How do you solve $\frac{2}{d} + \frac{1}{4} = \frac{11}{12}$ $2/d+1/4=11/12$ ?

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How do you solve $\frac{2}{d} + \frac{1}{4} = \frac{11}{12}$ $2/d+1/4=11/12$ ?

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Answer 1 · 2016-04-04T00:25:26

$d = 3$ $d=3$

Explanation:

$1$ $1$ . Subtract $\frac{1}{4}$ $color(red)(1/4)$ from both sides of the equation.

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

$\frac{2}{d} + \frac{1}{4}$ $2/d+1/4$ $- \frac{1}{4} = \frac{11}{12}$ $color(red)(-1/4)=11/12$ $- \frac{1}{4}$ $color(red)(-1/4)$

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

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

$\frac{2}{d} = \frac{11}{12} - \frac{1 \times 3}{4 \times 3}$ $2/d=11/12-(1color(red)(xx3))/(4color(red)(xx3))$

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

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

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

$3$ $3$ . Divide both sides by $\frac{8}{12}$ $color(red)(8/12)$ .

$\frac{2}{d} \div \frac{8}{12} = \frac{8}{12} \div \frac{8}{12}$ $2/dcolor(red)(-:8/12)=8/12color(red)(-:8/12)$

$\frac{2}{d}$ $2/d$ $\times \frac{12}{8} = \frac{8}{12} \times \frac{12}{8}$ $xx12/8=8/12xx12/8$

$color(darkorange)cancelcolor(black)2^1/d$ $xx12/color(darkorange)cancelcolor(black)8^4=color(red)cancelcolor(black)8^1/color(teal)cancelcolor(black)12^1xxcolor(teal)cancelcolor(black)12^1/color(red)cancelcolor(black)8^1$

$12/(4d)=1$

$4$ . Solve for $d$ .

$(12color(red)(-:4))/(4dcolor(red)(-:4))=1$

$3/d=1$

$color(green)(|bar(ul(color(white)(a/a)d=3color(white)(a/a)|)))$

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How do you solve $\frac{2}{d} + \frac{1}{4} = \frac{11}{12}$ $2/d+1/4=11/12$ ?

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Explanation:

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How do you solve $\frac{2}{d} + \frac{1}{4} = \frac{11}{12}$ $2/d+1/4=11/12$ ?