Solve 1x+2/11x-3=x-8/2x+3

$$\frac { 1 x + 2 } { 11 x - 3 } = \frac { x - 8 } { 2 x + 3 }$$

Solve for x

$$x = \frac{\sqrt{2239} + 49}{9} \approx 10.702008073$$

$$x=\frac{49-\sqrt{2239}}{9}\approx 0.186880816$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{3}{2},\frac{3}{11} since division by zero is not defined. Multiply both sides of the equation by \left(11x-3\right)\left(2x+3\right), the least common multiple of 11x-3,2x+3.

$$\left(2x+3\right)\left(1x+2\right)=\left(11x-3\right)\left(x-8\right)$$

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

$$2x^{2}+7x+6=\left(11x-3\right)\left(x-8\right)$$

Use the distributive property to multiply 11x-3 by x-8 and combine like terms.

$$2x^{2}+7x+6=11x^{2}-91x+24$$

Subtract 11x^{2} from both sides.

$$2x^{2}+7x+6-11x^{2}=-91x+24$$

Combine 2x^{2} and -11x^{2} to get -9x^{2}.

$$-9x^{2}+7x+6=-91x+24$$

Add 91x to both sides.

$$-9x^{2}+7x+6+91x=24$$

Combine 7x and 91x to get 98x.

$$-9x^{2}+98x+6=24$$

Subtract 24 from both sides.

$$-9x^{2}+98x+6-24=0$$

Subtract 24 from 6 to get -18.

$$-9x^{2}+98x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 98 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-98±\sqrt{98^{2}-4\left(-9\right)\left(-18\right)}}{2\left(-9\right)}$$

Square 98.

$$x=\frac{-98±\sqrt{9604-4\left(-9\right)\left(-18\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-98±\sqrt{9604+36\left(-18\right)}}{2\left(-9\right)}$$

Multiply 36 times -18.

$$x=\frac{-98±\sqrt{9604-648}}{2\left(-9\right)}$$

Add 9604 to -648.

$$x=\frac{-98±\sqrt{8956}}{2\left(-9\right)}$$

Take the square root of 8956.

$$x=\frac{-98±2\sqrt{2239}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-98±2\sqrt{2239}}{-18}$$

Now solve the equation x=\frac{-98±2\sqrt{2239}}{-18} when ± is plus. Add -98 to 2\sqrt{2239}.

$$x=\frac{2\sqrt{2239}-98}{-18}$$

Divide -98+2\sqrt{2239} by -18.

$$x=\frac{49-\sqrt{2239}}{9}$$

Now solve the equation x=\frac{-98±2\sqrt{2239}}{-18} when ± is minus. Subtract 2\sqrt{2239} from -98.

$$x=\frac{-2\sqrt{2239}-98}{-18}$$

Divide -98-2\sqrt{2239} by -18.

$$x=\frac{\sqrt{2239}+49}{9}$$

The equation is now solved.

$$x=\frac{49-\sqrt{2239}}{9} x=\frac{\sqrt{2239}+49}{9}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{3}{2},\frac{3}{11} since division by zero is not defined. Multiply both sides of the equation by \left(11x-3\right)\left(2x+3\right), the least common multiple of 11x-3,2x+3.

$$\left(2x+3\right)\left(1x+2\right)=\left(11x-3\right)\left(x-8\right)$$

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

$$2x^{2}+7x+6=\left(11x-3\right)\left(x-8\right)$$

Use the distributive property to multiply 11x-3 by x-8 and combine like terms.

$$2x^{2}+7x+6=11x^{2}-91x+24$$

Subtract 11x^{2} from both sides.

$$2x^{2}+7x+6-11x^{2}=-91x+24$$

Combine 2x^{2} and -11x^{2} to get -9x^{2}.

$$-9x^{2}+7x+6=-91x+24$$

Add 91x to both sides.

$$-9x^{2}+7x+6+91x=24$$

Combine 7x and 91x to get 98x.

$$-9x^{2}+98x+6=24$$

Subtract 6 from both sides.

$$-9x^{2}+98x=24-6$$

Subtract 6 from 24 to get 18.

$$-9x^{2}+98x=18$$

Divide both sides by -9.

$$\frac{-9x^{2}+98x}{-9}=\frac{18}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{98}{-9}x=\frac{18}{-9}$$

Divide 98 by -9.

$$x^{2}-\frac{98}{9}x=\frac{18}{-9}$$

Divide 18 by -9.

$$x^{2}-\frac{98}{9}x=-2$$

Divide -\frac{98}{9}, the coefficient of the x term, by 2 to get -\frac{49}{9}. Then add the square of -\frac{49}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{98}{9}x+\left(-\frac{49}{9}\right)^{2}=-2+\left(-\frac{49}{9}\right)^{2}$$

Square -\frac{49}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{98}{9}x+\frac{2401}{81}=-2+\frac{2401}{81}$$

Add -2 to \frac{2401}{81}.

$$x^{2}-\frac{98}{9}x+\frac{2401}{81}=\frac{2239}{81}$$

Factor x^{2}-\frac{98}{9}x+\frac{2401}{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}.

$$\left(x-\frac{49}{9}\right)^{2}=\frac{2239}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{49}{9}\right)^{2}}=\sqrt{\frac{2239}{81}}$$

Simplify.

$$x-\frac{49}{9}=\frac{\sqrt{2239}}{9} x-\frac{49}{9}=-\frac{\sqrt{2239}}{9}$$

Add \frac{49}{9} to both sides of the equation.

$$x=\frac{\sqrt{2239}+49}{9} x=\frac{49-\sqrt{2239}}{9}$$