Solve -(3x-1)*(x-1/3)/2=-2/3*x

$$- \frac { ( 3 x - 1 ) \cdot ( x - \frac { 1 } { 3 } ) } { 2 } = - \frac { 2 } { 3 } \cdot x$$

Solve for x

$$x=1$$

$$x=\frac{1}{9}\approx 0.111111111$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply -3 by 3x-1.

$$\left(-9x+3\right)\left(x-\frac{1}{3}\right)=-4x$$

Use the distributive property to multiply -9x+3 by x-\frac{1}{3} and combine like terms.

$$-9x^{2}+6x-1=-4x$$

Add 4x to both sides.

$$-9x^{2}+6x-1+4x=0$$

Combine 6x and 4x to get 10x.

$$-9x^{2}+10x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=10 ab=-9\left(-1\right)=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=9 b=1$$

Rewrite -9x^{2}+10x-1 as \left(-9x^{2}+9x\right)+\left(x-1\right).

$$\left(-9x^{2}+9x\right)+\left(x-1\right)$$

Factor out 9x in the first and -1 in the second group.

$$9x\left(-x+1\right)-\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(9x-1\right)$$

To find equation solutions, solve -x+1=0 and 9x-1=0.

$$x=1 x=\frac{1}{9}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply -3 by 3x-1.

$$\left(-9x+3\right)\left(x-\frac{1}{3}\right)=-4x$$

Use the distributive property to multiply -9x+3 by x-\frac{1}{3} and combine like terms.

$$-9x^{2}+6x-1=-4x$$

Add 4x to both sides.

$$-9x^{2}+6x-1+4x=0$$

Combine 6x and 4x to get 10x.

$$-9x^{2}+10x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 10 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-10±\sqrt{100+36\left(-1\right)}}{2\left(-9\right)}$$

Multiply 36 times -1.

$$x=\frac{-10±\sqrt{100-36}}{2\left(-9\right)}$$

Add 100 to -36.

$$x=\frac{-10±\sqrt{64}}{2\left(-9\right)}$$

Take the square root of 64.

$$x=\frac{-10±8}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-10±8}{-18}$$

Now solve the equation x=\frac{-10±8}{-18} when ± is plus. Add -10 to 8.

$$x=-\frac{2}{-18}$$

Reduce the fraction \frac{-2}{-18} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{9}$$

Now solve the equation x=\frac{-10±8}{-18} when ± is minus. Subtract 8 from -10.

$$x=-\frac{18}{-18}$$

Divide -18 by -18.

$$x=1$$

The equation is now solved.

$$x=\frac{1}{9} x=1$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 2,3.

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

Use the distributive property to multiply -3 by 3x-1.

$$\left(-9x+3\right)\left(x-\frac{1}{3}\right)=-4x$$

Use the distributive property to multiply -9x+3 by x-\frac{1}{3} and combine like terms.

$$-9x^{2}+6x-1=-4x$$

Add 4x to both sides.

$$-9x^{2}+6x-1+4x=0$$

Combine 6x and 4x to get 10x.

$$-9x^{2}+10x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$-9x^{2}+10x=1$$

Divide both sides by -9.

$$\frac{-9x^{2}+10x}{-9}=\frac{1}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{10}{-9}x=\frac{1}{-9}$$

Divide 10 by -9.

$$x^{2}-\frac{10}{9}x=\frac{1}{-9}$$

Divide 1 by -9.

$$x^{2}-\frac{10}{9}x=-\frac{1}{9}$$

Divide -\frac{10}{9}, the coefficient of the x term, by 2 to get -\frac{5}{9}. Then add the square of -\frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{9}x+\left(-\frac{5}{9}\right)^{2}=-\frac{1}{9}+\left(-\frac{5}{9}\right)^{2}$$

Square -\frac{5}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=-\frac{1}{9}+\frac{25}{81}$$

Add -\frac{1}{9} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=\frac{16}{81}$$

Factor x^{2}-\frac{10}{9}x+\frac{25}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{9}\right)^{2}=\frac{16}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{9}\right)^{2}}=\sqrt{\frac{16}{81}}$$

Simplify.

$$x-\frac{5}{9}=\frac{4}{9} x-\frac{5}{9}=-\frac{4}{9}$$

Add \frac{5}{9} to both sides of the equation.

$$x=1 x=\frac{1}{9}$$