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

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

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Answer 1 · 2016-02-02T08:54:05

$(\frac{x}{2} \cdot \frac{3}{3}) + (\frac{x}{3} \cdot \frac{2}{2}) - \frac{x}{6} = 4 \to \frac{4 x}{6} = 4$ $(x/2*3/3) + (x/3*2/2) - x/6 = 4 rarr (4x)/6 = 4$ , therefore, $x = 6$ $x = 6$

Explanation:

To add and subtract fractions they must have a common denominator. The common denominator for $2, 3$ $2, 3$ and $6$ $6$ is $6$ $6$ , therefore $\frac{x}{2}$ $x/2$ is multiplied by $\frac{3}{3}$ $3/3$ (any number over itself = $1$ $1$ , so you are not changing the value of the fraction by multiplying it by $1$ $1$ ) and $\frac{x}{3}$ $x/3$ is multiplied by $\frac{2}{2}$ $2/2$ , resulting in $\frac{3 x}{6} + \frac{2 x}{6}$ $(3x)/6 + (2x)/6$ , which equals $\frac{5 x}{6}$ $(5x)/6$ . Now we move all of the fractions to one side of the equation and the whole numbers to the other:

$\frac{5 x}{6} - \frac{x}{6} - 1 + 1 = \frac{x}{6} - \frac{x}{6} + 1 + 3$ $(5x)/6 - x/6 - 1 + 1 = x/6 - x/6 + 1 + 3$

$= \frac{4 x}{6} = 4$ $= (4x)/6 = 4$

$6 \cdot 4 = 4 x$ $6 * 4 = 4x$ , therefore, $x = 6$ $x = 6$

Answer 2 · 2016-02-02T12:52:07

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

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

$\to \frac{5 x}{6} - 1 = \frac{x}{6} + 3$ $rarr(5x)/6-1=x/6+3$

$\to \frac{5 x}{6} = \frac{x}{6} + 3 + 1$ $rarr(5x)/6=x/6+3+1$

$\to \frac{5 x}{6} = \frac{x}{6} + 4$ $rarr(5x)/6=x/6+4$

$\to \frac{5 x}{6} - \frac{x}{6} = 4$ $rarr(5x)/6-x/6=4$

$\to \frac{4 x}{6} = 4$ $rarr(4x)/6=4$

$\to \frac{2 x}{3} = 4$ $rarr(2x)/3=4$

$\to 2 x = 4 \cdot 3$ $rarr2x=4*3$

$\to 2 x = 12$ $rarr2x=12$

$\to x = \frac{12}{2} = 6$ $rarrx=12/2=6$

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