Solve 2/a-2=3a-1/2a+11

$$\frac { 2 } { a - 2 } = \frac { 3 a - 1 } { 2 a + 11 }$$

Solve for a

$$a=5$$

$$a = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Steps Using the Quadratic Formula

Variable a cannot be equal to any of the values -\frac{11}{2},2 since division by zero is not defined. Multiply both sides of the equation by \left(a-2\right)\left(2a+11\right), the least common multiple of a-2,2a+11.

$$\left(2a+11\right)\times 2=\left(a-2\right)\left(3a-1\right)$$

Use the distributive property to multiply 2a+11 by 2.

$$4a+22=\left(a-2\right)\left(3a-1\right)$$

Use the distributive property to multiply a-2 by 3a-1 and combine like terms.

$$4a+22=3a^{2}-7a+2$$

Subtract 3a^{2} from both sides.

$$4a+22-3a^{2}=-7a+2$$

Add 7a to both sides.

$$4a+22-3a^{2}+7a=2$$

Combine 4a and 7a to get 11a.

$$11a+22-3a^{2}=2$$

Subtract 2 from both sides.

$$11a+22-3a^{2}-2=0$$

Subtract 2 from 22 to get 20.

$$11a+20-3a^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-3a^{2}+11a+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 11 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-11±\sqrt{11^{2}-4\left(-3\right)\times 20}}{2\left(-3\right)}$$

Square 11.

$$a=\frac{-11±\sqrt{121-4\left(-3\right)\times 20}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$a=\frac{-11±\sqrt{121+12\times 20}}{2\left(-3\right)}$$

Multiply 12 times 20.

$$a=\frac{-11±\sqrt{121+240}}{2\left(-3\right)}$$

Add 121 to 240.

$$a=\frac{-11±\sqrt{361}}{2\left(-3\right)}$$

Take the square root of 361.

$$a=\frac{-11±19}{2\left(-3\right)}$$

Multiply 2 times -3.

$$a=\frac{-11±19}{-6}$$

Now solve the equation a=\frac{-11±19}{-6} when ± is plus. Add -11 to 19.

$$a=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

$$a=-\frac{4}{3}$$

Now solve the equation a=\frac{-11±19}{-6} when ± is minus. Subtract 19 from -11.

$$a=-\frac{30}{-6}$$

Divide -30 by -6.

$$a=5$$

The equation is now solved.

$$a=-\frac{4}{3} a=5$$

Steps for Completing the Square

Variable a cannot be equal to any of the values -\frac{11}{2},2 since division by zero is not defined. Multiply both sides of the equation by \left(a-2\right)\left(2a+11\right), the least common multiple of a-2,2a+11.

$$\left(2a+11\right)\times 2=\left(a-2\right)\left(3a-1\right)$$

Use the distributive property to multiply 2a+11 by 2.

$$4a+22=\left(a-2\right)\left(3a-1\right)$$

Use the distributive property to multiply a-2 by 3a-1 and combine like terms.

$$4a+22=3a^{2}-7a+2$$

Subtract 3a^{2} from both sides.

$$4a+22-3a^{2}=-7a+2$$

Add 7a to both sides.

$$4a+22-3a^{2}+7a=2$$

Combine 4a and 7a to get 11a.

$$11a+22-3a^{2}=2$$

Subtract 22 from both sides.

$$11a-3a^{2}=2-22$$

Subtract 22 from 2 to get -20.

$$11a-3a^{2}=-20$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-3a^{2}+11a=-20$$

Divide both sides by -3.

$$\frac{-3a^{2}+11a}{-3}=-\frac{20}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$a^{2}+\frac{11}{-3}a=-\frac{20}{-3}$$

Divide 11 by -3.

$$a^{2}-\frac{11}{3}a=-\frac{20}{-3}$$

Divide -20 by -3.

$$a^{2}-\frac{11}{3}a=\frac{20}{3}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{11}{3}a+\left(-\frac{11}{6}\right)^{2}=\frac{20}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{11}{3}a+\frac{121}{36}=\frac{20}{3}+\frac{121}{36}$$

Add \frac{20}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{11}{3}a+\frac{121}{36}=\frac{361}{36}$$

Factor a^{2}-\frac{11}{3}a+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-\frac{11}{6}\right)^{2}=\frac{361}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{11}{6}\right)^{2}}=\sqrt{\frac{361}{36}}$$

Simplify.

$$a-\frac{11}{6}=\frac{19}{6} a-\frac{11}{6}=-\frac{19}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$a=5 a=-\frac{4}{3}$$