Solve x^2-2x(x-1)/3-(x-1)(10x-1)/27=x

$$x ^ { 2 } - \frac { 2 x ( x - 1 ) } { 3 } - \frac { ( x - 1 ) ( 10 x - 1 ) } { 27 } = x$$

Solve for x

$$x=1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x by x-1.

$$x^{2}-\frac{2x^{2}-2x}{3}-\frac{\left(x-1\right)\left(10x-1\right)}{27}=x$$

Use the distributive property to multiply x-1 by 10x-1 and combine like terms.

$$x^{2}-\frac{2x^{2}-2x}{3}-\frac{10x^{2}-11x+1}{27}=x$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 27 is 27. Multiply \frac{2x^{2}-2x}{3} times \frac{9}{9}.

$$x^{2}-\frac{9\left(2x^{2}-2x\right)}{27}-\frac{10x^{2}-11x+1}{27}=x$$

Since -\frac{9\left(2x^{2}-2x\right)}{27} and \frac{10x^{2}-11x+1}{27} have the same denominator, subtract them by subtracting their numerators.

$$x^{2}+\frac{-9\left(2x^{2}-2x\right)-\left(10x^{2}-11x+1\right)}{27}=x$$

Do the multiplications in -9\left(2x^{2}-2x\right)-\left(10x^{2}-11x+1\right).

$$x^{2}+\frac{-18x^{2}+18x-10x^{2}+11x-1}{27}=x$$

Combine like terms in -18x^{2}+18x-10x^{2}+11x-1.

$$x^{2}+\frac{-28x^{2}+29x-1}{27}=x$$

Divide each term of -28x^{2}+29x-1 by 27 to get -\frac{28}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}.

$$x^{2}-\frac{28}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}=x$$

Combine x^{2} and -\frac{28}{27}x^{2} to get -\frac{1}{27}x^{2}.

$$-\frac{1}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}=x$$

Subtract x from both sides.

$$-\frac{1}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}-x=0$$

Combine \frac{29}{27}x and -x to get \frac{2}{27}x.

$$-\frac{1}{27}x^{2}+\frac{2}{27}x-\frac{1}{27}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{27} for a, \frac{2}{27} for b, and -\frac{1}{27} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{2}{27}±\sqrt{\left(\frac{2}{27}\right)^{2}-4\left(-\frac{1}{27}\right)\left(-\frac{1}{27}\right)}}{2\left(-\frac{1}{27}\right)}$$

Square \frac{2}{27} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{2}{27}±\sqrt{\frac{4}{729}-4\left(-\frac{1}{27}\right)\left(-\frac{1}{27}\right)}}{2\left(-\frac{1}{27}\right)}$$

Multiply -4 times -\frac{1}{27}.

$$x=\frac{-\frac{2}{27}±\sqrt{\frac{4}{729}+\frac{4}{27}\left(-\frac{1}{27}\right)}}{2\left(-\frac{1}{27}\right)}$$

Multiply \frac{4}{27} times -\frac{1}{27} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{2}{27}±\sqrt{\frac{4-4}{729}}}{2\left(-\frac{1}{27}\right)}$$

Add \frac{4}{729} to -\frac{4}{729} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{2}{27}±\sqrt{0}}{2\left(-\frac{1}{27}\right)}$$

Take the square root of 0.

$$x=-\frac{\frac{2}{27}}{2\left(-\frac{1}{27}\right)}$$

Multiply 2 times -\frac{1}{27}.

$$x=-\frac{\frac{2}{27}}{-\frac{2}{27}}$$

Divide -\frac{2}{27} by -\frac{2}{27} by multiplying -\frac{2}{27} by the reciprocal of -\frac{2}{27}.

$$x=1$$

Steps for Completing the Square

Use the distributive property to multiply 2x by x-1.

$$x^{2}-\frac{2x^{2}-2x}{3}-\frac{\left(x-1\right)\left(10x-1\right)}{27}=x$$

Use the distributive property to multiply x-1 by 10x-1 and combine like terms.

$$x^{2}-\frac{2x^{2}-2x}{3}-\frac{10x^{2}-11x+1}{27}=x$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 27 is 27. Multiply \frac{2x^{2}-2x}{3} times \frac{9}{9}.

$$x^{2}-\frac{9\left(2x^{2}-2x\right)}{27}-\frac{10x^{2}-11x+1}{27}=x$$

Since -\frac{9\left(2x^{2}-2x\right)}{27} and \frac{10x^{2}-11x+1}{27} have the same denominator, subtract them by subtracting their numerators.

$$x^{2}+\frac{-9\left(2x^{2}-2x\right)-\left(10x^{2}-11x+1\right)}{27}=x$$

Do the multiplications in -9\left(2x^{2}-2x\right)-\left(10x^{2}-11x+1\right).

$$x^{2}+\frac{-18x^{2}+18x-10x^{2}+11x-1}{27}=x$$

Combine like terms in -18x^{2}+18x-10x^{2}+11x-1.

$$x^{2}+\frac{-28x^{2}+29x-1}{27}=x$$

Divide each term of -28x^{2}+29x-1 by 27 to get -\frac{28}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}.

$$x^{2}-\frac{28}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}=x$$

Combine x^{2} and -\frac{28}{27}x^{2} to get -\frac{1}{27}x^{2}.

$$-\frac{1}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}=x$$

Subtract x from both sides.

$$-\frac{1}{27}x^{2}+\frac{29}{27}x-\frac{1}{27}-x=0$$

Combine \frac{29}{27}x and -x to get \frac{2}{27}x.

$$-\frac{1}{27}x^{2}+\frac{2}{27}x-\frac{1}{27}=0$$

Add \frac{1}{27} to both sides. Anything plus zero gives itself.

$$-\frac{1}{27}x^{2}+\frac{2}{27}x=\frac{1}{27}$$

Multiply both sides by -27.

$$\frac{-\frac{1}{27}x^{2}+\frac{2}{27}x}{-\frac{1}{27}}=\frac{\frac{1}{27}}{-\frac{1}{27}}$$

Dividing by -\frac{1}{27} undoes the multiplication by -\frac{1}{27}.

$$x^{2}+\frac{\frac{2}{27}}{-\frac{1}{27}}x=\frac{\frac{1}{27}}{-\frac{1}{27}}$$

Divide \frac{2}{27} by -\frac{1}{27} by multiplying \frac{2}{27} by the reciprocal of -\frac{1}{27}.

$$x^{2}-2x=\frac{\frac{1}{27}}{-\frac{1}{27}}$$

Divide \frac{1}{27} by -\frac{1}{27} by multiplying \frac{1}{27} by the reciprocal of -\frac{1}{27}.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Add -1 to 1.

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

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-1=0 x-1=0$$

Add 1 to both sides of the equation.

$$x=1 x=1$$

The equation is now solved. Solutions are the same.

$$x=1$$