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Solving:$$ \frac{3}{7}-\frac{x-1}{2}-(2-x)=\frac{3x-4}{14} $$

Could someone walk me through this, or what I'm doing wrong? I do the parenthesis, and get: $$ \frac{3}{7}-\frac{x-1}{2}-2+x=\frac{3x-4}{14} $$ Then I multiply everything by $14$, divide the fractions and have $6-7x-28+14x=3x-4$. Add up this and $7x-29=3x-4$. I end up with $4x=25$. Which doesn't make sense, since $x$ should equal $11/4$.

I've been going through the steps myself so long I think I'm going blind for my own mistake.

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3 Answers 3

You replace $x-1$ with $x$ in your calculation and did one of the signs wrong. The correct calculation should be

$$\frac{3}{7} - \frac{x-1}{2} - (2-x) = \frac{3x - 4}{14}$$

Multiplying through by $14$

$$6 - 7x + 7 + -28 + 14x = 3x - 4$$

Rearranging

$$-7x + 14x -3x = -4 -7 - 6 + 28$$

Summing up

$$4x = 11$$

$$x = \frac{11}{4}$$

as desired.

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You have

$$\frac{3}{7}-\frac{x-1}{2}-2+x=\frac{3x-4}{14}$$

Now multiply, as you said, by 14, and you get:

$$6-7(x-1)-28+14x=3x-4$$

Now, watch out for the signs on the left hand side:

$$6-7x+7-28+14x=3x-4$$

and this adds up to

$$-15+7x=3x-4$$

Now adding $15$ to both sides and subtracting $3x$ you get

$$4x=11$$

and that's exactly what you wanted.

Your error was basically forgetting one term and doing one sign wrong.

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Ah, thanks a lot! So when I multiply a fraction like the one above, I put x - 1 in parenthesis, and 14 in front before I divide? Then it all makes sense. –  John May 30 '12 at 12:29
    
That's it. The same way as you do with a $-$ in front of a fraction. Apply everything to the whole numerator. –  Gregor Bruns May 30 '12 at 12:44

When you multiply $$\frac{3}{7}-\frac{x-1}{2}-2+x=\frac{3x-4}{14}$$ by $14$ you should get $$6-7x+7-28+14x=3x-4$$ which becomes $$4x = 11.$$

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