Solve x/3+3/x=91/9

$$\frac { x } { 3 } + \frac { 3 } { x } = 9 \frac { 1 } { 9 }$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=27$$

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 9x, the least common multiple of 3,x,9.

$$3xx+9\times 3=x\left(9\times 9+1\right)$$

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

$$3x^{2}+9\times 3=x\left(9\times 9+1\right)$$

Multiply 9 and 3 to get 27.

$$3x^{2}+27=x\left(9\times 9+1\right)$$

Multiply 9 and 9 to get 81.

$$3x^{2}+27=x\left(81+1\right)$$

Add 81 and 1 to get 82.

$$3x^{2}+27=x\times 82$$

Subtract x\times 82 from both sides.

$$3x^{2}+27-x\times 82=0$$

Multiply -1 and 82 to get -82.

$$3x^{2}+27-82x=0$$

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

$$3x^{2}-82x+27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+27. To find a and b, set up a system to be solved.

$$a+b=-82 ab=3\times 27=81$$

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 81.

$$-1,-81 -3,-27 -9,-9$$

Calculate the sum for each pair.

$$-1-81=-82 -3-27=-30 -9-9=-18$$

The solution is the pair that gives sum -82.

$$a=-81 b=-1$$

Rewrite 3x^{2}-82x+27 as \left(3x^{2}-81x\right)+\left(-x+27\right).

$$\left(3x^{2}-81x\right)+\left(-x+27\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(x-27\right)-\left(x-27\right)$$

Factor out common term x-27 by using distributive property.

$$\left(x-27\right)\left(3x-1\right)$$

To find equation solutions, solve x-27=0 and 3x-1=0.

$$x=27 x=\frac{1}{3}$$

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 9x, the least common multiple of 3,x,9.

$$3xx+9\times 3=x\left(9\times 9+1\right)$$

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

$$3x^{2}+9\times 3=x\left(9\times 9+1\right)$$

Multiply 9 and 3 to get 27.

$$3x^{2}+27=x\left(9\times 9+1\right)$$

Multiply 9 and 9 to get 81.

$$3x^{2}+27=x\left(81+1\right)$$

Add 81 and 1 to get 82.

$$3x^{2}+27=x\times 82$$

Subtract x\times 82 from both sides.

$$3x^{2}+27-x\times 82=0$$

Multiply -1 and 82 to get -82.

$$3x^{2}+27-82x=0$$

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.

$$3x^{2}-82x+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -82 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-82\right)±\sqrt{\left(-82\right)^{2}-4\times 3\times 27}}{2\times 3}$$

Square -82.

$$x=\frac{-\left(-82\right)±\sqrt{6724-4\times 3\times 27}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-82\right)±\sqrt{6724-12\times 27}}{2\times 3}$$

Multiply -12 times 27.

$$x=\frac{-\left(-82\right)±\sqrt{6724-324}}{2\times 3}$$

Add 6724 to -324.

$$x=\frac{-\left(-82\right)±\sqrt{6400}}{2\times 3}$$

Take the square root of 6400.

$$x=\frac{-\left(-82\right)±80}{2\times 3}$$

The opposite of -82 is 82.

$$x=\frac{82±80}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{82±80}{6}$$

Now solve the equation x=\frac{82±80}{6} when ± is plus. Add 82 to 80.

$$x=\frac{162}{6}$$

Divide 162 by 6.

$$x=27$$

Now solve the equation x=\frac{82±80}{6} when ± is minus. Subtract 80 from 82.

$$x=\frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{3}$$

The equation is now solved.

$$x=27 x=\frac{1}{3}$$

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 9x, the least common multiple of 3,x,9.

$$3xx+9\times 3=x\left(9\times 9+1\right)$$

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

$$3x^{2}+9\times 3=x\left(9\times 9+1\right)$$

Multiply 9 and 3 to get 27.

$$3x^{2}+27=x\left(9\times 9+1\right)$$

Multiply 9 and 9 to get 81.

$$3x^{2}+27=x\left(81+1\right)$$

Add 81 and 1 to get 82.

$$3x^{2}+27=x\times 82$$

Subtract x\times 82 from both sides.

$$3x^{2}+27-x\times 82=0$$

Multiply -1 and 82 to get -82.

$$3x^{2}+27-82x=0$$

Subtract 27 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-82x=-27$$

Divide both sides by 3.

$$\frac{3x^{2}-82x}{3}=-\frac{27}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{82}{3}x=-\frac{27}{3}$$

Divide -27 by 3.

$$x^{2}-\frac{82}{3}x=-9$$

Divide -\frac{82}{3}, the coefficient of the x term, by 2 to get -\frac{41}{3}. Then add the square of -\frac{41}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{82}{3}x+\left(-\frac{41}{3}\right)^{2}=-9+\left(-\frac{41}{3}\right)^{2}$$

Square -\frac{41}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{82}{3}x+\frac{1681}{9}=-9+\frac{1681}{9}$$

Add -9 to \frac{1681}{9}.

$$x^{2}-\frac{82}{3}x+\frac{1681}{9}=\frac{1600}{9}$$

Factor x^{2}-\frac{82}{3}x+\frac{1681}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{41}{3}\right)^{2}=\frac{1600}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{3}\right)^{2}}=\sqrt{\frac{1600}{9}}$$

Simplify.

$$x-\frac{41}{3}=\frac{40}{3} x-\frac{41}{3}=-\frac{40}{3}$$

Add \frac{41}{3} to both sides of the equation.

$$x=27 x=\frac{1}{3}$$