# How do you solve 1/8x+3/2=3/4x-1?

Jun 26, 2017

See a solution process below:

#### Explanation:

First, multiply both sides of the equation by $\textcolor{red}{8}$ to eliminate the fractions. $\textcolor{red}{8}$ is the Lowest Common Denominator (LCD) of the three fractions:

$\textcolor{red}{8} \left(\frac{1}{8} x + \frac{3}{2}\right) = \textcolor{red}{8} \left(\frac{3}{4} x - 1\right)$

$\left(\textcolor{red}{8} \times \frac{1}{8} x\right) + \left(\textcolor{red}{8} \times \frac{3}{2}\right) = \left(\textcolor{red}{8} \times \frac{3}{4} x\right) - \left(\textcolor{red}{8} \times 1\right)$

$\left(\cancel{\textcolor{red}{8}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}} x\right) + \left(\cancel{\textcolor{red}{8}} 4 \times \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}\right) = \left(\cancel{\textcolor{red}{8}} 2 \times \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} x\right) - 8$

$1 x + 12 = 6 x - 8$

Next, subtract $\textcolor{red}{1 x}$ and add $\textcolor{red}{8}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:

$- \textcolor{red}{1 x} + 1 x + 12 + \textcolor{red}{8} = - \textcolor{red}{1 x} + 6 x - 8 + \textcolor{red}{8}$

$0 + 20 = \left(- \textcolor{red}{1} + 6\right) x - 0$

$20 = 5 x$

Now, divide each side of the equation by $\textcolor{red}{5}$ to solve for $x$ while keeping the equation balanced:

$\frac{20}{\textcolor{red}{5}} = \frac{5 x}{\textcolor{red}{5}}$

$4 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}}$

$4 = x$

$x = 4$

Jun 26, 2017

color(orange)(x!=4 is a extraneous solution

#### Explanation:

$\frac{1}{8} x + \frac{3}{2} = \frac{3}{4} x - 1$

multiply L.H.S AND R.H.S by $8$

$\therefore x + 12 = 6 x - 8$

$\therefore x - 6 x = - 8 - 12$

$\therefore - 5 x = - 20$

multiply L.H.S AND R.H.S by$- 1$

$\therefore 5 x = 20$

:.color(orange)(x=4

substitute color(orange)(x=4

$\therefore \frac{1}{8} \left(\textcolor{\mathmr{and} a n \ge}{4}\right) + \frac{3}{2} = \frac{3}{4} \left(\textcolor{\mathmr{and} a n \ge}{4}\right) - 1$

$\therefore \frac{4}{8} + \frac{3}{2} = \frac{12}{4} - 1$

multiply L.H.S AND R.H.S by$8$

$\therefore \frac{32}{8} + \frac{24}{2} = \frac{24}{4} - 8$

$\therefore 4 + 12 = 6 - 8$

:.color(orange)(16!=-2

:.color(orange)(x=4  is a extraneous solution