What is the answer to solvefor a, 1/a+1/b = 3/2020 ?

The answer to solvefor a, 1/a+1/b = 3/2020 is a=-(2020b)/(2020-3b);\quad b\ne 0,b\ne 2020/3

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solvefor a, 1/a+1/b = 3/2020

solve for

a, \frac{1}{a} + \frac{1}{b} = \frac{3}{2020}

a = - \frac{2020 b}{2020 - 3 b}; b \neq = 0, b \neq = \frac{2020}{3}

Solution steps

\frac{1}{a} + \frac{1}{b} = \frac{3}{2020}

Multiply both sides by

a

1 + \frac{a}{b} = \frac{3}{2020} a

Move

1

to the right side

\frac{a}{b} = \frac{3}{2020} a - 1

Multiply by LCM

2020 a = 3 ab - 2020 b; b \neq = 0

Move

3 ab

to the left side

2020 a - 3 ab = - 2020 b; b \neq = 0

Factor

2020 a - 3 ab : a (2020 - 3 b)

a (2020 - 3 b) = - 2020 b; b \neq = 0

Divide both sides by

2020 - 3 b; b \neq = 0, b \neq = \frac{2020}{3}

a = - \frac{2020 b}{2020 - 3 b}; b \neq = 0, b \neq = \frac{2020}{3}

\frac{( 216 x ^{3} ) + 1}{( 27 x ^{3} )} = 135

0.3 (x + 1) = \frac{1}{2 ( x - 1 )}

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

\frac{4 x}{5 + x + 2} = \frac{x}{2 + x + 5}

\frac{2}{( x + 2 )} + \frac{3}{( x - 1 )} = \frac{1}{( x + 2 )}

What is the answer to solvefor a, 1/a+1/b = 3/2020 ?
The answer to solvefor a, 1/a+1/b = 3/2020 is a=-(2020b)/(2020-3b);\quad b\ne 0,b\ne 2020/3