Solve 40*(-1/a+m/a)=2/3

$$40 \cdot ( - \frac { 1 } { a } + \frac { m } { a } ) = \frac { 2 } { 3 }$$

Solve for a

$$a=60\left(m-1\right)$$

$$m\neq 1$$

Solution Steps

Divide both sides by 40.

$$-\frac{1}{a}+\frac{m}{a}=\frac{\frac{2}{3}}{40}$$

Express \frac{\frac{2}{3}}{40} as a single fraction.

$$-\frac{1}{a}+\frac{m}{a}=\frac{2}{3\times 40}$$

Multiply 3 and 40 to get 120.

$$-\frac{1}{a}+\frac{m}{a}=\frac{2}{120}$$

Reduce the fraction \frac{2}{120} to lowest terms by extracting and canceling out 2.

$$-\frac{1}{a}+\frac{m}{a}=\frac{1}{60}$$

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 60a, the least common multiple of a,60.

$$-60+60m=a$$

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

$$a=-60+60m$$

Variable a cannot be equal to 0.

$$a=-60+60m\text{, }a\neq 0$$

Solve for m

$$m=\frac{a+60}{60}$$

$$a\neq 0$$

Steps for Solving Linear Equation

Divide both sides by 40.

$$-\frac{1}{a}+\frac{m}{a}=\frac{\frac{2}{3}}{40}$$

Express \frac{\frac{2}{3}}{40} as a single fraction.

$$-\frac{1}{a}+\frac{m}{a}=\frac{2}{3\times 40}$$

Multiply 3 and 40 to get 120.

$$-\frac{1}{a}+\frac{m}{a}=\frac{2}{120}$$

Reduce the fraction \frac{2}{120} to lowest terms by extracting and canceling out 2.

$$-\frac{1}{a}+\frac{m}{a}=\frac{1}{60}$$

Multiply both sides of the equation by 60a, the least common multiple of a,60.

$$-60+60m=a$$

Add 60 to both sides.

$$60m=a+60$$

Divide both sides by 60.

$$\frac{60m}{60}=\frac{a+60}{60}$$

Dividing by 60 undoes the multiplication by 60.

$$m=\frac{a+60}{60}$$

Divide a+60 by 60.

$$m=\frac{a}{60}+1$$