How do you solve $\frac{x - 1}{3} - \frac{x - 1}{2} = 6$ $(x-1) /3 - (x-1)/2 = 6$ ?

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How do you solve $\frac{x - 1}{3} - \frac{x - 1}{2} = 6$ $(x-1) /3 - (x-1)/2 = 6$ ?

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Answer 1 · 2016-10-16T22:55:26

$x = - 35$ $x = -35$

Explanation:

Start by making sure that you're working with fractions that have equal denominators.

You start with

$\frac{x - 1}{3} - \frac{x - 1}{2} = \frac{6}{1}$ $(x-1)/3 - (x-1)/2 = 6/1$

The common denominator here is $6$ $6$ , so multiply the first fraction by $1 = \frac{2}{2}$ $1 = 2/2$ , the second fraction by $1 = \frac{3}{3}$ $1 = 3/3$ , and the third fraction by $1 = \frac{6}{6}$ $1 = 6/6$ .

This will get you

$\frac{x - 1}{3} \cdot \frac{2}{2} - \frac{x - 1}{2} \cdot \frac{3}{3} = \frac{6}{1} \cdot \frac{6}{6}$ $(x-1)/3 * 2/2 - (x-1)/2 * 3/3 = 6/1 * 6/6$

$\frac{2 (x - 1)}{6} - \frac{3 (x - 1)}{6} = \frac{36}{6}$ $(2(x-1))/6 - (3(x-1))/6 = 36/6$

At this point, drop the denominators and focus exclusively on the numerators.

$2 (x - 1) - 3 (x - 1) = 36$ $2(x-1) - 3(x-1) = 36$

This will get you

$- (x - 1) = 36$ $-(x-1) = 36$

$- x + 1 = 36$ $-x + 1 = 36$

$x = - 35$ $x = - 35$

Do a quick check to make sure that the calculations are correct

$\frac{- 35 - 1}{3} - \frac{- 35 - 1}{2} = 6$ $(-35 - 1)/3 - (-35 - 1)/2 = 6$

$- \frac{36}{3} + \frac{36}{2} = 6$ $-36/3 + 36/2 = 6$

$- 12 + 18 = 6 \sqrt{}$ $-12 + 18 = 6 " "color(green)(sqrt())$

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How do you solve $\frac{x - 1}{3} - \frac{x - 1}{2} = 6$ $(x-1) /3 - (x-1)/2 = 6$ ?

1 Answer

Explanation:

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How do you solve $\frac{x - 1}{3} - \frac{x - 1}{2} = 6$ $(x-1) /3 - (x-1)/2 = 6$ ?