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

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

Solve for x

$$x=-10$$

$$x=3$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply x-5 by x+1 and combine like terms.

$$x^{2}-4x-5-\left(x+5\right)\left(x-2\right)=\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x+5 by x-2 and combine like terms.

$$x^{2}-4x-5-\left(x^{2}+3x-10\right)=\left(x-5\right)\left(x+5\right)$$

To find the opposite of x^{2}+3x-10, find the opposite of each term.

$$x^{2}-4x-5-x^{2}-3x+10=\left(x-5\right)\left(x+5\right)$$

Combine x^{2} and -x^{2} to get 0.

$$-4x-5-3x+10=\left(x-5\right)\left(x+5\right)$$

Combine -4x and -3x to get -7x.

$$-7x-5+10=\left(x-5\right)\left(x+5\right)$$

Add -5 and 10 to get 5.

$$-7x+5=\left(x-5\right)\left(x+5\right)$$

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$-7x+5=x^{2}-25$$

Subtract x^{2} from both sides.

$$-7x+5-x^{2}=-25$$

Add 25 to both sides.

$$-7x+5-x^{2}+25=0$$

Add 5 and 25 to get 30.

$$-7x+30-x^{2}=0$$

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.

$$-x^{2}-7x+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\times 30}}{2\left(-1\right)}$$

Multiply 4 times 30.

$$x=\frac{-\left(-7\right)±\sqrt{49+120}}{2\left(-1\right)}$$

Add 49 to 120.

$$x=\frac{-\left(-7\right)±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-7\right)±13}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{7±13}{-2}$$

Now solve the equation x=\frac{7±13}{-2} when ± is plus. Add 7 to 13.

$$x=\frac{20}{-2}$$

Divide 20 by -2.

$$x=-10$$

Now solve the equation x=\frac{7±13}{-2} when ± is minus. Subtract 13 from 7.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=-10 x=3$$

Steps for Completing the Square

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

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

Use the distributive property to multiply x-5 by x+1 and combine like terms.

$$x^{2}-4x-5-\left(x+5\right)\left(x-2\right)=\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x+5 by x-2 and combine like terms.

$$x^{2}-4x-5-\left(x^{2}+3x-10\right)=\left(x-5\right)\left(x+5\right)$$

To find the opposite of x^{2}+3x-10, find the opposite of each term.

$$x^{2}-4x-5-x^{2}-3x+10=\left(x-5\right)\left(x+5\right)$$

Combine x^{2} and -x^{2} to get 0.

$$-4x-5-3x+10=\left(x-5\right)\left(x+5\right)$$

Combine -4x and -3x to get -7x.

$$-7x-5+10=\left(x-5\right)\left(x+5\right)$$

Add -5 and 10 to get 5.

$$-7x+5=\left(x-5\right)\left(x+5\right)$$

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$-7x+5=x^{2}-25$$

Subtract x^{2} from both sides.

$$-7x+5-x^{2}=-25$$

Subtract 5 from both sides.

$$-7x-x^{2}=-25-5$$

Subtract 5 from -25 to get -30.

$$-7x-x^{2}=-30$$

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.

$$-x^{2}-7x=-30$$

Divide both sides by -1.

$$\frac{-x^{2}-7x}{-1}=-\frac{30}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{7}{-1}\right)x=-\frac{30}{-1}$$

Divide -7 by -1.

$$x^{2}+7x=-\frac{30}{-1}$$

Divide -30 by -1.

$$x^{2}+7x=30$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=30+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=30+\frac{49}{4}$$

Add 30 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{169}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{13}{2} x+\frac{7}{2}=-\frac{13}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=3 x=-10$$