How do you solve $\frac{3}{8 x} - \frac{1}{3 x + 6} = \frac{1}{6 x}$ $3/(8x)-1/(3x+6)=1/(6x)$ ?

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How do you solve $\frac{3}{8 x} - \frac{1}{3 x + 6} = \frac{1}{6 x}$ $3/(8x)-1/(3x+6)=1/(6x)$ ?

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Answer 1 · 2015-09-30T19:58:21

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$3/(8x)-1/(3x+6)=1/(6x)=>3/(8x)-1/(6x)=1/(3x+6)=> (3*6x-8x)/(48x^2)=1/(3x+6)=>(10x)/(48x^2)=1/(3x+6)=> 5/(24x)=1/(3x+6)=>5(3x+6)=24x=>15x+30=24x=> 9x=30=>x=10/3$

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How do you solve $\frac{3}{8 x} - \frac{1}{3 x + 6} = \frac{1}{6 x}$ $3/(8x)-1/(3x+6)=1/(6x)$ ?

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How do you solve $\frac{3}{8 x} - \frac{1}{3 x + 6} = \frac{1}{6 x}$ $3/(8x)-1/(3x+6)=1/(6x)$ ?