# How do you solve abs((2x+1) / (x-3))> 3?

Jun 17, 2015

Everything except 3.

#### Explanation:

This means that either:

$\frac{2 x + 1}{x - 3} > 3$ or $\frac{2 x + 1}{x - 3} < - 3$

Multiplying both sides by (x-3) gives us

$2 x + 1 > 3 \left(x - 3\right)$ or $2 x + 1 < - 3 \left(x - 3\right)$

Which simplifies to

$2 x + 1 > 3 x - 9$ or $2 x + 1 < - 3 x + 9$

And now we can algebraically reduce this to get a simple answer.

$2 x + 1 + 9 > 3 x - 9 + 9$ or $2 x + 1 - 1 < - 3 x + 9 - 1$

$2 x + 10 > 3 x$ or $2 x < - 3 x + 8$

$10 > x$ or $5 x < 8$

$10 > x$ or $x < \frac{5}{8}$

In simple terms (I like simple), $x$ can either be less than a big number, or greater than a small number. This means that $x$ can pretty much be anything, except for 3, since that would make the denominator of the equation 0, breaking all mathematical rules.

Jun 17, 2015

$\frac{8}{5} < x < 3$ or $3 < x < 10$

#### Explanation:

$\frac{2 x + 1}{x - 3} = \frac{2 x - 6 + 7}{x - 3} = \frac{2 \left(x - 3\right) + 7}{x - 3} = 2 + \frac{7}{x - 3}$

with exclusion $x \ne 3$

$\left\mid \frac{2 x + 1}{x - 3} \right\mid = \left\mid 2 + \frac{7}{x - 3} \right\mid > 3$

So we are looking for:

[1]: $2 + \frac{7}{x - 3} < - 3$

or

[2]: $2 + \frac{7}{x - 3} > 3$

In case [1] first subtract $2$ from both sides to get:

$\frac{7}{x - 3} < - 5$

If $x > 3$ then $\left(x - 3\right) > 0$ and $\frac{7}{x - 3} > 0$ so the inequality is not satisfied.

So $x < 3$ and $\left(x - 3\right) < 0$. Multiply both sides of the inequality by the negative value $\left(x - 3\right)$ and reverse the inequality to get:

$7 > - 5 \left(x - 3\right) = - 5 x + 15$

Add $5 x - 7$ to both sides to get:

$5 x > 8$

Divide both sides by $5$ to get:

$x > \frac{8}{5}$

So case [1] gives us the solution $\frac{8}{5} < x < 3$.

In case [2] first subtract $2$ from both sides to get:

$\frac{7}{x - 3} > 1$

We require $\frac{7}{x - 3}$ to be positive, so we need $\left(x - 3\right) > 0$ so $x > 3$.

Multiply both sides of the inequality by $\left(x - 3\right)$ (which is positive) to get:

$7 > x - 3$

Add $3$ to both sides to get:

$10 > x$

So case [2] gives us the solution $3 < x < 10$.

In general you can do any of the following to an inequality and preserve its truth:

(1) Add or subtract the same value on both sides of the inequality.
(2) Multiply or divide both sides of the inequality by the same positive value.
(3) Multiply or divide both sides of the inequality by the same negative value and reverse the inequality ($<$ becomes $>$, $\ge$ becomes $\le$, etc.).

graph{(y-abs((2x+1)/(x-3)))*(y-3) = 0 [-17.74, 22.26, -6.8, 13.2]}