# Unit Name: Polynomial and Rational Functions Solve the following inequality algebraically.

Unit Name: Polynomial and Rational Functions
Solve the following inequality algebraically. Explain your process.
$8x-3\le 2x+1\le 17x-8$
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Karen Robbins
Split the inequality into two:

Isolate x from each inequality using inverse operations.
$8x-3\le 2x+1$
$8x-2x\le 3+1$
$6x\le 4$
$x\le \frac{2}{3}$
AND
$2x+1\le 17x-8$
$2x-17x\le -8-1$
$-15x+15x\le -9+15x$
$0\le -9+15x$
$\frac{3}{5}\le x$
Therefore, the solution is: $\frac{3}{5}\le x\le \frac{2}{3}$
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Shannon Hodgkinson
Solve the following inequality algebraically.
$8x-3\le 2x+1\le 17x-8$
We've $8x-3\le 2x+1\le 17x-8$. (1)
$⇒8x-3-1\le 2x+1-1\le 17x-8-1$
$⇒8x-4\le 2x\le 17x-9$
Now dividing above equation by 2 we get
$⇒\frac{8x-4}{2}\le \frac{2x}{2}\le \frac{17}{2}x-\frac{9}{2}$
$⇒4x-2\le x\le \frac{17}{2}x-\frac{9}{2}$ (2)
Now from equation (2) we've two cases.
(1) $x\ge 4x-2$
$⇒$ adding 2 to both sides we get
$⇒x+2\ge 4x-2+2$
$⇒x+2\ge 4x$
Subtracting x from both sides
$⇒x+2-x\ge 4x-x$
$⇒2\ge 3x$
$⇒x\le \frac{2}{3}$
$x\le \frac{17}{2}x-\frac{9}{2}$
adding $\frac{9}{2}$ to both sides we get
$⇒x+\frac{9}{2}\le \frac{17}{2}x-\frac{9}{2}+\frac{9}{2}$
$⇒x+\frac{9}{2}\le \frac{17}{2}x$
Subtracting x from both sides
$⇒x+\frac{9}{2}-x\le \frac{17}{2}x-x$
$⇒\frac{q}{2}\le \frac{17x-2x}{2}$
$⇒\frac{9}{2}\le \frac{15x}{2}$
Multiplying 2 to both sides we get
$\text{Undefined control sequence \cancel}$
$⇒9\le 15x$
$⇒x\ge \frac{3}{5}$ (4)
From equation (3) and equation (4) we get
$\frac{3}{5}\le x\le \frac{2}{3}$
$⇒0.6\le x\le 0.6666$