$\displaystyle \frac {1}{a} + \frac{1}{b} = \frac{1}{c}$ When a = $\displaystyle \frac{4}{5}$ and c = $\displaystyle \frac{2}{3}$ How can i find b?
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Originally Posted by ADY $\displaystyle \frac {1}{a} + \frac{1}{b} = \frac{1}{c}$ When a = $\displaystyle \frac{4}{5}$ and c = $\displaystyle \frac{2}{3}$ How can i find b? Plug in the given values: $\displaystyle \dfrac1{\frac45} + \dfrac1b=\dfrac1{\frac23}$ Move the constants to the RHS of the equation: $\displaystyle \dfrac1b = \dfrac32 - \dfrac54=\dfrac14$ Thus $\displaystyle b = 4$
Originally Posted by ADY $\displaystyle \frac {1}{a} + \frac{1}{b} = \frac{1}{c}$ (a + b) / bc = 1 / c ac + bc = ab ab - bc = ac b(a - c) = ac b = ac / (a - c) Ya'll ok now ?
Originally Posted by Wilmer (a + b) / bc = 1 / c ac + bc = ab ab - bc = ac b(a - c) = ac b = ac / (a - c) Ya'll ok now ? I don't want to pick at you but shouldn't this line (a + b) / bc = 1 / c read (a + b) / ab = 1 / c
Originally Posted by earboth I don't want to pick at you but shouldn't this line (a + b) / bc = 1 / c read (a + b) / ab = 1 / c Yes, typo...thanks Mr E
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