Solve x^2+1/x-3+-2x/3-x=4x-12

$$\frac { x ^ { 2 } + 1 } { x - 3 } + \frac { - 2 x } { 3 - x } = 4 x - 12$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=7$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3, the least common multiple of x-3,3-x.

$$x^{2}+1-\left(-2x\right)=4x\left(x-3\right)+\left(x-3\right)\left(-12\right)$$

Multiply -1 and -2 to get 2.

$$x^{2}+1+2x=4x\left(x-3\right)+\left(x-3\right)\left(-12\right)$$

Use the distributive property to multiply 4x by x-3.

$$x^{2}+1+2x=4x^{2}-12x+\left(x-3\right)\left(-12\right)$$

Use the distributive property to multiply x-3 by -12.

$$x^{2}+1+2x=4x^{2}-12x-12x+36$$

Combine -12x and -12x to get -24x.

$$x^{2}+1+2x=4x^{2}-24x+36$$

Subtract 4x^{2} from both sides.

$$x^{2}+1+2x-4x^{2}=-24x+36$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+1+2x=-24x+36$$

Add 24x to both sides.

$$-3x^{2}+1+2x+24x=36$$

Combine 2x and 24x to get 26x.

$$-3x^{2}+1+26x=36$$

Subtract 36 from both sides.

$$-3x^{2}+1+26x-36=0$$

Subtract 36 from 1 to get -35.

$$-3x^{2}-35+26x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}+26x-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=26 ab=-3\left(-35\right)=105$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 105.

$$1,105 3,35 5,21 7,15$$

Calculate the sum for each pair.

$$1+105=106 3+35=38 5+21=26 7+15=22$$

The solution is the pair that gives sum 26.

$$a=21 b=5$$

Rewrite -3x^{2}+26x-35 as \left(-3x^{2}+21x\right)+\left(5x-35\right).

$$\left(-3x^{2}+21x\right)+\left(5x-35\right)$$

Factor out 3x in the first and -5 in the second group.

$$3x\left(-x+7\right)-5\left(-x+7\right)$$

Factor out common term -x+7 by using distributive property.

$$\left(-x+7\right)\left(3x-5\right)$$

To find equation solutions, solve -x+7=0 and 3x-5=0.

$$x=7 x=\frac{5}{3}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3, the least common multiple of x-3,3-x.

$$x^{2}+1-\left(-2x\right)=4x\left(x-3\right)+\left(x-3\right)\left(-12\right)$$

Multiply -1 and -2 to get 2.

$$x^{2}+1+2x=4x\left(x-3\right)+\left(x-3\right)\left(-12\right)$$

Use the distributive property to multiply 4x by x-3.

$$x^{2}+1+2x=4x^{2}-12x+\left(x-3\right)\left(-12\right)$$

Use the distributive property to multiply x-3 by -12.

$$x^{2}+1+2x=4x^{2}-12x-12x+36$$

Combine -12x and -12x to get -24x.

$$x^{2}+1+2x=4x^{2}-24x+36$$

Subtract 4x^{2} from both sides.

$$x^{2}+1+2x-4x^{2}=-24x+36$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+1+2x=-24x+36$$

Add 24x to both sides.

$$-3x^{2}+1+2x+24x=36$$

Combine 2x and 24x to get 26x.

$$-3x^{2}+1+26x=36$$

Subtract 36 from both sides.

$$-3x^{2}+1+26x-36=0$$

Subtract 36 from 1 to get -35.

$$-3x^{2}-35+26x=0$$

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.

$$-3x^{2}+26x-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 26 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\left(-3\right)\left(-35\right)}}{2\left(-3\right)}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\left(-3\right)\left(-35\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-26±\sqrt{676+12\left(-35\right)}}{2\left(-3\right)}$$

Multiply 12 times -35.

$$x=\frac{-26±\sqrt{676-420}}{2\left(-3\right)}$$

Add 676 to -420.

$$x=\frac{-26±\sqrt{256}}{2\left(-3\right)}$$

Take the square root of 256.

$$x=\frac{-26±16}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-26±16}{-6}$$

Now solve the equation x=\frac{-26±16}{-6} when ± is plus. Add -26 to 16.

$$x=-\frac{10}{-6}$$

Reduce the fraction \frac{-10}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{-26±16}{-6} when ± is minus. Subtract 16 from -26.

$$x=-\frac{42}{-6}$$

Divide -42 by -6.

$$x=7$$

The equation is now solved.

$$x=\frac{5}{3} x=7$$

Steps for Completing the Square

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3, the least common multiple of x-3,3-x.

$$x^{2}+1-\left(-2x\right)=4x\left(x-3\right)+\left(x-3\right)\left(-12\right)$$

Multiply -1 and -2 to get 2.

$$x^{2}+1+2x=4x\left(x-3\right)+\left(x-3\right)\left(-12\right)$$

Use the distributive property to multiply 4x by x-3.

$$x^{2}+1+2x=4x^{2}-12x+\left(x-3\right)\left(-12\right)$$

Use the distributive property to multiply x-3 by -12.

$$x^{2}+1+2x=4x^{2}-12x-12x+36$$

Combine -12x and -12x to get -24x.

$$x^{2}+1+2x=4x^{2}-24x+36$$

Subtract 4x^{2} from both sides.

$$x^{2}+1+2x-4x^{2}=-24x+36$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+1+2x=-24x+36$$

Add 24x to both sides.

$$-3x^{2}+1+2x+24x=36$$

Combine 2x and 24x to get 26x.

$$-3x^{2}+1+26x=36$$

Subtract 1 from both sides.

$$-3x^{2}+26x=36-1$$

Subtract 1 from 36 to get 35.

$$-3x^{2}+26x=35$$

Divide both sides by -3.

$$\frac{-3x^{2}+26x}{-3}=\frac{35}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{26}{-3}x=\frac{35}{-3}$$

Divide 26 by -3.

$$x^{2}-\frac{26}{3}x=\frac{35}{-3}$$

Divide 35 by -3.

$$x^{2}-\frac{26}{3}x=-\frac{35}{3}$$

Divide -\frac{26}{3}, the coefficient of the x term, by 2 to get -\frac{13}{3}. Then add the square of -\frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{26}{3}x+\left(-\frac{13}{3}\right)^{2}=-\frac{35}{3}+\left(-\frac{13}{3}\right)^{2}$$

Square -\frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=-\frac{35}{3}+\frac{169}{9}$$

Add -\frac{35}{3} to \frac{169}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{26}{3}x+\frac{169}{9}=\frac{64}{9}$$

Factor x^{2}-\frac{26}{3}x+\frac{169}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x-\frac{13}{3}=\frac{8}{3} x-\frac{13}{3}=-\frac{8}{3}$$

Add \frac{13}{3} to both sides of the equation.

$$x=7 x=\frac{5}{3}$$