Solve 2x-7/x-4-x+2/x+1=x+6/(x-4)(x+1)

$$\frac { 2 x - 7 } { x - 4 } - \frac { x + 2 } { x + 1 } = \frac { x + 6 } { ( x - 4 ) ( x + 1 ) }$$

Solve for x

$$x=5$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+1\right), the least common multiple of x-4,x+1,\left(x-4\right)\left(x+1\right).

$$\left(x+1\right)\left(2x-7\right)-\left(x-4\right)\left(x+2\right)=x+6$$

Use the distributive property to multiply x+1 by 2x-7 and combine like terms.

$$2x^{2}-5x-7-\left(x-4\right)\left(x+2\right)=x+6$$

Use the distributive property to multiply x-4 by x+2 and combine like terms.

$$2x^{2}-5x-7-\left(x^{2}-2x-8\right)=x+6$$

To find the opposite of x^{2}-2x-8, find the opposite of each term.

$$2x^{2}-5x-7-x^{2}+2x+8=x+6$$

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

$$x^{2}-5x-7+2x+8=x+6$$

Combine -5x and 2x to get -3x.

$$x^{2}-3x-7+8=x+6$$

Add -7 and 8 to get 1.

$$x^{2}-3x+1=x+6$$

Subtract x from both sides.

$$x^{2}-3x+1-x=6$$

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

$$x^{2}-4x+1=6$$

Subtract 6 from both sides.

$$x^{2}-4x+1-6=0$$

Subtract 6 from 1 to get -5.

$$x^{2}-4x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$x=\frac{-\left(-4\right)±\sqrt{16+20}}{2}$$

Add 16 to 20.

$$x=\frac{-\left(-4\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-\left(-4\right)±6}{2}$$

The opposite of -4 is 4.

$$x=\frac{4±6}{2}$$

Now solve the equation x=\frac{4±6}{2} when ± is plus. Add 4 to 6.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{4±6}{2} when ± is minus. Subtract 6 from 4.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=5 x=-1$$

Variable x cannot be equal to -1.

$$x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,4 since division by zero is not defined. Multiply both sides of the equation by \left(x-4\right)\left(x+1\right), the least common multiple of x-4,x+1,\left(x-4\right)\left(x+1\right).

$$\left(x+1\right)\left(2x-7\right)-\left(x-4\right)\left(x+2\right)=x+6$$

Use the distributive property to multiply x+1 by 2x-7 and combine like terms.

$$2x^{2}-5x-7-\left(x-4\right)\left(x+2\right)=x+6$$

Use the distributive property to multiply x-4 by x+2 and combine like terms.

$$2x^{2}-5x-7-\left(x^{2}-2x-8\right)=x+6$$

To find the opposite of x^{2}-2x-8, find the opposite of each term.

$$2x^{2}-5x-7-x^{2}+2x+8=x+6$$

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

$$x^{2}-5x-7+2x+8=x+6$$

Combine -5x and 2x to get -3x.

$$x^{2}-3x-7+8=x+6$$

Add -7 and 8 to get 1.

$$x^{2}-3x+1=x+6$$

Subtract x from both sides.

$$x^{2}-3x+1-x=6$$

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

$$x^{2}-4x+1=6$$

Subtract 1 from both sides.

$$x^{2}-4x=6-1$$

Subtract 1 from 6 to get 5.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add 5 to 4.

$$x^{2}-4x+4=9$$

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=3 x-2=-3$$

Add 2 to both sides of the equation.

$$x=5 x=-1$$

Variable x cannot be equal to -1.

$$x=5$$