# How do you solve x /( 2x + 1 ) + ( 1 / 4 ) = 2 / (2x + 1) and find any extraneous solutions?

Nov 7, 2017

Solution: $1 \frac{1}{6}$ and no extraneous root.

#### Explanation:

$\frac{x}{2 x + 1} + \frac{1}{4} = \frac{2}{2 x + 1}$ Multiplying by $4 \left(2 x + 1\right)$ on

both sides we get , $4 x + \left(2 x + 1\right) = 8 \mathmr{and} 6 x = 7 \mathmr{and} x = \frac{7}{6}$

Check: L.H.S $= \frac{x}{2 x + 1} + \frac{1}{4} = \frac{\frac{7}{6}}{2 \cdot \frac{7}{6} + 1} + \frac{1}{4}$

$= \frac{\frac{7}{\cancel{6}}}{\frac{20}{\cancel{6}}} + \frac{1}{4}$

$= \frac{7}{20} + \frac{1}{4} = \frac{12}{20} = \frac{3}{5}$

R.H.S $= \frac{2}{2 x + 1} = \frac{2}{2 \cdot \frac{7}{6} + 1} = \frac{2}{\frac{20}{6}} = \frac{12}{20} = \frac{3}{5}$

L.H.S=R.H.S , no extraneous root.

Solution: $x = \frac{7}{6} = 1 \frac{1}{6}$ [Ans]

Nov 7, 2017

$x = \frac{7}{6}$

#### Explanation:

$\text{since fractions on both sides of the equation have a}$
$\textcolor{b l u e}{\text{common denominator "" we can combine them}}$

$\text{subtract "2/(2x+1)" from both sides}$

$\frac{x}{2 x + 1} - \frac{2}{2 x + 1} + \frac{1}{4} = 0$

$\text{subtract "1/4" from both sides}$

$\frac{x}{2 x + 1} - \frac{2}{2 x + 1} \cancel{+ \frac{1}{4}} \cancel{- \frac{1}{4}} = 0 - \frac{1}{4}$

$\Rightarrow \frac{x - 2}{2 x + 1} = - \frac{1}{4} \leftarrow \textcolor{b l u e}{\text{combining fractions}}$

$\text{using the method of "color(blue)"cross-multiplying}$

•color(white)(x)a/b=c/drArraxxd=bxxc

$\text{attaching the negative sign to 1 }$

$\Rightarrow - 1 \left(2 x + 1\right) = 4 \left(x - 2\right)$

$\Rightarrow - 2 x - 1 = 4 x - 8 \leftarrow \textcolor{b l u e}{\text{distributing}}$

$\text{subtract 4x from both sides}$

$- 2 x - 4 x - 1 = \cancel{4 x} \cancel{- 4 x} - 8$

$\Rightarrow - 6 x - 1 = - 8$

$\text{add 1 to both sides}$

$- 6 x \cancel{- 1} \cancel{+ 1} = - 8 + 1$

$\Rightarrow - 6 x = - 7$

$\text{divide both sides by } - 6$

$\frac{\cancel{- 6} x}{\cancel{- 6}} = \frac{- 7}{- 6}$

$\Rightarrow x = \frac{7}{6} \text{ is the solution}$

$\text{there are no extraneous solutions}$