# How do you solve |2x+5| = 3x+4 and find any extraneous solutions?

Apr 4, 2017

Non-extraneous solution $x = 1$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \left\mid 2 x + 5 \right\mid = 3 x + 4$

One method of solving this would be to square both sides:
$\textcolor{w h i t e}{\text{XXX}} 4 {x}^{2} + 20 x + 25 = 9 {x}^{2} + 24 x + 16$

Simplifying
$\textcolor{w h i t e}{\text{XXX}} 5 {x}^{2} + 4 x - 9 = 0$

Factoring
$\textcolor{w h i t e}{\text{XXX}} \left(5 x + 9\right) \left(x - 1\right) = 0$

Giving potential solutions
color(white)("XXX"){:(x=-9/5,color(white)("XX")andcolor(white)("XX"),x=+1):}

Checking these against the original equation:
color(white)("XXX"){: (ul(color(white)("XXXXXX")),ul(x=-9/5),color(white)("XXX"),ul(x=1)), (abs(2x+5),=abs(-18/5+25/5),,=abs(2+5)), (,=7/5,,=7), (ul(color(white)("XXXXXX")),ul(color(white)("XXXXXX")),,ul(color(white)("XXXXXX"))), (3x+4,=-27/5+20/5,,=3+4), (,=-7/5,,=7) :}

Since $\left\mid 2 x + 5 \right\mid \ne 3 x + 4$ when $x = - \frac{9}{5}$
this solution is extraneous.

Apr 4, 2017

$x = 1 , \text{ extraneous solution is } x = - \frac{9}{5}$

#### Explanation:

This equation has 2 solutions.

$\Rightarrow 2 x + 5 = \textcolor{red}{\pm} 3 x + 4$

• " solve " 2x+5=3x+4

subtract 2x from both sides.

$\cancel{2 x} \cancel{- 2 x} + 5 = 3 x - 2 x + 4$

$\Rightarrow 5 = x + 4$

subtract 4 from both sides.

$5 - 4 = x \cancel{+ 4} \cancel{- 4}$

$\Rightarrow x = 1 \leftarrow \textcolor{red}{\text{ to be verified}}$

• " solve " 2x+5=color(red)(-)(3x+4)

$\Rightarrow 2 x + 5 = - 3 x - 4$

$2 x + 3 x + 5 = \cancel{- 3 x} \cancel{+ 3 x} - 4$

$\Rightarrow 5 x + 5 = - 4$

subtract 5 from both sides.

$5 x \cancel{+ 5} \cancel{- 5} = - 4 - 5$

$\Rightarrow 5 x = - 9$

divide both sides by 5

$\frac{\cancel{5} x}{\cancel{5}} = \frac{- 9}{5}$

$\Rightarrow x = - \frac{9}{5} \leftarrow \textcolor{red}{\text{ to be verified}}$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the equation and if the left side equals the right side then they are the solutions.

$x = 1 \to | 2 + 5 | = 3 + 4 \to 7 = 7 \to \text{ true}$

$x = - \frac{9}{5} \to | - \frac{18}{5} + \frac{25}{5} | = - \frac{27}{5} + \frac{20}{5} \to \frac{7}{5} = - \frac{7}{5} \to \text{ false}$

$\Rightarrow x = 1 \text{ is the solution}$

$\Rightarrow x = - \frac{9}{5} \text{ is an extraneous solution}$