Solve 1/r+3=r+4/r-2+6/r-2

$$\frac { 1 } { r + 3 } = \frac { r + 4 } { r - 2 } + \frac { 6 } { r - 2 }$$

Solve for r

$$r=-8$$

$$r=-4$$

Steps Using the Quadratic Formula

Variable r cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(r-2\right)\left(r+3\right), the least common multiple of r+3,r-2.

$$r-2=\left(r+3\right)\left(r+4\right)+\left(r+3\right)\times 6$$

Use the distributive property to multiply r+3 by r+4 and combine like terms.

$$r-2=r^{2}+7r+12+\left(r+3\right)\times 6$$

Use the distributive property to multiply r+3 by 6.

$$r-2=r^{2}+7r+12+6r+18$$

Combine 7r and 6r to get 13r.

$$r-2=r^{2}+13r+12+18$$

Add 12 and 18 to get 30.

$$r-2=r^{2}+13r+30$$

Subtract r^{2} from both sides.

$$r-2-r^{2}=13r+30$$

Subtract 13r from both sides.

$$r-2-r^{2}-13r=30$$

Combine r and -13r to get -12r.

$$-12r-2-r^{2}=30$$

Subtract 30 from both sides.

$$-12r-2-r^{2}-30=0$$

Subtract 30 from -2 to get -32.

$$-12r-32-r^{2}=0$$

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.

$$-r^{2}-12r-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -12 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\left(-1\right)\left(-32\right)}}{2\left(-1\right)}$$

Square -12.

$$r=\frac{-\left(-12\right)±\sqrt{144-4\left(-1\right)\left(-32\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$r=\frac{-\left(-12\right)±\sqrt{144+4\left(-32\right)}}{2\left(-1\right)}$$

Multiply 4 times -32.

$$r=\frac{-\left(-12\right)±\sqrt{144-128}}{2\left(-1\right)}$$

Add 144 to -128.

$$r=\frac{-\left(-12\right)±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$r=\frac{-\left(-12\right)±4}{2\left(-1\right)}$$

The opposite of -12 is 12.

$$r=\frac{12±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$r=\frac{12±4}{-2}$$

Now solve the equation r=\frac{12±4}{-2} when ± is plus. Add 12 to 4.

$$r=\frac{16}{-2}$$

Divide 16 by -2.

$$r=-8$$

Now solve the equation r=\frac{12±4}{-2} when ± is minus. Subtract 4 from 12.

$$r=\frac{8}{-2}$$

Divide 8 by -2.

$$r=-4$$

The equation is now solved.

$$r=-8 r=-4$$

Steps for Completing the Square

Variable r cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(r-2\right)\left(r+3\right), the least common multiple of r+3,r-2.

$$r-2=\left(r+3\right)\left(r+4\right)+\left(r+3\right)\times 6$$

Use the distributive property to multiply r+3 by r+4 and combine like terms.

$$r-2=r^{2}+7r+12+\left(r+3\right)\times 6$$

Use the distributive property to multiply r+3 by 6.

$$r-2=r^{2}+7r+12+6r+18$$

Combine 7r and 6r to get 13r.

$$r-2=r^{2}+13r+12+18$$

Add 12 and 18 to get 30.

$$r-2=r^{2}+13r+30$$

Subtract r^{2} from both sides.

$$r-2-r^{2}=13r+30$$

Subtract 13r from both sides.

$$r-2-r^{2}-13r=30$$

Combine r and -13r to get -12r.

$$-12r-2-r^{2}=30$$

Add 2 to both sides.

$$-12r-r^{2}=30+2$$

Add 30 and 2 to get 32.

$$-12r-r^{2}=32$$

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.

$$-r^{2}-12r=32$$

Divide both sides by -1.

$$\frac{-r^{2}-12r}{-1}=\frac{32}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$r^{2}+\left(-\frac{12}{-1}\right)r=\frac{32}{-1}$$

Divide -12 by -1.

$$r^{2}+12r=\frac{32}{-1}$$

Divide 32 by -1.

$$r^{2}+12r=-32$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+12r+6^{2}=-32+6^{2}$$

Square 6.

$$r^{2}+12r+36=-32+36$$

Add -32 to 36.

$$r^{2}+12r+36=4$$

Factor r^{2}+12r+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}.

$$\left(r+6\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+6\right)^{2}}=\sqrt{4}$$

Simplify.

$$r+6=2 r+6=-2$$

Subtract 6 from both sides of the equation.

$$r=-4 r=-8$$