# How do you solve 2/d+1/4=11/12?

Apr 4, 2016

$d = 3$

#### Explanation:

$1$. Subtract $\textcolor{red}{\frac{1}{4}}$ from both sides of the equation.

$\frac{2}{d} + \frac{1}{4} = \frac{11}{12}$

$\frac{2}{d} + \frac{1}{4}$ $\textcolor{red}{- \frac{1}{4}} = \frac{11}{12}$ $\textcolor{red}{- \frac{1}{4}}$

$\frac{2}{d} = \frac{11}{12} - \frac{1}{4}$

$2$. Simplify the right side of the equation.

$\frac{2}{d} = \frac{11}{12} - \frac{1 \textcolor{red}{\times 3}}{4 \textcolor{red}{\times 3}}$

$\frac{2}{d} = \frac{11}{12} - \frac{3}{12}$

$\frac{2}{d} = \frac{11 - 3}{12}$

$\frac{2}{d} = \frac{8}{12}$

$3$. Divide both sides by $\textcolor{red}{\frac{8}{12}}$.

$\frac{2}{\mathrm{dc}} o l \mathmr{and} \left(red\right) \left(\div \frac{8}{12}\right) = \frac{8}{12} \textcolor{red}{\div \frac{8}{12}}$

$\frac{2}{d}$$\times \frac{12}{8} = \frac{8}{12} \times \frac{12}{8}$

${\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\cancel{\textcolor{b l a c k}{2}}}}^{1} / d$$\times \frac{12}{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\cancel{\textcolor{b l a c k}{8}}}} ^ 4 = {\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}^{1} / {\textcolor{t e a l}{\cancel{\textcolor{b l a c k}{12}}}}^{1} \times {\textcolor{t e a l}{\cancel{\textcolor{b l a c k}{12}}}}^{1} / {\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}^{1}$

$\frac{12}{4 d} = 1$

$4$. Solve for $d$.

$\frac{12 \textcolor{red}{\div 4}}{4 \mathrm{dc} o l \mathmr{and} \left(red\right) \left(\div 4\right)} = 1$

$\frac{3}{d} = 1$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} d = 3 \textcolor{w h i t e}{\frac{a}{a}} |}}}$