Solve x+9/20x=27/20

$$x+ \frac{ 9 }{ 20x } = \frac{ 27 }{ 20 }$$

Solve for x

$$x=\frac{3}{5}=0.6$$

$$x=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 20x, the least common multiple of 20x,20.

$$20xx+9=27x$$

Multiply x and x to get x^{2}.

$$20x^{2}+9=27x$$

Subtract 27x from both sides.

$$20x^{2}+9-27x=0$$

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

$$20x^{2}-27x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-27 ab=20\times 9=180$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 180.

$$-1,-180 -2,-90 -3,-60 -4,-45 -5,-36 -6,-30 -9,-20 -10,-18 -12,-15$$

Calculate the sum for each pair.

$$-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27$$

The solution is the pair that gives sum -27.

$$a=-15 b=-12$$

Rewrite 20x^{2}-27x+9 as \left(20x^{2}-15x\right)+\left(-12x+9\right).

$$\left(20x^{2}-15x\right)+\left(-12x+9\right)$$

Factor out 5x in the first and -3 in the second group.

$$5x\left(4x-3\right)-3\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(5x-3\right)$$

To find equation solutions, solve 4x-3=0 and 5x-3=0.

$$x=\frac{3}{4} x=\frac{3}{5}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 20x, the least common multiple of 20x,20.

$$20xx+9=27x$$

Multiply x and x to get x^{2}.

$$20x^{2}+9=27x$$

Subtract 27x from both sides.

$$20x^{2}+9-27x=0$$

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.

$$20x^{2}-27x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -27 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-27\right)±\sqrt{\left(-27\right)^{2}-4\times 20\times 9}}{2\times 20}$$

Square -27.

$$x=\frac{-\left(-27\right)±\sqrt{729-4\times 20\times 9}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-\left(-27\right)±\sqrt{729-80\times 9}}{2\times 20}$$

Multiply -80 times 9.

$$x=\frac{-\left(-27\right)±\sqrt{729-720}}{2\times 20}$$

Add 729 to -720.

$$x=\frac{-\left(-27\right)±\sqrt{9}}{2\times 20}$$

Take the square root of 9.

$$x=\frac{-\left(-27\right)±3}{2\times 20}$$

The opposite of -27 is 27.

$$x=\frac{27±3}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{27±3}{40}$$

Now solve the equation x=\frac{27±3}{40} when ± is plus. Add 27 to 3.

$$x=\frac{30}{40}$$

Reduce the fraction \frac{30}{40} to lowest terms by extracting and canceling out 10.

$$x=\frac{3}{4}$$

Now solve the equation x=\frac{27±3}{40} when ± is minus. Subtract 3 from 27.

$$x=\frac{24}{40}$$

Reduce the fraction \frac{24}{40} to lowest terms by extracting and canceling out 8.

$$x=\frac{3}{5}$$

The equation is now solved.

$$x=\frac{3}{4} x=\frac{3}{5}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 20x, the least common multiple of 20x,20.

$$20xx+9=27x$$

Multiply x and x to get x^{2}.

$$20x^{2}+9=27x$$

Subtract 27x from both sides.

$$20x^{2}+9-27x=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$20x^{2}-27x=-9$$

Divide both sides by 20.

$$\frac{20x^{2}-27x}{20}=-\frac{9}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}-\frac{27}{20}x=-\frac{9}{20}$$

Divide -\frac{27}{20}, the coefficient of the x term, by 2 to get -\frac{27}{40}. Then add the square of -\frac{27}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{27}{20}x+\left(-\frac{27}{40}\right)^{2}=-\frac{9}{20}+\left(-\frac{27}{40}\right)^{2}$$

Square -\frac{27}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{27}{20}x+\frac{729}{1600}=-\frac{9}{20}+\frac{729}{1600}$$

Add -\frac{9}{20} to \frac{729}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{27}{20}x+\frac{729}{1600}=\frac{9}{1600}$$

Factor x^{2}-\frac{27}{20}x+\frac{729}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{40}\right)^{2}=\frac{9}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{40}\right)^{2}}=\sqrt{\frac{9}{1600}}$$

Simplify.

$$x-\frac{27}{40}=\frac{3}{40} x-\frac{27}{40}=-\frac{3}{40}$$

Add \frac{27}{40} to both sides of the equation.

$$x=\frac{3}{4} x=\frac{3}{5}$$