Inequalities

Example 0..1   Solve the inequality

$$\frac{1}{2}(1 + x) \leq 6$$

SOLUTION To solve this inequality, we first multiply both sides of inequality by 2.

$$1 + x \leq 12$$

Then subtract 1 from both sides of inequality to obtain

$$x \leq 11$$

Then the solution is $(-\infty, 11]$?D $\ \blacksquare$

Example 0..2   Solve the inequality?D

$$\frac{1}{5}(x^{2} - 4x + 3) < 0$$

SOLUTION To solve this inequality, we first multiply both sides of inequality by 5.

$$x^{2} - 4x + 3 < 0$$

Then factor the left-hand side to obtain

$$(x-1)(x-3) < 0$$

Note that $(x-1)(x-3)$ becomes 0 at 1 and 3?DThen we make circle on the number line at 1 and 3. Then we have 3 separated parts?D

$$(-\infty, 1),\ (1,3), \ (3, \infty)$$

In the given interval?Cthe sign of the product $(x-1)(x-3)$ does not change. Thus we have

$(-\infty, 1)$	${\rm sgn}[(x-1)(x-3)] = (-)(-) = +$
$(1,3)$	${\rm sgn}[(x-1)(x-3)] = (+)(-) = -$
$(3, \infty)$	${\rm sgn}[(x-1)(x-3)] = (+)(+) = +$

From this we obtain $(1,3)$?D $\ \blacksquare$

Example 0..3   Solve the inequality?D

$$\frac{x+2}{1 - x} \geq 1$$

SOLUTION To solve this inequality, we add $-1$ to both sides of the inequality to get rid of 1.

$$\frac{x+2}{1 - x} - 1 \geq 0$$

Simplifying

$$\frac{x+2 - (1 - x)}{1 - x} \geq$$ 0

$$\frac{2x + 1}{1-x} \geq 0$$

Now we would like to get rid of the denominator $1-x$. But if we multiply $1-x$ to get rid of the denominator, then we have to be careful about the sign of $1-x$. Thus instead?Cwe multiply $(1-x)^{2}$. Then we have

$$(2x+1)(1-x) \geq 0$$

Now as above, the product $(2x+1)(1-x)$ becomes 0 at $-\frac{1}{2}$ and 1?DThen we make circle on the number line at $-\frac{1}{2}$ and 1?D Next note that the equality only holds at the point where the numerator is 0. From this we fill the circle at $-\frac{1}{2}$?DFrom this the number line can be separated by 3 parts?D

$$(-\infty, -\frac{1}{2}],\ [-\frac{1}{2},1), \ (1, \infty)$$

$(-\infty, -\frac{1}{2}]$	${\rm sgn}[(2x+1)(1-x)] = (-)(+) = -$
$[-\frac{1}{2},1)$	${\rm sgn}[(2x+1)(1-x)] = (+)(+) = +$
$(1, \infty)$	${\rm sgn}[(2x+1)(1-x)] = (+)(-) = -$

Therefore, the solution is $[-\frac{1}{2},1)$?D $\ \blacksquare$