What is the answer to solvefor c,d=1-(s/c) 1/n ?

The answer to solvefor c,d=1-(s/c) 1/n is c=-s/(n(d-1));\quad d\ne 1

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solvefor c,d=1-(s/c) 1/n

solve for

c, d = 1 - (\frac{s}{c}) \frac{1}{n}

c = - \frac{s}{n ( d - 1 )}; d \neq = 1

Solution steps

d = 1 - (\frac{s}{c}) \frac{1}{n}

Move

1

to the left side

d - 1 = - \frac{s}{c} \cdot \frac{1}{n}

Simplify

- \frac{s}{c} \cdot \frac{1}{n} : - \frac{s}{n c}

d - 1 = - \frac{s}{n c}

Multiply both sides by

c

d c - c = - \frac{s}{n}

Factor

d c - c : c (d - 1)

c (d - 1) = - \frac{s}{n}

Divide both sides by

d - 1; d \neq = 1

c = - \frac{s}{n ( d - 1 )}; d \neq = 1

\frac{( 3 - 1 )}{( 3 + t )} = a - b \cdot 3

\frac{1}{( x - 2 )} + \frac{1}{( x - 3 )} = \frac{1}{( x + 1 )} + \frac{1}{( x - 6 )}

\frac{31}{2 + x} = \frac{51}{4}

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

\frac{6 x ^{2} - ( 3 a )}{2 \cdot x - a ^{2}} = 0

What is the answer to solvefor c,d=1-(s/c) 1/n ?
The answer to solvefor c,d=1-(s/c) 1/n is c=-s/(n(d-1));\quad d\ne 1