What is the answer to solvefor t,((s-2))/t = 1/p+1/r ?

The answer to solvefor t,((s-2))/t = 1/p+1/r is t=(spr-2pr)/(r+p);\quad pr\ne 0,r+p\ne 0

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solvefor t,((s-2))/t = 1/p+1/r

solve for

t, \frac{( s - 2 )}{t} = \frac{1}{p} + \frac{1}{r}

t = \frac{s p r - 2 p r}{r + p}; p r \neq = 0, r + p \neq = 0

Solution steps

\frac{( s - 2 )}{t} = \frac{1}{p} + \frac{1}{r}

Simplify

\frac{( s - 2 )}{t} : \frac{s - 2}{t}

\frac{s - 2}{t} = \frac{1}{p} + \frac{1}{r}

Multiply both sides by

t

s - 2 = \frac{t}{p} + \frac{t}{r}

Switch sides

\frac{t}{p} + \frac{t}{r} = s - 2

Multiply by LCM

t r + tp = s p r - 2 p r; p r \neq = 0

Factor

t r + tp : t (r + p)

t (r + p) = s p r - 2 p r; p r \neq = 0

Divide both sides by

r + p; p r \neq = 0, r + p \neq = 0

t = \frac{s p r - 2 p r}{r + p}; p r \neq = 0, r + p \neq = 0

\frac{x ^{2} + 1}{( x ^{2} ) + 1} = 0

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

\frac{∣ x + 2 ∣}{∣ x - 3 ∣} = 4

2 (\frac{x ^{2} + 1}{x ^{2}}) - 3 (\frac{x + 1}{x}) - 1 = 0

\frac{( a ^{3} + 3 a )}{( 3 a ^{2} + 1 )} = \frac{14}{13}

What is the answer to solvefor t,((s-2))/t = 1/p+1/r ?
The answer to solvefor t,((s-2))/t = 1/p+1/r is t=(spr-2pr)/(r+p);\quad pr\ne 0,r+p\ne 0