w-3):

$5(w-3)\left[\frac{6}{5}+ \frac{3}{w-3}\right]= 5(w-3)\left[\frac{9}{5(w- 3)}\right]$
$6(w-3)+ 15= 9$
immediately.
That's what the poster was shown earlier in the thread. I'm guessing maybe the poster isn't allowed to skip any steps at all...?

12. I wouldn't consider that "skipping" any steps at all.