How do you solve $- 5 < \frac{8 x - 5}{6} < 7$ $-5< \frac{8x - 5}{6} < 7$ ?

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Answer 1 · 2018-06-05T05:54:46

Undo the operations surrounding $x$ $x$ by applying the inverses to both sides in the reverse order.

Multiply all sides by $6$ $6$ :

$- 30 < 8 x - 5 < 42$ $-30 < 8x - 5 < 42$

Add $5$ $5$ :

$- 25 < 8 x < 47$ $-25 < 8x < 47$

Divide all sides by $8$ $8$ :

$- \frac{25}{8} < x < \frac{47}{8}$ $-25/8 < x < 47/8$

How do you solve -5< \frac{8x - 5}{6} < 7−5<8x−56<7-5< \frac{8x - 5}{6} < 7?