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 $(-\infty, 11]$ $\blacksquare$

Basic Technique To solve inequality, first eliminate the outside factor $\frac{1}{2}$

Example 0..2   Solve the inequality

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

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 3Then we make circle on the number line at 1 and 3. Then we have 3 separated parts

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

In the given intervalthe 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 $(1,3)$ $\blacksquare$

Example 0..3   Solve the inequality

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

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 insteadwe multiply $(1-x)^{2}$. Then we have

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

Inequality in Fraction To solve the inequality with the fraction, first transpose an expression to get one side of inequality is 0. Then eliminate the denominator by multiplying the square of the denominator

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

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

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

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