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The equation in question is:

$$\frac{(2x+1)}{3} - \frac {5x-2}{7} = x- 1$$

My solution is:

$\frac {7(2x+1)}{3\cdot7} - \frac {3(5x-2)}{3\cdot7} = \frac {3\cdot7(x-1)}{3\cdot7}$

$\frac {(14x+7)}{21} - \frac {(15x-6)}{21} = \frac {21x-21}{21}$

$\frac {29x+1}{21}=\frac{21x-21}{21}$

$\frac{(29x+1)}{21} - \frac {(21x-21)}{21} = 0$

$\frac {(8x+22)}{21}$

$\frac {2(4x+11)}{21}$

$4x+11 = 0$

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

But it seems that I'm wrong, as Maxima gave $\frac {17}{11}$ as the answer. What's the problem with my solution? What point I've missed?

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  • $\begingroup$ You have a sign error passing from line 2 to line 3. You have 14x - 15x = -x, not 29x, and similarly 7 - (-6) = 13, not 1. $\endgroup$
    – Alon Amit
    Jul 14, 2015 at 4:44

1 Answer 1

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In the second step, [(14x+7)-(15x+6)]= (-x+13) ,not, (29x+1). Except that, process is perfect. Rectify this to get the desired answer.

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