What is 1/8 <= (1-8t)/9 < 8/9 ?

The solution to 1/8 <= (1-8t)/9 < 8/9 is -7/8 <t<=-1/64

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1/8 <= (1-8t)/9 < 8/9

\frac{1}{8} \leq \frac{1 - 8 t}{9} < \frac{8}{9}

- \frac{7}{8} < t \leq - \frac{1}{64}

Solution steps

\frac{1}{8} \leq \frac{1 - 8 t}{9} < \frac{8}{9}

a \leq u < b

then

a \leq u and u < b

\frac{1}{8} \leq \frac{1 - 8 t}{9} and \frac{1 - 8 t}{9} < \frac{8}{9}

\frac{1}{8} \leq \frac{1 - 8 t}{9} : t \leq - \frac{1}{64}

\frac{1 - 8 t}{9} < \frac{8}{9} : t > - \frac{7}{8}

Combine the intervals

t \leq - \frac{1}{64} and t > - \frac{7}{8}

Merge Overlapping Intervals

- \frac{7}{8} < t \leq - \frac{1}{64}

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

∣ 3 x + 2 ∣ < 10

4 x - 8 = 5

2 x + 3 = 5

(- 4 x^{2})^{3}

What is 1/8 <= (1-8t)/9 < 8/9 ?
The solution to 1/8 <= (1-8t)/9 < 8/9 is -7/8 <t<=-1/64