Solve 1/a-1/b=1/c-2

$$\frac { 1 } { a } - \frac { 1 } { b } = \frac { 1 } { c - 2 }$$

Solve for a

$$a=\frac{b\left(2-c\right)}{2-c-b}$$

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

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\left(c-2\right), the least common multiple of a,b,c-2.

$$b\left(c-2\right)-a\left(c-2\right)=ab$$

Use the distributive property to multiply b by c-2.

$$bc-2b-a\left(c-2\right)=ab$$

Use the distributive property to multiply a by c-2.

$$bc-2b-\left(ac-2a\right)=ab$$

To find the opposite of ac-2a, find the opposite of each term.

$$bc-2b-ac+2a=ab$$

Subtract ab from both sides.

$$bc-2b-ac+2a-ab=0$$

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

$$-2b-ac+2a-ab=-bc$$

Add 2b to both sides.

$$-ac+2a-ab=-bc+2b$$

Reorder the terms.

$$-ab-ac+2a=-bc+2b$$

Combine all terms containing a.

$$\left(-b-c+2\right)a=-bc+2b$$

The equation is in standard form.

$$\left(2-c-b\right)a=2b-bc$$

Divide both sides by -b-c+2.

$$\frac{\left(2-c-b\right)a}{2-c-b}=\frac{b\left(2-c\right)}{2-c-b}$$

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

$$a=\frac{b\left(2-c\right)}{2-c-b}$$

Variable a cannot be equal to 0.

$$a=\frac{b\left(2-c\right)}{2-c-b}\text{, }a\neq 0$$

Solve for b

$$b=-\frac{a\left(2-c\right)}{-a+c-2}$$

$$c\neq 2\text{ and }a\neq 0\text{ and }a\neq -\left(2-c\right)$$

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 ab\left(c-2\right), the least common multiple of a,b,c-2.

$$b\left(c-2\right)-a\left(c-2\right)=ab$$

Use the distributive property to multiply b by c-2.

$$bc-2b-a\left(c-2\right)=ab$$

Use the distributive property to multiply a by c-2.

$$bc-2b-\left(ac-2a\right)=ab$$

To find the opposite of ac-2a, find the opposite of each term.

$$bc-2b-ac+2a=ab$$

Subtract ab from both sides.

$$bc-2b-ac+2a-ab=0$$

Add ac to both sides. Anything plus zero gives itself.

$$bc-2b+2a-ab=ac$$

Subtract 2a from both sides.

$$bc-2b-ab=ac-2a$$

Combine all terms containing b.

$$\left(c-2-a\right)b=ac-2a$$

The equation is in standard form.

$$\left(-a+c-2\right)b=ac-2a$$

Divide both sides by c-2-a.

$$\frac{\left(-a+c-2\right)b}{-a+c-2}=\frac{a\left(c-2\right)}{-a+c-2}$$

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

$$b=\frac{a\left(c-2\right)}{-a+c-2}$$

Variable b cannot be equal to 0.

$$b=\frac{a\left(c-2\right)}{-a+c-2}\text{, }b\neq 0$$