Solve 4k/k^2+x=10

$$\frac { 4 k } { k ^ { 2 } } + x = 10$$

Solve for x

$$x=10-\frac{4}{k}$$

$$k\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by k^{2}.

$$4k+k^{2}x=10k^{2}$$

Subtract 4k from both sides.

$$k^{2}x=10k^{2}-4k$$

Divide both sides by k^{2}.

$$\frac{k^{2}x}{k^{2}}=\frac{2k\left(5k-2\right)}{k^{2}}$$

Dividing by k^{2} undoes the multiplication by k^{2}.

$$x=\frac{2k\left(5k-2\right)}{k^{2}}$$

Divide 2k\left(-2+5k\right) by k^{2}.

$$x=10-\frac{4}{k}$$

Solve for k

$$k=-\frac{4}{x-10}$$

$$x\neq 10$$