Solve 5/a+3/b=1/a*b

$$\frac { 5 } { a } + \frac { 3 } { b } = \frac { 1 } { a \cdot b }$$

Solve for a

$$a=\frac{1-5b}{3}$$

$$b\neq \frac{1}{5}\text{ and }b\neq 0$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ab, the least common multiple of a,b,ab.

$$b\times 5+a\times 3=1$$

Subtract b\times 5 from both sides.

$$a\times 3=1-b\times 5$$

Multiply -1 and 5 to get -5.

$$a\times 3=1-5b$$

The equation is in standard form.

$$3a=1-5b$$

Divide both sides by 3.

$$\frac{3a}{3}=\frac{1-5b}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a=\frac{1-5b}{3}$$

Variable a cannot be equal to 0.

$$a=\frac{1-5b}{3}\text{, }a\neq 0$$

Solve for b

$$b=\frac{1-3a}{5}$$

$$a\neq \frac{1}{3}\text{ and }a\neq 0$$

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 ab, the least common multiple of a,b,ab.

$$b\times 5+a\times 3=1$$

Subtract a\times 3 from both sides.

$$b\times 5=1-a\times 3$$

Multiply -1 and 3 to get -3.

$$b\times 5=1-3a$$

The equation is in standard form.

$$5b=1-3a$$

Divide both sides by 5.

$$\frac{5b}{5}=\frac{1-3a}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$b=\frac{1-3a}{5}$$

Variable b cannot be equal to 0.

$$b=\frac{1-3a}{5}\text{, }b\neq 0$$