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

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

Solve for x

$$x=-2$$

$$x=\frac{5}{7}\approx 0.714285714$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x-1\right)\left(x+1\right), the least common multiple of 3x,x-1,x^{2}+x.

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

Multiply x and x to get x^{2}.

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

Use the distributive property to multiply 3x^{2} by x+1.

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

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

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

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

$$4x^{2}-1+3x^{3}=3x^{3}-9x-3x^{2}+9$$

Subtract 3x^{3} from both sides.

$$4x^{2}-1+3x^{3}-3x^{3}=-9x-3x^{2}+9$$

Combine 3x^{3} and -3x^{3} to get 0.

$$4x^{2}-1=-9x-3x^{2}+9$$

Add 9x to both sides.

$$4x^{2}-1+9x=-3x^{2}+9$$

Add 3x^{2} to both sides.

$$4x^{2}-1+9x+3x^{2}=9$$

Combine 4x^{2} and 3x^{2} to get 7x^{2}.

$$7x^{2}-1+9x=9$$

Subtract 9 from both sides.

$$7x^{2}-1+9x-9=0$$

Subtract 9 from -1 to get -10.

$$7x^{2}-10+9x=0$$

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

$$7x^{2}+9x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=9 ab=7\left(-10\right)=-70$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -70.

$$-1,70 -2,35 -5,14 -7,10$$

Calculate the sum for each pair.

$$-1+70=69 -2+35=33 -5+14=9 -7+10=3$$

The solution is the pair that gives sum 9.

$$a=-5 b=14$$

Rewrite 7x^{2}+9x-10 as \left(7x^{2}-5x\right)+\left(14x-10\right).

$$\left(7x^{2}-5x\right)+\left(14x-10\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(7x-5\right)+2\left(7x-5\right)$$

Factor out common term 7x-5 by using distributive property.

$$\left(7x-5\right)\left(x+2\right)$$

To find equation solutions, solve 7x-5=0 and x+2=0.

$$x=\frac{5}{7} x=-2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x-1\right)\left(x+1\right), the least common multiple of 3x,x-1,x^{2}+x.

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

Multiply x and x to get x^{2}.

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

Use the distributive property to multiply 3x^{2} by x+1.

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

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

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

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

$$4x^{2}-1+3x^{3}=3x^{3}-9x-3x^{2}+9$$

Subtract 3x^{3} from both sides.

$$4x^{2}-1+3x^{3}-3x^{3}=-9x-3x^{2}+9$$

Combine 3x^{3} and -3x^{3} to get 0.

$$4x^{2}-1=-9x-3x^{2}+9$$

Add 9x to both sides.

$$4x^{2}-1+9x=-3x^{2}+9$$

Add 3x^{2} to both sides.

$$4x^{2}-1+9x+3x^{2}=9$$

Combine 4x^{2} and 3x^{2} to get 7x^{2}.

$$7x^{2}-1+9x=9$$

Subtract 9 from both sides.

$$7x^{2}-1+9x-9=0$$

Subtract 9 from -1 to get -10.

$$7x^{2}-10+9x=0$$

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.

$$7x^{2}+9x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7 for a, 9 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 7\left(-10\right)}}{2\times 7}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 7\left(-10\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-9±\sqrt{81-28\left(-10\right)}}{2\times 7}$$

Multiply -28 times -10.

$$x=\frac{-9±\sqrt{81+280}}{2\times 7}$$

Add 81 to 280.

$$x=\frac{-9±\sqrt{361}}{2\times 7}$$

Take the square root of 361.

$$x=\frac{-9±19}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-9±19}{14}$$

Now solve the equation x=\frac{-9±19}{14} when ± is plus. Add -9 to 19.

$$x=\frac{10}{14}$$

Reduce the fraction \frac{10}{14} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-9±19}{14} when ± is minus. Subtract 19 from -9.

$$x=-\frac{28}{14}$$

Divide -28 by 14.

$$x=-2$$

The equation is now solved.

$$x=\frac{5}{7} x=-2$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by 3x\left(x-1\right)\left(x+1\right), the least common multiple of 3x,x-1,x^{2}+x.

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

Multiply x and x to get x^{2}.

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

Use the distributive property to multiply 3x^{2} by x+1.

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

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

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

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

$$4x^{2}-1+3x^{3}=3x^{3}-9x-3x^{2}+9$$

Subtract 3x^{3} from both sides.

$$4x^{2}-1+3x^{3}-3x^{3}=-9x-3x^{2}+9$$

Combine 3x^{3} and -3x^{3} to get 0.

$$4x^{2}-1=-9x-3x^{2}+9$$

Add 9x to both sides.

$$4x^{2}-1+9x=-3x^{2}+9$$

Add 3x^{2} to both sides.

$$4x^{2}-1+9x+3x^{2}=9$$

Combine 4x^{2} and 3x^{2} to get 7x^{2}.

$$7x^{2}-1+9x=9$$

Add 1 to both sides.

$$7x^{2}+9x=9+1$$

Add 9 and 1 to get 10.

$$7x^{2}+9x=10$$

Divide both sides by 7.

$$\frac{7x^{2}+9x}{7}=\frac{10}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}+\frac{9}{7}x=\frac{10}{7}$$

Divide \frac{9}{7}, the coefficient of the x term, by 2 to get \frac{9}{14}. Then add the square of \frac{9}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{7}x+\left(\frac{9}{14}\right)^{2}=\frac{10}{7}+\left(\frac{9}{14}\right)^{2}$$

Square \frac{9}{14} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{7}x+\frac{81}{196}=\frac{10}{7}+\frac{81}{196}$$

Add \frac{10}{7} to \frac{81}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{7}x+\frac{81}{196}=\frac{361}{196}$$

Factor x^{2}+\frac{9}{7}x+\frac{81}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{14}\right)^{2}=\frac{361}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{14}\right)^{2}}=\sqrt{\frac{361}{196}}$$

Simplify.

$$x+\frac{9}{14}=\frac{19}{14} x+\frac{9}{14}=-\frac{19}{14}$$

Subtract \frac{9}{14} from both sides of the equation.

$$x=\frac{5}{7} x=-2$$