Solve a/b+3=c

$$\frac { a } { b } + 3 = c$$

Solve for a

$$a=b\left(c-3\right)$$

$$b\neq 0$$

Solution Steps

Multiply both sides of the equation by b.

$$a+b\times 3=cb$$

Subtract b\times 3 from both sides.

$$a=cb-b\times 3$$

Multiply -1 and 3 to get -3.

$$a=cb-3b$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{a}{3-c}\text{, }&a\neq 0\text{ and }c\neq 3\\b\neq 0\text{, }&c=3\text{ and }a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$a+b\times 3=cb$$

Subtract cb from both sides.

$$a+b\times 3-cb=0$$

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

$$b\times 3-cb=-a$$

Combine all terms containing b.

$$\left(3-c\right)b=-a$$

Divide both sides by 3-c.

$$\frac{\left(3-c\right)b}{3-c}=-\frac{a}{3-c}$$

Dividing by 3-c undoes the multiplication by 3-c.

$$b=-\frac{a}{3-c}$$

Variable b cannot be equal to 0.

$$b=-\frac{a}{3-c}\text{, }b\neq 0$$