Calculing X...

**Miishu** · March 25th 2009, 08:54 PM

I have to solve 5 problems, but I'm stuck with these two:

25/9-3/2.(1-5/3x)=2/3x-1-(1/3x-1/9)

and

5/6.(x-12/10)+1/3x+1=35/10.(2x-3)

...I'd appreciate your help, since I have classes in a few hours, and I'm working on another project right now.

**mollymcf2009** · March 25th 2009, 09:27 PM

Originally Posted by Miishu

I have to solve 5 problems, but I'm stuck with these two:

25/9-3/2.(1-5/3x)=2/3x-1-(1/3x-1/9)

and

5/6.(x-12/10)+1/3x+1=35/10.(2x-3)

...I'd appreciate your help, since I have classes in a few hours, and I'm working on another project right now.

Ok, let me clarify what you've got here:

1) $\frac{25}{9} - \frac{3}{2} \cdot 1-(\frac{5}{3x}) = \frac{2}{3x} - 1 -(\frac{1}{3x} - \frac{1}{9})$

Wow, what a mess. Is that right for the first one??

**ADARSH** · March 25th 2009, 09:36 PM

Originally Posted by Miishu

I have to solve 5 problems, but I'm stuck with these two:

25/9-3/2.(1-5/3x)=2/3x-1-(1/3x-1/9)

and

5/6.(x-12/10)+1/3x+1=35/10.(2x-3)

...I'd appreciate your help, since I have classes in a few hours, and I'm working on another project right now.

Possibility no. 2 of Q1
$\frac{25}{9} - \frac{3}{2(1-\frac{5}{3x})} = \frac{2}{3x-1} - (\frac{1}{3x} - \frac{1}{9})$

No 3 of Q1
$\frac{25}{9} - \frac{3}{2(1-\frac{5}{3x})} = \frac{2}{3x}-1 - (\frac{1}{3x} - \frac{1}{9})$

**Miishu** · March 25th 2009, 09:47 PM

Originally Posted by mollymcf2009

Ok, let me clarify what you've got here:

1) $\frac{25}{9} - \frac{3}{2} \cdot 1-(\frac{5}{3x}) = \frac{2}{3x} - 1 -(\frac{1}{3x} - \frac{1}{9})$

Wow, what a mess. Is that right for the first one??

X goes with the whole fraction.

**ADARSH** · March 25th 2009, 10:11 PM

Originally Posted by mollymcf2009

Ok, let me clarify what you've got here:

1) $\frac{25}{9} - \frac{3}{2} \cdot 1-(\frac{5}{3x}) = \frac{2}{3x} - 1 -(\frac{1}{3x} - \frac{1}{9})$

Wow, what a mess. Is that right for the first one??

Probability no. 4

$\frac{25}{9} - \frac{3}{2} \cdot (1-\frac{5x}{3}) = \frac{2x}{3} - 1 -(\frac{x}{3} - \frac{1}{9})$

Probability no 5

$\frac{25}{9} - \frac{3}{2} \cdot 1-(\frac{5x}{3}) = \frac{2x}{3} - 1 -(\frac{x}{3} - \frac{1}{9})$

Hey! Miishu which one was correct?

**Miishu** · March 25th 2009, 10:20 PM

Originally Posted by ADARSH

Probability no. 4

$\frac{25}{9} - \frac{3}{2} \cdot (1-\frac{5x}{3}) = \frac{2x}{3} - 1 -(\frac{x}{3} - \frac{1}{9})$

Probability no 5

$\frac{25}{9} - \frac{3}{2} \cdot 1-(\frac{5x}{3}) = \frac{2x}{3} - 1 -(\frac{x}{3} - \frac{1}{9})$

Hey! Miishu which one was correct?

The 1st one.

**ADARSH** · March 25th 2009, 10:51 PM

Originally Posted by ADARSH

Probability no. 4

$\frac{25}{9} - \frac{3}{2} \cdot (1-\frac{5x}{3}) = \frac{2x}{3} - 1 -(\frac{x}{3} - \frac{1}{9})$

$\frac{25}{9} - (\frac{3}{2} -\frac{5x}{2}) = \frac{6x}{9} - \frac{9}{9} -(\frac{3x}{9} - \frac{1}{9})$

$\frac{25}{9} - (\frac{3-5x}{2} ) = \frac{6x}{9} - \frac{9}{9} -(\frac{3x-1}{9} )$

$\frac{25 \times 2}{18} - (\frac{9(3-5x)}{18} ) = \frac{6x\times 2}{18} - \frac{9\times 2}{18} -\frac{(3x-1)\times 2}{18}$

$50 - 9(3-5x) = 12x - 18 -(6x-2)$

Go ahead !

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