Solve C=5+3x/20+4x

$$C = \frac { 5 + 3 x } { 20 + 4 x }$$

Solve for x

$$x=-\frac{5\left(4C-1\right)}{4C-3}$$

$$C\neq \frac{3}{4}$$

Steps for Solving Linear Equation

Variable x cannot be equal to -5 since division by zero is not defined. Multiply both sides of the equation by 4\left(x+5\right).

$$C\times 4\left(x+5\right)=5+3x$$

Use the distributive property to multiply C\times 4 by x+5.

$$4Cx+5C\times 4=5+3x$$

Multiply 5 and 4 to get 20.

$$4Cx+20C=5+3x$$

Subtract 3x from both sides.

$$4Cx+20C-3x=5$$

Subtract 20C from both sides.

$$4Cx-3x=5-20C$$

Combine all terms containing x.

$$\left(4C-3\right)x=5-20C$$

Divide both sides by 4C-3.

$$\frac{\left(4C-3\right)x}{4C-3}=\frac{5-20C}{4C-3}$$

Dividing by 4C-3 undoes the multiplication by 4C-3.

$$x=\frac{5-20C}{4C-3}$$

Divide 5-20C by 4C-3.

$$x=\frac{5\left(1-4C\right)}{4C-3}$$

Variable x cannot be equal to -5.

$$x=\frac{5\left(1-4C\right)}{4C-3}\text{, }x\neq -5$$

Solve for C

$$C=\frac{3x+5}{4\left(x+5\right)}$$

$$x\neq -5$$