Solve 1/2x+5=x/11x+20

$$\frac { 1 } { 2 x + 5 } = \frac { x } { 11 x + 20 }$$

Solve for x

$$x=-2$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -\frac{5}{2},-\frac{20}{11} since division by zero is not defined. Multiply both sides of the equation by \left(2x+5\right)\left(11x+20\right), the least common multiple of 2x+5,11x+20.

$$11x+20=\left(2x+5\right)x$$

Use the distributive property to multiply 2x+5 by x.

$$11x+20=2x^{2}+5x$$

Subtract 2x^{2} from both sides.

$$11x+20-2x^{2}=5x$$

Subtract 5x from both sides.

$$11x+20-2x^{2}-5x=0$$

Combine 11x and -5x to get 6x.

$$6x+20-2x^{2}=0$$

Divide both sides by 2.

$$3x+10-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+3x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=3 ab=-10=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=5 b=-2$$

Rewrite -x^{2}+3x+10 as \left(-x^{2}+5x\right)+\left(-2x+10\right).

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

Factor out -x in the first and -2 in the second group.

$$-x\left(x-5\right)-2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(-x-2\right)$$

To find equation solutions, solve x-5=0 and -x-2=0.

$$x=5 x=-2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{5}{2},-\frac{20}{11} since division by zero is not defined. Multiply both sides of the equation by \left(2x+5\right)\left(11x+20\right), the least common multiple of 2x+5,11x+20.

$$11x+20=\left(2x+5\right)x$$

Use the distributive property to multiply 2x+5 by x.

$$11x+20=2x^{2}+5x$$

Subtract 2x^{2} from both sides.

$$11x+20-2x^{2}=5x$$

Subtract 5x from both sides.

$$11x+20-2x^{2}-5x=0$$

Combine 11x and -5x to get 6x.

$$6x+20-2x^{2}=0$$

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.

$$-2x^{2}+6x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 6 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)\times 20}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-6±\sqrt{36+8\times 20}}{2\left(-2\right)}$$

Multiply 8 times 20.

$$x=\frac{-6±\sqrt{36+160}}{2\left(-2\right)}$$

Add 36 to 160.

$$x=\frac{-6±\sqrt{196}}{2\left(-2\right)}$$

Take the square root of 196.

$$x=\frac{-6±14}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-6±14}{-4}$$

Now solve the equation x=\frac{-6±14}{-4} when ± is plus. Add -6 to 14.

$$x=\frac{8}{-4}$$

Divide 8 by -4.

$$x=-2$$

Now solve the equation x=\frac{-6±14}{-4} when ± is minus. Subtract 14 from -6.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

The equation is now solved.

$$x=-2 x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{5}{2},-\frac{20}{11} since division by zero is not defined. Multiply both sides of the equation by \left(2x+5\right)\left(11x+20\right), the least common multiple of 2x+5,11x+20.

$$11x+20=\left(2x+5\right)x$$

Use the distributive property to multiply 2x+5 by x.

$$11x+20=2x^{2}+5x$$

Subtract 2x^{2} from both sides.

$$11x+20-2x^{2}=5x$$

Subtract 5x from both sides.

$$11x+20-2x^{2}-5x=0$$

Combine 11x and -5x to get 6x.

$$6x+20-2x^{2}=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$6x-2x^{2}=-20$$

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.

$$-2x^{2}+6x=-20$$

Divide both sides by -2.

$$\frac{-2x^{2}+6x}{-2}=-\frac{20}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{6}{-2}x=-\frac{20}{-2}$$

Divide 6 by -2.

$$x^{2}-3x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x^{2}-3x=10$$

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.

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

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

$$x^{2}-3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{49}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{7}{2} x-\frac{3}{2}=-\frac{7}{2}$$

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

$$x=5 x=-2$$