How do you solve $\frac{9}{8} x - 1 \leq 5 x - \frac{7}{4}$ $\frac { 9} { 8} x - 1\leq 5x - \frac { 7} { 4}$ ?

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Answer 1 · 2018-03-22T11:56:45

$x \geq \frac{6}{31}$ $x>=6/31$

As always do some cleanup:

$\frac{9}{8} x - 5 x \leq - \frac{7}{4} + 1$ $9/8x-5x <= -7/4+ 1$

$- \frac{31}{8} x \leq - \frac{3}{4}$ $-31/8x<=-3/4$

Now we multiply by $- 8$ $-8$ . Therefore the $\leq$ $<=$ has to be changed to $\geq$ $>=$ , resulting in

$31 x \geq 6$ $31x>=6$ .

Division by 31 solves the problem:

$x \geq \frac{6}{31}$ $x>=6/31$

How do you solve 98x−1≤5x−74\frac { 9} { 8} x - 1\leq 5x - \frac { 7} { 4}?