Solve 2(2x-1/x+3)-3(x+3/2x-1)=5

$$2 ( \frac { 2 x - 1 } { x + 3 } ) - 3 ( \frac { x + 3 } { 2 x - 1 } ) = 5$$

Solve for x

$$x=-10$$

$$x=-\frac{1}{5}=-0.2$$

Steps Using the Quadratic Formula

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

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

Multiply 2x-1 and 2x-1 to get \left(2x-1\right)^{2}.

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

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$2\left(4x^{2}-4x+1\right)-3\left(x+3\right)^{2}=5\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply 2 by 4x^{2}-4x+1.

$$8x^{2}-8x+2-3\left(x+3\right)^{2}=5\left(2x-1\right)\left(x+3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$8x^{2}-8x+2-3\left(x^{2}+6x+9\right)=5\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply -3 by x^{2}+6x+9.

$$8x^{2}-8x+2-3x^{2}-18x-27=5\left(2x-1\right)\left(x+3\right)$$

Combine 8x^{2} and -3x^{2} to get 5x^{2}.

$$5x^{2}-8x+2-18x-27=5\left(2x-1\right)\left(x+3\right)$$

Combine -8x and -18x to get -26x.

$$5x^{2}-26x+2-27=5\left(2x-1\right)\left(x+3\right)$$

Subtract 27 from 2 to get -25.

$$5x^{2}-26x-25=5\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply 5 by 2x-1.

$$5x^{2}-26x-25=\left(10x-5\right)\left(x+3\right)$$

Use the distributive property to multiply 10x-5 by x+3 and combine like terms.

$$5x^{2}-26x-25=10x^{2}+25x-15$$

Subtract 10x^{2} from both sides.

$$5x^{2}-26x-25-10x^{2}=25x-15$$

Combine 5x^{2} and -10x^{2} to get -5x^{2}.

$$-5x^{2}-26x-25=25x-15$$

Subtract 25x from both sides.

$$-5x^{2}-26x-25-25x=-15$$

Combine -26x and -25x to get -51x.

$$-5x^{2}-51x-25=-15$$

Add 15 to both sides.

$$-5x^{2}-51x-25+15=0$$

Add -25 and 15 to get -10.

$$-5x^{2}-51x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -51 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-51\right)±\sqrt{\left(-51\right)^{2}-4\left(-5\right)\left(-10\right)}}{2\left(-5\right)}$$

Square -51.

$$x=\frac{-\left(-51\right)±\sqrt{2601-4\left(-5\right)\left(-10\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-51\right)±\sqrt{2601+20\left(-10\right)}}{2\left(-5\right)}$$

Multiply 20 times -10.

$$x=\frac{-\left(-51\right)±\sqrt{2601-200}}{2\left(-5\right)}$$

Add 2601 to -200.

$$x=\frac{-\left(-51\right)±\sqrt{2401}}{2\left(-5\right)}$$

Take the square root of 2401.

$$x=\frac{-\left(-51\right)±49}{2\left(-5\right)}$$

The opposite of -51 is 51.

$$x=\frac{51±49}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{51±49}{-10}$$

Now solve the equation x=\frac{51±49}{-10} when ± is plus. Add 51 to 49.

$$x=\frac{100}{-10}$$

Divide 100 by -10.

$$x=-10$$

Now solve the equation x=\frac{51±49}{-10} when ± is minus. Subtract 49 from 51.

$$x=\frac{2}{-10}$$

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

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=-10 x=-\frac{1}{5}$$

Steps for Completing the Square

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

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

Multiply 2x-1 and 2x-1 to get \left(2x-1\right)^{2}.

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

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

$$2\left(4x^{2}-4x+1\right)-3\left(x+3\right)^{2}=5\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply 2 by 4x^{2}-4x+1.

$$8x^{2}-8x+2-3\left(x+3\right)^{2}=5\left(2x-1\right)\left(x+3\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$8x^{2}-8x+2-3\left(x^{2}+6x+9\right)=5\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply -3 by x^{2}+6x+9.

$$8x^{2}-8x+2-3x^{2}-18x-27=5\left(2x-1\right)\left(x+3\right)$$

Combine 8x^{2} and -3x^{2} to get 5x^{2}.

$$5x^{2}-8x+2-18x-27=5\left(2x-1\right)\left(x+3\right)$$

Combine -8x and -18x to get -26x.

$$5x^{2}-26x+2-27=5\left(2x-1\right)\left(x+3\right)$$

Subtract 27 from 2 to get -25.

$$5x^{2}-26x-25=5\left(2x-1\right)\left(x+3\right)$$

Use the distributive property to multiply 5 by 2x-1.

$$5x^{2}-26x-25=\left(10x-5\right)\left(x+3\right)$$

Use the distributive property to multiply 10x-5 by x+3 and combine like terms.

$$5x^{2}-26x-25=10x^{2}+25x-15$$

Subtract 10x^{2} from both sides.

$$5x^{2}-26x-25-10x^{2}=25x-15$$

Combine 5x^{2} and -10x^{2} to get -5x^{2}.

$$-5x^{2}-26x-25=25x-15$$

Subtract 25x from both sides.

$$-5x^{2}-26x-25-25x=-15$$

Combine -26x and -25x to get -51x.

$$-5x^{2}-51x-25=-15$$

Add 25 to both sides.

$$-5x^{2}-51x=-15+25$$

Add -15 and 25 to get 10.

$$-5x^{2}-51x=10$$

Divide both sides by -5.

$$\frac{-5x^{2}-51x}{-5}=\frac{10}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{51}{-5}\right)x=\frac{10}{-5}$$

Divide -51 by -5.

$$x^{2}+\frac{51}{5}x=\frac{10}{-5}$$

Divide 10 by -5.

$$x^{2}+\frac{51}{5}x=-2$$

Divide \frac{51}{5}, the coefficient of the x term, by 2 to get \frac{51}{10}. Then add the square of \frac{51}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{51}{5}x+\left(\frac{51}{10}\right)^{2}=-2+\left(\frac{51}{10}\right)^{2}$$

Square \frac{51}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{51}{5}x+\frac{2601}{100}=-2+\frac{2601}{100}$$

Add -2 to \frac{2601}{100}.

$$x^{2}+\frac{51}{5}x+\frac{2601}{100}=\frac{2401}{100}$$

Factor x^{2}+\frac{51}{5}x+\frac{2601}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{51}{10}\right)^{2}=\frac{2401}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{51}{10}\right)^{2}}=\sqrt{\frac{2401}{100}}$$

Simplify.

$$x+\frac{51}{10}=\frac{49}{10} x+\frac{51}{10}=-\frac{49}{10}$$

Subtract \frac{51}{10} from both sides of the equation.

$$x=-\frac{1}{5} x=-10$$