Solve 4/d+cd/2=5c

$$\frac { 4 } { d } + \frac { c d } { 2 } = 5 c$$

Solve for c

$$c=-\frac{8}{d\left(d-10\right)}$$

$$d\neq 10\text{ and }d\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2d, the least common multiple of d,2.

$$2\times 4+dcd=5c\times 2d$$

Multiply 2 and 4 to get 8.

$$8+dcd=5c\times 2d$$

Multiply d and d to get d^{2}.

$$8+d^{2}c=5c\times 2d$$

Multiply 5 and 2 to get 10.

$$8+d^{2}c=10cd$$

Subtract 10cd from both sides.

$$8+d^{2}c-10cd=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$d^{2}c-10cd=-8$$

Combine all terms containing c.

$$\left(d^{2}-10d\right)c=-8$$

Divide both sides by d^{2}-10d.

$$\frac{\left(d^{2}-10d\right)c}{d^{2}-10d}=-\frac{8}{d^{2}-10d}$$

Dividing by d^{2}-10d undoes the multiplication by d^{2}-10d.

$$c=-\frac{8}{d^{2}-10d}$$

Divide -8 by d^{2}-10d.

$$c=-\frac{8}{d\left(d-10\right)}$$

Solve for d

$$d=\frac{\sqrt{c\left(25c-8\right)}+5c}{c}$$

$$d=\frac{-\sqrt{c\left(25c-8\right)}+5c}{c}\text{, }c<0\text{ or }c\geq \frac{8}{25}$$