Solve 5/x+1+9/x-3=2

$$\frac { 5 } { x + 1 } + \frac { 9 } { x - 3 } = 2$$

Solve for x

$$x=9$$

$$x=0$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+1\right), the least common multiple of x+1,x-3.

$$\left(x-3\right)\times 5+\left(x+1\right)\times 9=2\left(x-3\right)\left(x+1\right)$$

Use the distributive property to multiply x-3 by 5.

$$5x-15+\left(x+1\right)\times 9=2\left(x-3\right)\left(x+1\right)$$

Use the distributive property to multiply x+1 by 9.

$$5x-15+9x+9=2\left(x-3\right)\left(x+1\right)$$

Combine 5x and 9x to get 14x.

$$14x-15+9=2\left(x-3\right)\left(x+1\right)$$

Add -15 and 9 to get -6.

$$14x-6=2\left(x-3\right)\left(x+1\right)$$

Use the distributive property to multiply 2 by x-3.

$$14x-6=\left(2x-6\right)\left(x+1\right)$$

Use the distributive property to multiply 2x-6 by x+1 and combine like terms.

$$14x-6=2x^{2}-4x-6$$

Subtract 2x^{2} from both sides.

$$14x-6-2x^{2}=-4x-6$$

Add 4x to both sides.

$$14x-6-2x^{2}+4x=-6$$

Combine 14x and 4x to get 18x.

$$18x-6-2x^{2}=-6$$

Add 6 to both sides.

$$18x-6-2x^{2}+6=0$$

Add -6 and 6 to get 0.

$$18x-2x^{2}=0$$

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.

$$-2x^{2}+18x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 18 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}}}{2\left(-2\right)}$$

Take the square root of 18^{2}.

$$x=\frac{-18±18}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-18±18}{-4}$$

Now solve the equation x=\frac{-18±18}{-4} when ± is plus. Add -18 to 18.

$$x=\frac{0}{-4}$$

Divide 0 by -4.

$$x=0$$

Now solve the equation x=\frac{-18±18}{-4} when ± is minus. Subtract 18 from -18.

$$x=-\frac{36}{-4}$$

Divide -36 by -4.

$$x=9$$

The equation is now solved.

$$x=0 x=9$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+1\right), the least common multiple of x+1,x-3.

$$\left(x-3\right)\times 5+\left(x+1\right)\times 9=2\left(x-3\right)\left(x+1\right)$$

Use the distributive property to multiply x-3 by 5.

$$5x-15+\left(x+1\right)\times 9=2\left(x-3\right)\left(x+1\right)$$

Use the distributive property to multiply x+1 by 9.

$$5x-15+9x+9=2\left(x-3\right)\left(x+1\right)$$

Combine 5x and 9x to get 14x.

$$14x-15+9=2\left(x-3\right)\left(x+1\right)$$

Add -15 and 9 to get -6.

$$14x-6=2\left(x-3\right)\left(x+1\right)$$

Use the distributive property to multiply 2 by x-3.

$$14x-6=\left(2x-6\right)\left(x+1\right)$$

Use the distributive property to multiply 2x-6 by x+1 and combine like terms.

$$14x-6=2x^{2}-4x-6$$

Subtract 2x^{2} from both sides.

$$14x-6-2x^{2}=-4x-6$$

Add 4x to both sides.

$$14x-6-2x^{2}+4x=-6$$

Combine 14x and 4x to get 18x.

$$18x-6-2x^{2}=-6$$

Add 6 to both sides.

$$18x-2x^{2}=-6+6$$

Add -6 and 6 to get 0.

$$18x-2x^{2}=0$$

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.

$$-2x^{2}+18x=0$$

Divide both sides by -2.

$$\frac{-2x^{2}+18x}{-2}=\frac{0}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{18}{-2}x=\frac{0}{-2}$$

Divide 18 by -2.

$$x^{2}-9x=\frac{0}{-2}$$

Divide 0 by -2.

$$x^{2}-9x=0$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=\frac{81}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{9}{2} x-\frac{9}{2}=-\frac{9}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=9 x=0$$