What is the answer to solvefor k,g^{-1}(4k)=k+1 ?

The answer to solvefor k,g^{-1}(4k)=k+1 is k= g/(4-g);\quad g\ne 0,g\ne 4

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solvefor k,g^{-1}(4k)=k+1

solve for

k, g^{- 1} (4 k) = k + 1

k = \frac{g}{4 - g}; g \neq = 0, g \neq = 4

Solution steps

g^{- 1} (4 k) = k + 1

Expand

g^{- 1} (4 k) : \frac{4 k}{g}

\frac{4 k}{g} = k + 1

Multiply both sides by

g

4 k = k g + g; g \neq = 0

Move

k g

to the left side

4 k - k g = g; g \neq = 0

Factor

4 k - k g : k (4 - g)

k (4 - g) = g; g \neq = 0

Divide both sides by

4 - g; g \neq = 0, g \neq = 4

k = \frac{g}{4 - g}; g \neq = 0, g \neq = 4

5 (a + 3) + 4 a - 5 = - (2 a - 4)

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

3 (2 - x) + 4 x = 6 - [3 - (2 - 2 x)]

M = \frac{( 1 + i ) ^{1} 1}{( 32 ( 1 - i ) )}

c, 3 c + 4 d = 5

What is the answer to solvefor k,g^{-1}(4k)=k+1 ?
The answer to solvefor k,g^{-1}(4k)=k+1 is k= g/(4-g);\quad g\ne 0,g\ne 4