Solve 1/c(2x+1)=x+3

$$\frac { 1 } { c } ( 2 x + 1 ) = x + 3$$

Solve for c

$$c=\frac{2x+1}{x+3}$$

$$x\neq -\frac{1}{2}\text{ and }x\neq -3$$

Steps for Solving Linear Equation

Variable c cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by c.

$$1\left(2x+1\right)=cx+c\times 3$$

Use the distributive property to multiply 1 by 2x+1.

$$2x+1=cx+c\times 3$$

Swap sides so that all variable terms are on the left hand side.

$$cx+c\times 3=2x+1$$

Combine all terms containing c.

$$\left(x+3\right)c=2x+1$$

Divide both sides by x+3.

$$\frac{\left(x+3\right)c}{x+3}=\frac{2x+1}{x+3}$$

Dividing by x+3 undoes the multiplication by x+3.

$$c=\frac{2x+1}{x+3}$$

Variable c cannot be equal to 0.

$$c=\frac{2x+1}{x+3}\text{, }c\neq 0$$

Solve for x

$$x=-\frac{1-3c}{2-c}$$

$$c\neq 2\text{ and }c\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by c.

$$1\left(2x+1\right)=cx+c\times 3$$

Use the distributive property to multiply 1 by 2x+1.

$$2x+1=cx+c\times 3$$

Subtract cx from both sides.

$$2x+1-cx=c\times 3$$

Subtract 1 from both sides.

$$2x-cx=c\times 3-1$$

Combine all terms containing x.

$$\left(2-c\right)x=c\times 3-1$$

The equation is in standard form.

$$\left(2-c\right)x=3c-1$$

Divide both sides by 2-c.

$$\frac{\left(2-c\right)x}{2-c}=\frac{3c-1}{2-c}$$

Dividing by 2-c undoes the multiplication by 2-c.

$$x=\frac{3c-1}{2-c}$$