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

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

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

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

$$x\times 3x-\left(x-1\right)\times 4=3$$

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

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

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

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

To find the opposite of 4x-4, find the opposite of each term.

$$x^{2}\times 3-4x+4=3$$

Subtract 3 from both sides.

$$x^{2}\times 3-4x+4-3=0$$

Subtract 3 from 4 to get 1.

$$x^{2}\times 3-4x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=3\times 1=3$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-3 b=-1$$

Rewrite 3x^{2}-4x+1 as \left(3x^{2}-3x\right)+\left(-x+1\right).

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

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-1\right)-\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(3x-1\right)$$

To find equation solutions, solve x-1=0 and 3x-1=0.

$$x=1 x=\frac{1}{3}$$

Variable x cannot be equal to 1.

$$x=\frac{1}{3}$$

Steps Using the Quadratic Formula

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

$$x\times 3x-\left(x-1\right)\times 4=3$$

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

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

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

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

To find the opposite of 4x-4, find the opposite of each term.

$$x^{2}\times 3-4x+4=3$$

Subtract 3 from both sides.

$$x^{2}\times 3-4x+4-3=0$$

Subtract 3 from 4 to get 1.

$$x^{2}\times 3-4x+1=0$$

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.

$$3x^{2}-4x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-4\right)±\sqrt{16-12}}{2\times 3}$$

Add 16 to -12.

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

Take the square root of 4.

$$x=\frac{-\left(-4\right)±2}{2\times 3}$$

The opposite of -4 is 4.

$$x=\frac{4±2}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{4±2}{6}$$

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

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

Divide 6 by 6.

$$x=1$$

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

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

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

$$x=\frac{1}{3}$$

The equation is now solved.

$$x=1 x=\frac{1}{3}$$

Variable x cannot be equal to 1.

$$x=\frac{1}{3}$$

Steps for Completing the Square

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

$$x\times 3x-\left(x-1\right)\times 4=3$$

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

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

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

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

To find the opposite of 4x-4, find the opposite of each term.

$$x^{2}\times 3-4x+4=3$$

Subtract 4 from both sides.

$$x^{2}\times 3-4x=3-4$$

Subtract 4 from 3 to get -1.

$$x^{2}\times 3-4x=-1$$

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.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

$$x=1 x=\frac{1}{3}$$

Variable x cannot be equal to 1.

$$x=\frac{1}{3}$$