Question

find x such that 3x−3 ⩾ x+5 and 1÷x−1 ⩾ 1÷8​

Answer 1

Step-by-step explanation:

Given:

$1)3x - 3 \geqslant x + 5 \\ \\ 2) \frac{1}{x - 1} \geqslant \frac{1}{8}$

To find: Value of x in both cases.

Solution:

$\bold{1)3x - 3 \geqslant x + 5}$

Subtract x from both sides

$3x \red{- x} - 3 \geqslant x \red{- x} + 5 \\ \\ 2x - 3 \geqslant 5$

Add 3 both sides

$2x - 3 \red{+ 3} \geqslant 5 \red{+ 3} \\ \\ 2x \geqslant 8$

Divide by 2 both sides

$\frac{2x}{\red{2}} \geqslant \frac{8}{\red{2}} \\ \\ \bold{\green{x \geqslant 4}}$

$\bold{2) \frac{1}{x - 1} \geqslant \frac{1}{8}}$

Cross multiply

$8 \geqslant x - 1$

Add 1 both sides

$8 + 1 \geqslant x - 1 + 1 \\ \\ 9 \geqslant x$

Final answer:

$1)\bold{x \geqslant 4} \\ 2)\bold{x \leqslant 9}$

Hope it helps you.