Solve 6-x/2x-3=x-9/x+2

$$\frac { 6 - x } { 2 x - 3 } = \frac { x - 9 } { x + 2 }$$

Solve for x

$$x = \frac{\sqrt{445} + 25}{6} \approx 7.682503852$$

$$x=\frac{25-\sqrt{445}}{6}\approx 0.650829482$$

Steps Using the Quadratic Formula

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

$$\left(x+2\right)\left(6-x\right)=\left(2x-3\right)\left(x-9\right)$$

Use the distributive property to multiply x+2 by 6-x and combine like terms.

$$4x-x^{2}+12=\left(2x-3\right)\left(x-9\right)$$

Use the distributive property to multiply 2x-3 by x-9 and combine like terms.

$$4x-x^{2}+12=2x^{2}-21x+27$$

Subtract 2x^{2} from both sides.

$$4x-x^{2}+12-2x^{2}=-21x+27$$

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

$$4x-3x^{2}+12=-21x+27$$

Add 21x to both sides.

$$4x-3x^{2}+12+21x=27$$

Combine 4x and 21x to get 25x.

$$25x-3x^{2}+12=27$$

Subtract 27 from both sides.

$$25x-3x^{2}+12-27=0$$

Subtract 27 from 12 to get -15.

$$25x-3x^{2}-15=0$$

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.

$$-3x^{2}+25x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 25 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-3\right)\left(-15\right)}}{2\left(-3\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-3\right)\left(-15\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-25±\sqrt{625+12\left(-15\right)}}{2\left(-3\right)}$$

Multiply 12 times -15.

$$x=\frac{-25±\sqrt{625-180}}{2\left(-3\right)}$$

Add 625 to -180.

$$x=\frac{-25±\sqrt{445}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-25±\sqrt{445}}{-6}$$

Now solve the equation x=\frac{-25±\sqrt{445}}{-6} when ± is plus. Add -25 to \sqrt{445}.

$$x=\frac{\sqrt{445}-25}{-6}$$

Divide -25+\sqrt{445} by -6.

$$x=\frac{25-\sqrt{445}}{6}$$

Now solve the equation x=\frac{-25±\sqrt{445}}{-6} when ± is minus. Subtract \sqrt{445} from -25.

$$x=\frac{-\sqrt{445}-25}{-6}$$

Divide -25-\sqrt{445} by -6.

$$x=\frac{\sqrt{445}+25}{6}$$

The equation is now solved.

$$x=\frac{25-\sqrt{445}}{6} x=\frac{\sqrt{445}+25}{6}$$

Steps for Completing the Square

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

$$\left(x+2\right)\left(6-x\right)=\left(2x-3\right)\left(x-9\right)$$

Use the distributive property to multiply x+2 by 6-x and combine like terms.

$$4x-x^{2}+12=\left(2x-3\right)\left(x-9\right)$$

Use the distributive property to multiply 2x-3 by x-9 and combine like terms.

$$4x-x^{2}+12=2x^{2}-21x+27$$

Subtract 2x^{2} from both sides.

$$4x-x^{2}+12-2x^{2}=-21x+27$$

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

$$4x-3x^{2}+12=-21x+27$$

Add 21x to both sides.

$$4x-3x^{2}+12+21x=27$$

Combine 4x and 21x to get 25x.

$$25x-3x^{2}+12=27$$

Subtract 12 from both sides.

$$25x-3x^{2}=27-12$$

Subtract 12 from 27 to get 15.

$$25x-3x^{2}=15$$

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.

$$-3x^{2}+25x=15$$

Divide both sides by -3.

$$\frac{-3x^{2}+25x}{-3}=\frac{15}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{25}{-3}x=\frac{15}{-3}$$

Divide 25 by -3.

$$x^{2}-\frac{25}{3}x=\frac{15}{-3}$$

Divide 15 by -3.

$$x^{2}-\frac{25}{3}x=-5$$

Divide -\frac{25}{3}, the coefficient of the x term, by 2 to get -\frac{25}{6}. Then add the square of -\frac{25}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{3}x+\left(-\frac{25}{6}\right)^{2}=-5+\left(-\frac{25}{6}\right)^{2}$$

Square -\frac{25}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{3}x+\frac{625}{36}=-5+\frac{625}{36}$$

Add -5 to \frac{625}{36}.

$$x^{2}-\frac{25}{3}x+\frac{625}{36}=\frac{445}{36}$$

Factor x^{2}-\frac{25}{3}x+\frac{625}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{6}\right)^{2}=\frac{445}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{6}\right)^{2}}=\sqrt{\frac{445}{36}}$$

Simplify.

$$x-\frac{25}{6}=\frac{\sqrt{445}}{6} x-\frac{25}{6}=-\frac{\sqrt{445}}{6}$$

Add \frac{25}{6} to both sides of the equation.

$$x=\frac{\sqrt{445}+25}{6} x=\frac{25-\sqrt{445}}{6}$$