Solve x-4/x+1-10/x^2-1=4/9

$$\frac { x - 4 } { x + 1 } - \frac { 10 } { x ^ { 2 } - 1 } = \frac { 4 } { 9 }$$

Solve for x

$$x=10$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-1\right)\left(x+1\right), the least common multiple of x+1,x^{2}-1,9.

$$\left(9x-9\right)\left(x-4\right)-9\times 10=4\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 9x-9 by x-4 and combine like terms.

$$9x^{2}-45x+36-9\times 10=4\left(x-1\right)\left(x+1\right)$$

Multiply -9 and 10 to get -90.

$$9x^{2}-45x+36-90=4\left(x-1\right)\left(x+1\right)$$

Subtract 90 from 36 to get -54.

$$9x^{2}-45x-54=4\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 4 by x-1.

$$9x^{2}-45x-54=\left(4x-4\right)\left(x+1\right)$$

Use the distributive property to multiply 4x-4 by x+1 and combine like terms.

$$9x^{2}-45x-54=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$9x^{2}-45x-54-4x^{2}=-4$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-45x-54=-4$$

Add 4 to both sides.

$$5x^{2}-45x-54+4=0$$

Add -54 and 4 to get -50.

$$5x^{2}-45x-50=0$$

Divide both sides by 5.

$$x^{2}-9x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\left(-10\right)=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -9.

$$a=-10 b=1$$

Rewrite x^{2}-9x-10 as \left(x^{2}-10x\right)+\left(x-10\right).

$$\left(x^{2}-10x\right)+\left(x-10\right)$$

Factor out x in x^{2}-10x.

$$x\left(x-10\right)+x-10$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+1\right)$$

To find equation solutions, solve x-10=0 and x+1=0.

$$x=10 x=-1$$

Variable x cannot be equal to -1.

$$x=10$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-1\right)\left(x+1\right), the least common multiple of x+1,x^{2}-1,9.

$$\left(9x-9\right)\left(x-4\right)-9\times 10=4\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 9x-9 by x-4 and combine like terms.

$$9x^{2}-45x+36-9\times 10=4\left(x-1\right)\left(x+1\right)$$

Multiply -9 and 10 to get -90.

$$9x^{2}-45x+36-90=4\left(x-1\right)\left(x+1\right)$$

Subtract 90 from 36 to get -54.

$$9x^{2}-45x-54=4\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 4 by x-1.

$$9x^{2}-45x-54=\left(4x-4\right)\left(x+1\right)$$

Use the distributive property to multiply 4x-4 by x+1 and combine like terms.

$$9x^{2}-45x-54=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$9x^{2}-45x-54-4x^{2}=-4$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-45x-54=-4$$

Add 4 to both sides.

$$5x^{2}-45x-54+4=0$$

Add -54 and 4 to get -50.

$$5x^{2}-45x-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -45 for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-45\right)±\sqrt{\left(-45\right)^{2}-4\times 5\left(-50\right)}}{2\times 5}$$

Square -45.

$$x=\frac{-\left(-45\right)±\sqrt{2025-4\times 5\left(-50\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-45\right)±\sqrt{2025-20\left(-50\right)}}{2\times 5}$$

Multiply -20 times -50.

$$x=\frac{-\left(-45\right)±\sqrt{2025+1000}}{2\times 5}$$

Add 2025 to 1000.

$$x=\frac{-\left(-45\right)±\sqrt{3025}}{2\times 5}$$

Take the square root of 3025.

$$x=\frac{-\left(-45\right)±55}{2\times 5}$$

The opposite of -45 is 45.

$$x=\frac{45±55}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{45±55}{10}$$

Now solve the equation x=\frac{45±55}{10} when ± is plus. Add 45 to 55.

$$x=\frac{100}{10}$$

Divide 100 by 10.

$$x=10$$

Now solve the equation x=\frac{45±55}{10} when ± is minus. Subtract 55 from 45.

$$x=-\frac{10}{10}$$

Divide -10 by 10.

$$x=-1$$

The equation is now solved.

$$x=10 x=-1$$

Variable x cannot be equal to -1.

$$x=10$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,1 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-1\right)\left(x+1\right), the least common multiple of x+1,x^{2}-1,9.

$$\left(9x-9\right)\left(x-4\right)-9\times 10=4\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 9x-9 by x-4 and combine like terms.

$$9x^{2}-45x+36-9\times 10=4\left(x-1\right)\left(x+1\right)$$

Multiply -9 and 10 to get -90.

$$9x^{2}-45x+36-90=4\left(x-1\right)\left(x+1\right)$$

Subtract 90 from 36 to get -54.

$$9x^{2}-45x-54=4\left(x-1\right)\left(x+1\right)$$

Use the distributive property to multiply 4 by x-1.

$$9x^{2}-45x-54=\left(4x-4\right)\left(x+1\right)$$

Use the distributive property to multiply 4x-4 by x+1 and combine like terms.

$$9x^{2}-45x-54=4x^{2}-4$$

Subtract 4x^{2} from both sides.

$$9x^{2}-45x-54-4x^{2}=-4$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-45x-54=-4$$

Add 54 to both sides.

$$5x^{2}-45x=-4+54$$

Add -4 and 54 to get 50.

$$5x^{2}-45x=50$$

Divide both sides by 5.

$$\frac{5x^{2}-45x}{5}=\frac{50}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{45}{5}\right)x=\frac{50}{5}$$

Divide -45 by 5.

$$x^{2}-9x=\frac{50}{5}$$

Divide 50 by 5.

$$x^{2}-9x=10$$

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.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=10+\left(-\frac{9}{2}\right)^{2}$$

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

$$x^{2}-9x+\frac{81}{4}=10+\frac{81}{4}$$

Add 10 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{121}{4}$$

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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{11}{2} x-\frac{9}{2}=-\frac{11}{2}$$

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

$$x=10 x=-1$$

Variable x cannot be equal to -1.

$$x=10$$