# How do you solve 4 |4-3x |= 4x + 6 and find any extraneous solutions?

Jun 28, 2016

$x = \frac{5}{8} \mathmr{and} x = \frac{11}{4}$

#### Explanation:

The equation is equivalent to:

$\left(4 - 3 x \ge 0 \mathmr{and} 4 \left(4 - 3 x\right) = 4 x + 6\right) \mathmr{and} \left(4 - 3 x < 0 \mathmr{and} 4 \left(- 4 + 3 x\right) = 4 x + 6\right)$

that becomes:

$\left(x \le \frac{4}{3} \mathmr{and} 16 - 12 x = 4 x + 6\right) \mathmr{and} \left(x > \frac{4}{3} \mathmr{and} - 16 + 12 x = 4 x + 6\right)$

$\left(x \le \frac{4}{3} \mathmr{and} 16 x = 10\right) \mathmr{and} \left(x > \frac{4}{3} \mathmr{and} 8 x = 22\right)$

$\left(x \le \frac{4}{3} \mathmr{and} x = \frac{5}{8}\right) \mathmr{and} \left(x > \frac{4}{3} \mathmr{and} x = \frac{11}{4}\right)$

so

$x = \frac{5}{8} \mathmr{and} x = \frac{11}{4}$

Jun 28, 2016

$x = \frac{5}{6} \mathmr{and} \frac{11}{4}$. There is no extraneous solution.

#### Explanation:

Divide by 4 and make it

$| 4 - 3 x | = x + \frac{3}{2} \ge 0$ So, the solution(s) should satisfy $x \ge - \frac{3}{2}$.

The given equation is equivalent to the pair

$4 - 3 x = x + 1.5$, when $x \le \frac{4}{3}$.and

3x-4=x+1.5, when $x \ge \frac{4}{3.}$

Solving both,

x=5/8<4/3 and x=11/4>4/3.

Both are valid..

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