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

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

Solve for x

$$x = \frac{\sqrt{3} + 3}{2} \approx 2.366025404$$

$$x=\frac{3-\sqrt{3}}{2}\approx 0.633974596$$

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 3\left(x-3\right), the least common multiple of 3,x-3.

$$\left(x-3\right)\left(2x+1\right)+3\times 2=\left(x-3\right)\left(1-2x\right)$$

Use the distributive property to multiply x-3 by 2x+1 and combine like terms.

$$2x^{2}-5x-3+3\times 2=\left(x-3\right)\left(1-2x\right)$$

Multiply 3 and 2 to get 6.

$$2x^{2}-5x-3+6=\left(x-3\right)\left(1-2x\right)$$

Add -3 and 6 to get 3.

$$2x^{2}-5x+3=\left(x-3\right)\left(1-2x\right)$$

Use the distributive property to multiply x-3 by 1-2x and combine like terms.

$$2x^{2}-5x+3=7x-2x^{2}-3$$

Subtract 7x from both sides.

$$2x^{2}-5x+3-7x=-2x^{2}-3$$

Combine -5x and -7x to get -12x.

$$2x^{2}-12x+3=-2x^{2}-3$$

Add 2x^{2} to both sides.

$$2x^{2}-12x+3+2x^{2}=-3$$

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

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

Add 3 to both sides.

$$4x^{2}-12x+3+3=0$$

Add 3 and 3 to get 6.

$$4x^{2}-12x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 6}}{2\times 4}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-\left(-12\right)±\sqrt{144-96}}{2\times 4}$$

Add 144 to -96.

$$x=\frac{-\left(-12\right)±\sqrt{48}}{2\times 4}$$

Take the square root of 48.

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

The opposite of -12 is 12.

$$x=\frac{12±4\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{12±4\sqrt{3}}{8}$$

Now solve the equation x=\frac{12±4\sqrt{3}}{8} when ± is plus. Add 12 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+12}{8}$$

Divide 12+4\sqrt{3} by 8.

$$x=\frac{\sqrt{3}+3}{2}$$

Now solve the equation x=\frac{12±4\sqrt{3}}{8} when ± is minus. Subtract 4\sqrt{3} from 12.

$$x=\frac{12-4\sqrt{3}}{8}$$

Divide 12-4\sqrt{3} by 8.

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

The equation is now solved.

$$x=\frac{\sqrt{3}+3}{2} x=\frac{3-\sqrt{3}}{2}$$

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 3\left(x-3\right), the least common multiple of 3,x-3.

$$\left(x-3\right)\left(2x+1\right)+3\times 2=\left(x-3\right)\left(1-2x\right)$$

Use the distributive property to multiply x-3 by 2x+1 and combine like terms.

$$2x^{2}-5x-3+3\times 2=\left(x-3\right)\left(1-2x\right)$$

Multiply 3 and 2 to get 6.

$$2x^{2}-5x-3+6=\left(x-3\right)\left(1-2x\right)$$

Add -3 and 6 to get 3.

$$2x^{2}-5x+3=\left(x-3\right)\left(1-2x\right)$$

Use the distributive property to multiply x-3 by 1-2x and combine like terms.

$$2x^{2}-5x+3=7x-2x^{2}-3$$

Subtract 7x from both sides.

$$2x^{2}-5x+3-7x=-2x^{2}-3$$

Combine -5x and -7x to get -12x.

$$2x^{2}-12x+3=-2x^{2}-3$$

Add 2x^{2} to both sides.

$$2x^{2}-12x+3+2x^{2}=-3$$

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

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

Subtract 3 from both sides.

$$4x^{2}-12x=-3-3$$

Subtract 3 from -3 to get -6.

$$4x^{2}-12x=-6$$

Divide both sides by 4.

$$\frac{4x^{2}-12x}{4}=-\frac{6}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{12}{4}\right)x=-\frac{6}{4}$$

Divide -12 by 4.

$$x^{2}-3x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

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

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

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

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

Add -\frac{3}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$x=\frac{\sqrt{3}+3}{2} x=\frac{3-\sqrt{3}}{2}$$