What is |(8x)/(2x-2)|<= 1 ?

The solution to |(8x)/(2x-2)|<= 1 is -1/3 <= x<= 1/5

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|(8x)/(2x-2)|<= 1

\frac{8 x}{2 x - 2} \leq 1

- \frac{1}{3} \leq x \leq \frac{1}{5}

Solution steps

\frac{8 x}{2 x - 2} \leq 1

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 1 \leq \frac{8 x}{2 x - 2} \leq 1

\frac{8 x}{2 x - 2} \geq - 1 and \frac{8 x}{2 x - 2} \leq 1

(x \leq \frac{1}{5} or x > 1) and - \frac{1}{3} \leq x < 1

Merge Overlapping Intervals

- \frac{1}{3} \leq x \leq \frac{1}{5}

- 8 \leq \frac{2}{5} (k - 2)

8 - 6 m = 4 m + 48

b^{2} + 6 b = 0

6 x^{2} - 18 x

a (a) + 14 a + 45

What is |(8x)/(2x-2)|<= 1 ?
The solution to |(8x)/(2x-2)|<= 1 is -1/3 <= x<= 1/5