Solve 5k^2-26k+5/25k^3-k=k-5/k(5k+1)

$$\frac { 5 k ^ { 2 } - 26 k + 5 } { 25 k ^ { 3 } - k } = \frac { k - 5 } { k ( 5 k + 1 ) }$$

Solve for k (complex solution)

$$k\in \mathrm{C}\setminus -\frac{1}{5},\frac{1}{5},0$$

Solution Steps

Variable k cannot be equal to any of the values -\frac{1}{5},0,\frac{1}{5} since division by zero is not defined. Multiply both sides of the equation by k\left(5k-1\right)\left(5k+1\right), the least common multiple of 25k^{3}-k,k\left(5k+1\right).

$$5k^{2}-26k+5=\left(5k-1\right)\left(k-5\right)$$

Use the distributive property to multiply 5k-1 by k-5 and combine like terms.

$$5k^{2}-26k+5=5k^{2}-26k+5$$

Subtract 5k^{2} from both sides.

$$5k^{2}-26k+5-5k^{2}=-26k+5$$

Combine 5k^{2} and -5k^{2} to get 0.

$$-26k+5=-26k+5$$

Add 26k to both sides.

$$-26k+5+26k=5$$

Combine -26k and 26k to get 0.

$$5=5$$

Compare 5 and 5.

$$\text{true}$$

This is true for any k.

$$k\in \mathrm{C}$$

Variable k cannot be equal to any of the values -\frac{1}{5},\frac{1}{5},0.

$$k\in \mathrm{C}\setminus -\frac{1}{5},0,\frac{1}{5}$$

Solve for k

$$k\in \mathrm{R}\setminus \frac{1}{5},-\frac{1}{5},0$$

Solution Steps

Variable k cannot be equal to any of the values -\frac{1}{5},0,\frac{1}{5} since division by zero is not defined. Multiply both sides of the equation by k\left(5k-1\right)\left(5k+1\right), the least common multiple of 25k^{3}-k,k\left(5k+1\right).

$$5k^{2}-26k+5=\left(5k-1\right)\left(k-5\right)$$

Use the distributive property to multiply 5k-1 by k-5 and combine like terms.

$$5k^{2}-26k+5=5k^{2}-26k+5$$

Subtract 5k^{2} from both sides.

$$5k^{2}-26k+5-5k^{2}=-26k+5$$

Combine 5k^{2} and -5k^{2} to get 0.

$$-26k+5=-26k+5$$

Add 26k to both sides.

$$-26k+5+26k=5$$

Combine -26k and 26k to get 0.

$$5=5$$

Compare 5 and 5.

$$\text{true}$$

This is true for any k.

$$k\in \mathrm{R}$$

Variable k cannot be equal to any of the values -\frac{1}{5},\frac{1}{5},0.

$$k\in \mathrm{R}\setminus -\frac{1}{5},0,\frac{1}{5}$$