How do you solve $\frac{1}{3 m} + \frac{6 m - 9}{3 m} = \frac{3 m - 3}{4 m}$ $1/(3m)+(6m-9)/(3m)=(3m-3)/(4m)$ ?

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How do you solve $\frac{1}{3 m} + \frac{6 m - 9}{3 m} = \frac{3 m - 3}{4 m}$ $1/(3m)+(6m-9)/(3m)=(3m-3)/(4m)$ ?

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Answer 1 · 2016-11-27T14:51:32

$m = 0 O r m = \frac{23}{15}$ $m=0 " "Or" "m=23/15$

Explanation:

$\frac{1}{3 m} + \frac{6 m - 9}{3 m} = \frac{3 m - 3}{4 m}$ $1/(3m)+(6m-9)/(3m) = (3m-3)/(4m)$
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$\Rightarrow \frac{1 + 6 m - 9}{3 m} = \frac{3 m - 3}{4 m}$ $rArr(1+6m-9)/(3m)=(3m-3)/(4m)$
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$\Rightarrow \frac{- 8 + 6 m}{3 m} = \frac{3 m - 3}{4 m}$ $rArr(-8+6m)/(3m)=(3m-3)/(4m)$
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$\Rightarrow (- 8 + 6 m) \times 4 m = (3 m - 3) \times 3 m$ $rArrcolor(blue)((-8+6m)xx4m = (3m-3)xx3m)$
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$\Rightarrow - 32 m + 24 m^{2} = 9 m^{2} - 9 m$ $rArr-32m + 24m^2 = 9m^2-9m$
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$\Rightarrow - 32 m + 24 m^{2} - 9 m^{2} + 9 m = 0$ $rArr-32m + 24m^2 - 9m^2+9m=0$
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$\Rightarrow 15 m^{2} - 23 m = 0$ $rArr15m^2-23m=0$
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$\Rightarrow m (15 m - 23) = 0$ $rArrm(15m-23)=0$
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$m = 0$ $m=0$
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Or
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$15 m - 23 = 0 \Rightarrow 15 m = 23 \Rightarrow m = \frac{23}{15}$ $15m - 23 =0rArr15m = 23rArrm=23/15$

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How do you solve $\frac{1}{3 m} + \frac{6 m - 9}{3 m} = \frac{3 m - 3}{4 m}$ $1/(3m)+(6m-9)/(3m)=(3m-3)/(4m)$ ?

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Explanation:

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How do you solve $\frac{1}{3 m} + \frac{6 m - 9}{3 m} = \frac{3 m - 3}{4 m}$ $1/(3m)+(6m-9)/(3m)=(3m-3)/(4m)$ ?