Solve 2/5(n-5)=1/30+1/(n-5)(n-6)

$$\frac { 2 } { 5 ( n - 5 ) } = \frac { 1 } { 30 } + \frac { 1 } { ( n - 5 ) ( n - 6 ) }$$

Solve for n

$$n=11$$

$$n=12$$

Steps Using the Quadratic Formula

Variable n cannot be equal to any of the values 5,6 since division by zero is not defined. Multiply both sides of the equation by 30\left(n-6\right)\left(n-5\right), the least common multiple of 5\left(n-5\right),30,\left(n-5\right)\left(n-6\right).

$$\left(6n-36\right)\times 2=30\left(n-6\right)\left(n-5\right)\times \frac{1}{30}+30$$

Use the distributive property to multiply 6n-36 by 2.

$$12n-72=30\left(n-6\right)\left(n-5\right)\times \frac{1}{30}+30$$

Multiply 30 and \frac{1}{30} to get 1.

$$12n-72=\left(n-6\right)\left(n-5\right)+30$$

Use the distributive property to multiply n-6 by n-5 and combine like terms.

$$12n-72=n^{2}-11n+30+30$$

Add 30 and 30 to get 60.

$$12n-72=n^{2}-11n+60$$

Subtract n^{2} from both sides.

$$12n-72-n^{2}=-11n+60$$

Add 11n to both sides.

$$12n-72-n^{2}+11n=60$$

Combine 12n and 11n to get 23n.

$$23n-72-n^{2}=60$$

Subtract 60 from both sides.

$$23n-72-n^{2}-60=0$$

Subtract 60 from -72 to get -132.

$$23n-132-n^{2}=0$$

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.

$$-n^{2}+23n-132=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 23 for b, and -132 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-23±\sqrt{23^{2}-4\left(-1\right)\left(-132\right)}}{2\left(-1\right)}$$

Square 23.

$$n=\frac{-23±\sqrt{529-4\left(-1\right)\left(-132\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$n=\frac{-23±\sqrt{529+4\left(-132\right)}}{2\left(-1\right)}$$

Multiply 4 times -132.

$$n=\frac{-23±\sqrt{529-528}}{2\left(-1\right)}$$

Add 529 to -528.

$$n=\frac{-23±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$n=\frac{-23±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$n=\frac{-23±1}{-2}$$

Now solve the equation n=\frac{-23±1}{-2} when ± is plus. Add -23 to 1.

$$n=-\frac{22}{-2}$$

Divide -22 by -2.

$$n=11$$

Now solve the equation n=\frac{-23±1}{-2} when ± is minus. Subtract 1 from -23.

$$n=-\frac{24}{-2}$$

Divide -24 by -2.

$$n=12$$

The equation is now solved.

$$n=11 n=12$$

Steps for Completing the Square

Variable n cannot be equal to any of the values 5,6 since division by zero is not defined. Multiply both sides of the equation by 30\left(n-6\right)\left(n-5\right), the least common multiple of 5\left(n-5\right),30,\left(n-5\right)\left(n-6\right).

$$\left(6n-36\right)\times 2=30\left(n-6\right)\left(n-5\right)\times \frac{1}{30}+30$$

Use the distributive property to multiply 6n-36 by 2.

$$12n-72=30\left(n-6\right)\left(n-5\right)\times \frac{1}{30}+30$$

Multiply 30 and \frac{1}{30} to get 1.

$$12n-72=\left(n-6\right)\left(n-5\right)+30$$

Use the distributive property to multiply n-6 by n-5 and combine like terms.

$$12n-72=n^{2}-11n+30+30$$

Add 30 and 30 to get 60.

$$12n-72=n^{2}-11n+60$$

Subtract n^{2} from both sides.

$$12n-72-n^{2}=-11n+60$$

Add 11n to both sides.

$$12n-72-n^{2}+11n=60$$

Combine 12n and 11n to get 23n.

$$23n-72-n^{2}=60$$

Add 72 to both sides.

$$23n-n^{2}=60+72$$

Add 60 and 72 to get 132.

$$23n-n^{2}=132$$

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.

$$-n^{2}+23n=132$$

Divide both sides by -1.

$$\frac{-n^{2}+23n}{-1}=\frac{132}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$n^{2}+\frac{23}{-1}n=\frac{132}{-1}$$

Divide 23 by -1.

$$n^{2}-23n=\frac{132}{-1}$$

Divide 132 by -1.

$$n^{2}-23n=-132$$

Divide -23, the coefficient of the x term, by 2 to get -\frac{23}{2}. Then add the square of -\frac{23}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-23n+\left(-\frac{23}{2}\right)^{2}=-132+\left(-\frac{23}{2}\right)^{2}$$

Square -\frac{23}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-23n+\frac{529}{4}=-132+\frac{529}{4}$$

Add -132 to \frac{529}{4}.

$$n^{2}-23n+\frac{529}{4}=\frac{1}{4}$$

Factor n^{2}-23n+\frac{529}{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}.

$$\left(n-\frac{23}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{23}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$n-\frac{23}{2}=\frac{1}{2} n-\frac{23}{2}=-\frac{1}{2}$$

Add \frac{23}{2} to both sides of the equation.

$$n=12 n=11$$