Solve 1/3x+1+2/x+1=3

$$\frac { 1 } { 3 x + 1 } + \frac { 2 } { x + 1 } = 3$$

Solve for x

$$x=-\frac{5}{9}\approx -0.555555556$$

$$x=0$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -1,-\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(3x+1\right), the least common multiple of 3x+1,x+1.

$$x+1+\left(3x+1\right)\times 2=3\left(x+1\right)\left(3x+1\right)$$

Use the distributive property to multiply 3x+1 by 2.

$$x+1+6x+2=3\left(x+1\right)\left(3x+1\right)$$

Combine x and 6x to get 7x.

$$7x+1+2=3\left(x+1\right)\left(3x+1\right)$$

Add 1 and 2 to get 3.

$$7x+3=3\left(x+1\right)\left(3x+1\right)$$

Use the distributive property to multiply 3 by x+1.

$$7x+3=\left(3x+3\right)\left(3x+1\right)$$

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

$$7x+3=9x^{2}+12x+3$$

Subtract 9x^{2} from both sides.

$$7x+3-9x^{2}=12x+3$$

Subtract 12x from both sides.

$$7x+3-9x^{2}-12x=3$$

Combine 7x and -12x to get -5x.

$$-5x+3-9x^{2}=3$$

Subtract 3 from both sides.

$$-5x+3-9x^{2}-3=0$$

Subtract 3 from 3 to get 0.

$$-5x-9x^{2}=0$$

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.

$$-9x^{2}-5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, -5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\left(-9\right)}$$

Take the square root of \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\left(-9\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±5}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{5±5}{-18}$$

Now solve the equation x=\frac{5±5}{-18} when ± is plus. Add 5 to 5.

$$x=\frac{10}{-18}$$

Reduce the fraction \frac{10}{-18} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{9}$$

Now solve the equation x=\frac{5±5}{-18} when ± is minus. Subtract 5 from 5.

$$x=\frac{0}{-18}$$

Divide 0 by -18.

$$x=0$$

The equation is now solved.

$$x=-\frac{5}{9} x=0$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -1,-\frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)\left(3x+1\right), the least common multiple of 3x+1,x+1.

$$x+1+\left(3x+1\right)\times 2=3\left(x+1\right)\left(3x+1\right)$$

Use the distributive property to multiply 3x+1 by 2.

$$x+1+6x+2=3\left(x+1\right)\left(3x+1\right)$$

Combine x and 6x to get 7x.

$$7x+1+2=3\left(x+1\right)\left(3x+1\right)$$

Add 1 and 2 to get 3.

$$7x+3=3\left(x+1\right)\left(3x+1\right)$$

Use the distributive property to multiply 3 by x+1.

$$7x+3=\left(3x+3\right)\left(3x+1\right)$$

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

$$7x+3=9x^{2}+12x+3$$

Subtract 9x^{2} from both sides.

$$7x+3-9x^{2}=12x+3$$

Subtract 12x from both sides.

$$7x+3-9x^{2}-12x=3$$

Combine 7x and -12x to get -5x.

$$-5x+3-9x^{2}=3$$

Subtract 3 from both sides.

$$-5x-9x^{2}=3-3$$

Subtract 3 from 3 to get 0.

$$-5x-9x^{2}=0$$

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.

$$-9x^{2}-5x=0$$

Divide both sides by -9.

$$\frac{-9x^{2}-5x}{-9}=\frac{0}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\left(-\frac{5}{-9}\right)x=\frac{0}{-9}$$

Divide -5 by -9.

$$x^{2}+\frac{5}{9}x=\frac{0}{-9}$$

Divide 0 by -9.

$$x^{2}+\frac{5}{9}x=0$$

Divide \frac{5}{9}, the coefficient of the x term, by 2 to get \frac{5}{18}. Then add the square of \frac{5}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{9}x+\left(\frac{5}{18}\right)^{2}=\left(\frac{5}{18}\right)^{2}$$

Square \frac{5}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{9}x+\frac{25}{324}=\frac{25}{324}$$

Factor x^{2}+\frac{5}{9}x+\frac{25}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{18}\right)^{2}=\frac{25}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{18}\right)^{2}}=\sqrt{\frac{25}{324}}$$

Simplify.

$$x+\frac{5}{18}=\frac{5}{18} x+\frac{5}{18}=-\frac{5}{18}$$

Subtract \frac{5}{18} from both sides of the equation.

$$x=0 x=-\frac{5}{9}$$