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

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

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Answer 1 · 2018-06-27T17:25:25

$∣ ∣ (\frac{x}{2}) + 3 ∣ ∣ = 3 x - 3$ $|(x/2)+3|=3x-3$
$\frac{x}{2} + 3 = \pm (3 x - 3)$ $x/2 + 3=+-(3x-3)$
Therefore, $\frac{x}{2} + 3 = (3 x - 3) or (3 - 3 x)$ $x/2 + 3=(3x-3) or (3-3x)$
When, $\frac{x}{2} + 3 = 3 x - 3$ $x/2 + 3= 3x-3$
$6 = \frac{5 x}{2}$ $6=(5x)/2$
$12 = 5 x$ $12=5x$
$x = \frac{12}{5}$ $x=12/5$
^1st answer

When, $\frac{x}{2} + 3 = 3 - 3 x$ $x/2 +3=3-3x$
$\frac{7 x}{2} = 0$ $(7x)/2=0$
$7 x = 0$ $7x=0$
$x = 0$ $x=0$

Therefore, $x = 0 or x = \frac{12}{5}$ $x=0 or x= 12/5$

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

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