What is the answer to solvefor a, 1/a = 1/b+1/c ?

The answer to solvefor a, 1/a = 1/b+1/c is a=(bc)/(c+b);\quad bc\ne 0,b\ne-c

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solvefor a, 1/a = 1/b+1/c

solve for

a, \frac{1}{a} = \frac{1}{b} + \frac{1}{c}

a = \frac{b c}{c + b}; b c \neq = 0, b \neq = - c

Solution steps

\frac{1}{a} = \frac{1}{b} + \frac{1}{c}

Multiply both sides by

a

1 = \frac{a}{b} + \frac{a}{c}

Switch sides

\frac{a}{b} + \frac{a}{c} = 1

Multiply by LCM

a c + ab = b c; b c \neq = 0

Factor

a c + ab : a (c + b)

a (c + b) = b c; b c \neq = 0

Divide both sides by

c + b; b c \neq = 0, b \neq = - c

a = \frac{b c}{c + b}; b c \neq = 0, b \neq = - c

\frac{3}{x - 1} + \frac{2}{x} = 8

\frac{3 x ^{2} - 5 x}{3 x - 2} - 1 = \frac{2}{2 - 3 x}

G (x) = \frac{x ^{2} + 3 x - 5}{x + 6}

\frac{7}{2 x} - \frac{5}{3 x} = \frac{22}{3}

1 + \frac{4}{x - 6} = - \frac{x}{x - 6}

What is the answer to solvefor a, 1/a = 1/b+1/c ?
The answer to solvefor a, 1/a = 1/b+1/c is a=(bc)/(c+b);\quad bc\ne 0,b\ne-c