What is |-3x-4|<= 12 ?

The solution to |-3x-4|<= 12 is -16/3 <= x<= 8/3

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|-3x-4|<= 12

∣ - 3 x - 4 ∣ \leq 12

- \frac{16}{3} \leq x \leq \frac{8}{3}

Solution steps

∣ - 3 x - 4 ∣ \leq 12

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 12 \leq - 3 x - 4 \leq 12

- 3 x - 4 \geq - 12 and - 3 x - 4 \leq 12

x \leq \frac{8}{3} and x \geq - \frac{16}{3}

Merge Overlapping Intervals

- \frac{16}{3} \leq x \leq \frac{8}{3}

- (x^{- 5} y^{\frac{1}{3}})^{- \frac{3}{5}}

\frac{( 4 - 2 i )}{7 + 3 i}

x^{2} + x - 4 = 0

2 0^{\circ} = π

x^{2} + 6 x + 8 > 0