Solve 1-3/2-a=a-1/5-2

$$\frac { 1 - 3 } { 2 - a } = \frac { a - 1 } { 5 - 2 }$$

Solve for a

$$a=-1$$

$$a=4$$

Steps Using the Quadratic Formula

Variable a cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by -a+2.

$$1-3=\left(-\frac{1}{3}a+\frac{2}{3}\right)\left(a-1\right)$$

Subtract 3 from 1 to get -2.

$$-2=\left(-\frac{1}{3}a+\frac{2}{3}\right)\left(a-1\right)$$

Use the distributive property to multiply -\frac{1}{3}a+\frac{2}{3} by a-1 and combine like terms.

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

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

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

Add 2 to both sides.

$$-\frac{1}{3}a^{2}+a-\frac{2}{3}+2=0$$

Add -\frac{2}{3} and 2 to get \frac{4}{3}.

$$-\frac{1}{3}a^{2}+a+\frac{4}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{3} for a, 1 for b, and \frac{4}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-1±\sqrt{1^{2}-4\left(-\frac{1}{3}\right)\times \frac{4}{3}}}{2\left(-\frac{1}{3}\right)}$$

Square 1.

$$a=\frac{-1±\sqrt{1-4\left(-\frac{1}{3}\right)\times \frac{4}{3}}}{2\left(-\frac{1}{3}\right)}$$

Multiply -4 times -\frac{1}{3}.

$$a=\frac{-1±\sqrt{1+\frac{4}{3}\times \frac{4}{3}}}{2\left(-\frac{1}{3}\right)}$$

Multiply \frac{4}{3} times \frac{4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=\frac{-1±\sqrt{1+\frac{16}{9}}}{2\left(-\frac{1}{3}\right)}$$

Add 1 to \frac{16}{9}.

$$a=\frac{-1±\sqrt{\frac{25}{9}}}{2\left(-\frac{1}{3}\right)}$$

Take the square root of \frac{25}{9}.

$$a=\frac{-1±\frac{5}{3}}{2\left(-\frac{1}{3}\right)}$$

Multiply 2 times -\frac{1}{3}.

$$a=\frac{-1±\frac{5}{3}}{-\frac{2}{3}}$$

Now solve the equation a=\frac{-1±\frac{5}{3}}{-\frac{2}{3}} when ± is plus. Add -1 to \frac{5}{3}.

$$a=\frac{\frac{2}{3}}{-\frac{2}{3}}$$

Divide \frac{2}{3} by -\frac{2}{3} by multiplying \frac{2}{3} by the reciprocal of -\frac{2}{3}.

$$a=-1$$

Now solve the equation a=\frac{-1±\frac{5}{3}}{-\frac{2}{3}} when ± is minus. Subtract \frac{5}{3} from -1.

$$a=-\frac{\frac{8}{3}}{-\frac{2}{3}}$$

Divide -\frac{8}{3} by -\frac{2}{3} by multiplying -\frac{8}{3} by the reciprocal of -\frac{2}{3}.

$$a=4$$

The equation is now solved.

$$a=-1 a=4$$

Steps for Completing the Square

Variable a cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by -a+2.

$$1-3=\left(-\frac{1}{3}a+\frac{2}{3}\right)\left(a-1\right)$$

Subtract 3 from 1 to get -2.

$$-2=\left(-\frac{1}{3}a+\frac{2}{3}\right)\left(a-1\right)$$

Use the distributive property to multiply -\frac{1}{3}a+\frac{2}{3} by a-1 and combine like terms.

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

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

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

Add \frac{2}{3} to both sides.

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

Add -2 and \frac{2}{3} to get -\frac{4}{3}.

$$-\frac{1}{3}a^{2}+a=-\frac{4}{3}$$

Multiply both sides by -3.

$$\frac{-\frac{1}{3}a^{2}+a}{-\frac{1}{3}}=-\frac{\frac{4}{3}}{-\frac{1}{3}}$$

Dividing by -\frac{1}{3} undoes the multiplication by -\frac{1}{3}.

$$a^{2}+\frac{1}{-\frac{1}{3}}a=-\frac{\frac{4}{3}}{-\frac{1}{3}}$$

Divide 1 by -\frac{1}{3} by multiplying 1 by the reciprocal of -\frac{1}{3}.

$$a^{2}-3a=-\frac{\frac{4}{3}}{-\frac{1}{3}}$$

Divide -\frac{4}{3} by -\frac{1}{3} by multiplying -\frac{4}{3} by the reciprocal of -\frac{1}{3}.

$$a^{2}-3a=4$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-3a+\left(-\frac{3}{2}\right)^{2}=4+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-3a+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$a^{2}-3a+\frac{9}{4}=\frac{25}{4}$$

Factor a^{2}-3a+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$a-\frac{3}{2}=\frac{5}{2} a-\frac{3}{2}=-\frac{5}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$a=4 a=-1$$