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

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

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Answer 1 · 2018-05-23T03:11:15

$x = - 11$ $color(orange)(x = -11$

Explanation:

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

Order of operations - BEDMAS

$(\frac{1}{2}) \cdot x - (\frac{1}{2}) \cdot 3 = (\frac{1}{3}) \cdot 2 x + (\frac{1}{3}) \cdot 1$ $(1/2)*x - (1/2) *3 = (1/3)*2x +(1/3)*1$ , removing braces.

$\frac{x}{2} - \frac{3}{2} = (\frac{2}{3}) x + \frac{1}{3}$ $x/2 - 3/2 = (2/3)x + 1/3$ , multiplying.

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

$\frac{4 x - 3 x}{6} = \frac{- 9 - 2}{6}$ $(4x-3x)/6 = (-9-2)/6$ , taking L C M.

$x/cancel6 = -11/cancel6$ , simplifying.

$color(orange)(x = -11$

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

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Explanation:

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