Solve 1+1/alpha/1-frac{1{beta}}=3

$$\frac { 1 + \frac { 1 } { \alpha } } { 1 - \frac { 1 } { \beta } } = 3$$

Solve for α

$$\alpha =-\frac{\beta }{3-2\beta }$$

$$\beta \neq 0\text{ and }\beta \neq \frac{3}{2}\text{ and }\beta \neq 1$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\alpha }{\alpha }.

$$\frac{\frac{\alpha }{\alpha }+\frac{1}{\alpha }}{1-\frac{1}{\beta }}=3$$

Since \frac{\alpha }{\alpha } and \frac{1}{\alpha } have the same denominator, add them by adding their numerators.

$$\frac{\frac{\alpha +1}{\alpha }}{1-\frac{1}{\beta }}=3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\beta }{\beta }.

$$\frac{\frac{\alpha +1}{\alpha }}{\frac{\beta }{\beta }-\frac{1}{\beta }}=3$$

Since \frac{\beta }{\beta } and \frac{1}{\beta } have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\alpha +1}{\alpha }}{\frac{\beta -1}{\beta }}=3$$

Divide \frac{\alpha +1}{\alpha } by \frac{\beta -1}{\beta } by multiplying \frac{\alpha +1}{\alpha } by the reciprocal of \frac{\beta -1}{\beta }.

$$\frac{\left(\alpha +1\right)\beta }{\alpha \left(\beta -1\right)}=3$$

Use the distributive property to multiply \alpha +1 by \beta .

$$\frac{\alpha \beta +\beta }{\alpha \left(\beta -1\right)}=3$$

Use the distributive property to multiply \alpha by \beta -1.

$$\frac{\alpha \beta +\beta }{\alpha \beta -\alpha }=3$$

Variable \alpha cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by \alpha \left(\beta -1\right).

$$\alpha \beta +\beta =3\alpha \left(\beta -1\right)$$

Use the distributive property to multiply 3\alpha by \beta -1.

$$\alpha \beta +\beta =3\alpha \beta -3\alpha$$

Subtract 3\alpha \beta from both sides.

$$\alpha \beta +\beta -3\alpha \beta =-3\alpha$$

Combine \alpha \beta and -3\alpha \beta to get -2\alpha \beta .

$$-2\alpha \beta +\beta =-3\alpha$$

Add 3\alpha to both sides.

$$-2\alpha \beta +\beta +3\alpha =0$$

Subtract \beta from both sides. Anything subtracted from zero gives its negation.

$$-2\alpha \beta +3\alpha =-\beta$$

Combine all terms containing \alpha .

$$\left(-2\beta +3\right)\alpha =-\beta$$

The equation is in standard form.

$$\left(3-2\beta \right)\alpha =-\beta$$

Divide both sides by 3-2\beta .

$$\frac{\left(3-2\beta \right)\alpha }{3-2\beta }=-\frac{\beta }{3-2\beta }$$

Dividing by 3-2\beta undoes the multiplication by 3-2\beta .

$$\alpha =-\frac{\beta }{3-2\beta }$$

Variable \alpha cannot be equal to 0.

$$\alpha =-\frac{\beta }{3-2\beta }\text{, }\alpha \neq 0$$

Solve for β

$$\beta =-\frac{3\alpha }{1-2\alpha }$$

$$\alpha \neq -1\text{ and }\alpha \neq \frac{1}{2}\text{ and }\alpha \neq 0$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\alpha }{\alpha }.

$$\frac{\frac{\alpha }{\alpha }+\frac{1}{\alpha }}{1-\frac{1}{\beta }}=3$$

Since \frac{\alpha }{\alpha } and \frac{1}{\alpha } have the same denominator, add them by adding their numerators.

$$\frac{\frac{\alpha +1}{\alpha }}{1-\frac{1}{\beta }}=3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\beta }{\beta }.

$$\frac{\frac{\alpha +1}{\alpha }}{\frac{\beta }{\beta }-\frac{1}{\beta }}=3$$

Since \frac{\beta }{\beta } and \frac{1}{\beta } have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\alpha +1}{\alpha }}{\frac{\beta -1}{\beta }}=3$$

Variable \beta cannot be equal to 0 since division by zero is not defined. Divide \frac{\alpha +1}{\alpha } by \frac{\beta -1}{\beta } by multiplying \frac{\alpha +1}{\alpha } by the reciprocal of \frac{\beta -1}{\beta }.

$$\frac{\left(\alpha +1\right)\beta }{\alpha \left(\beta -1\right)}=3$$

Use the distributive property to multiply \alpha +1 by \beta .

$$\frac{\alpha \beta +\beta }{\alpha \left(\beta -1\right)}=3$$

Use the distributive property to multiply \alpha by \beta -1.

$$\frac{\alpha \beta +\beta }{\alpha \beta -\alpha }=3$$

Variable \beta cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by \alpha \left(\beta -1\right).

$$\alpha \beta +\beta =3\alpha \left(\beta -1\right)$$

Use the distributive property to multiply 3\alpha by \beta -1.

$$\alpha \beta +\beta =3\alpha \beta -3\alpha$$

Subtract 3\alpha \beta from both sides.

$$\alpha \beta +\beta -3\alpha \beta =-3\alpha$$

Combine \alpha \beta and -3\alpha \beta to get -2\alpha \beta .

$$-2\alpha \beta +\beta =-3\alpha$$

Combine all terms containing \beta .

$$\left(-2\alpha +1\right)\beta =-3\alpha$$

The equation is in standard form.

$$\left(1-2\alpha \right)\beta =-3\alpha$$

Divide both sides by -2\alpha +1.

$$\frac{\left(1-2\alpha \right)\beta }{1-2\alpha }=-\frac{3\alpha }{1-2\alpha }$$

Dividing by -2\alpha +1 undoes the multiplication by -2\alpha +1.

$$\beta =-\frac{3\alpha }{1-2\alpha }$$

Variable \beta cannot be equal to any of the values 1,0.

$$\beta =-\frac{3\alpha }{1-2\alpha }\text{, }\beta \neq 1\text{ and }\beta \neq 0$$