How do you solve for x given $∣ ∣ ∣ \frac{2 x}{7} + 9 ∣ ∣ ∣ > \frac{5}{7}$ $|\frac{2x}{7}+9 | > frac{5}{7}$ ?

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Answer 1 · 2018-04-04T14:55:25

Either $x > - 29$ $x> -29$ or $x < - 34$ $x< -34$

$∣ ∣ ∣ \frac{2 x}{7} + 9 ∣ ∣ ∣ > \frac{5}{7}$ $|(2x)/7+9|>5/7$

may mean :

either
$\frac{2 x}{7} + 9 > \frac{5}{7} \Rightarrow 2 x + 63 > 5 \Rightarrow 2 x > 5 - 63 = - 58 \Rightarrow x > - 29$ $(2x)/7+9>5/7 implies 2x+63>5implies 2x>5-63=-58implies x > -29$
or
$\frac{2 x}{7} + 9 < - \frac{5}{7} \Rightarrow 2 x + 63 < - 5 \Rightarrow 2 x < - 5 - 63 = - 68 \Rightarrow x < - 34$ $(2x)/7+9<-5/7 implies 2x+63< -5 implies 2x< - 5-63=-68implies x < -34$

So, either $x > - 29$ $x> -29$ or $x < - 34$ $x<-34$

How do you solve for x given |\frac{2x}{7}+9 | > frac{5}{7}∣∣∣2x7+9∣∣∣>57|\frac{2x}{7}+9 | > frac{5}{7}?