Solve 3+4/x+1/3+frac{1{x+2}}=5/6

$$\frac { 3 + \frac { 4 } { x + 1 } } { 3 + \frac { 1 } { x + 2 } } = \frac { 5 } { 6 }$$

Solve for x

$$x=-7$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x+1}{x+1}.

$$\frac{\frac{3\left(x+1\right)}{x+1}+\frac{4}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

Since \frac{3\left(x+1\right)}{x+1} and \frac{4}{x+1} have the same denominator, add them by adding their numerators.

$$\frac{\frac{3\left(x+1\right)+4}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

Do the multiplications in 3\left(x+1\right)+4.

$$\frac{\frac{3x+3+4}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

Combine like terms in 3x+3+4.

$$\frac{\frac{3x+7}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x+2}{x+2}.

$$\frac{\frac{3x+7}{x+1}}{\frac{3\left(x+2\right)}{x+2}+\frac{1}{x+2}}=\frac{5}{6}$$

Since \frac{3\left(x+2\right)}{x+2} and \frac{1}{x+2} have the same denominator, add them by adding their numerators.

$$\frac{\frac{3x+7}{x+1}}{\frac{3\left(x+2\right)+1}{x+2}}=\frac{5}{6}$$

Do the multiplications in 3\left(x+2\right)+1.

$$\frac{\frac{3x+7}{x+1}}{\frac{3x+6+1}{x+2}}=\frac{5}{6}$$

Combine like terms in 3x+6+1.

$$\frac{\frac{3x+7}{x+1}}{\frac{3x+7}{x+2}}=\frac{5}{6}$$

Variable x cannot be equal to -2 since division by zero is not defined. Divide \frac{3x+7}{x+1} by \frac{3x+7}{x+2} by multiplying \frac{3x+7}{x+1} by the reciprocal of \frac{3x+7}{x+2}.

$$\frac{\left(3x+7\right)\left(x+2\right)}{\left(x+1\right)\left(3x+7\right)}=\frac{5}{6}$$

Use the distributive property to multiply 3x+7 by x+2 and combine like terms.

$$\frac{3x^{2}+13x+14}{\left(x+1\right)\left(3x+7\right)}=\frac{5}{6}$$

Use the distributive property to multiply x+1 by 3x+7 and combine like terms.

$$\frac{3x^{2}+13x+14}{3x^{2}+10x+7}=\frac{5}{6}$$

Subtract \frac{5}{6} from both sides.

$$\frac{3x^{2}+13x+14}{3x^{2}+10x+7}-\frac{5}{6}=0$$

Factor 3x^{2}+10x+7.

$$\frac{3x^{2}+13x+14}{\left(x+1\right)\left(3x+7\right)}-\frac{5}{6}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(x+1\right)\left(3x+7\right) and 6 is 6\left(x+1\right)\left(3x+7\right). Multiply \frac{3x^{2}+13x+14}{\left(x+1\right)\left(3x+7\right)} times \frac{6}{6}. Multiply \frac{5}{6} times \frac{\left(x+1\right)\left(3x+7\right)}{\left(x+1\right)\left(3x+7\right)}.

$$\frac{6\left(3x^{2}+13x+14\right)}{6\left(x+1\right)\left(3x+7\right)}-\frac{5\left(x+1\right)\left(3x+7\right)}{6\left(x+1\right)\left(3x+7\right)}=0$$

Since \frac{6\left(3x^{2}+13x+14\right)}{6\left(x+1\right)\left(3x+7\right)} and \frac{5\left(x+1\right)\left(3x+7\right)}{6\left(x+1\right)\left(3x+7\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6\left(3x^{2}+13x+14\right)-5\left(x+1\right)\left(3x+7\right)}{6\left(x+1\right)\left(3x+7\right)}=0$$

Do the multiplications in 6\left(3x^{2}+13x+14\right)-5\left(x+1\right)\left(3x+7\right).

$$\frac{18x^{2}+78x+84-15x^{2}-35x-15x-35}{6\left(x+1\right)\left(3x+7\right)}=0$$

Combine like terms in 18x^{2}+78x+84-15x^{2}-35x-15x-35.

$$\frac{3x^{2}+28x+49}{6\left(x+1\right)\left(3x+7\right)}=0$$

Variable x cannot be equal to any of the values -\frac{7}{3},-1 since division by zero is not defined. Multiply both sides of the equation by 6\left(x+1\right)\left(3x+7\right).

$$3x^{2}+28x+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 28 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\times 3\times 49}}{2\times 3}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 3\times 49}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-28±\sqrt{784-12\times 49}}{2\times 3}$$

Multiply -12 times 49.

$$x=\frac{-28±\sqrt{784-588}}{2\times 3}$$

Add 784 to -588.

$$x=\frac{-28±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$x=\frac{-28±14}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-28±14}{6}$$

Now solve the equation x=\frac{-28±14}{6} when ± is plus. Add -28 to 14.

$$x=-\frac{14}{6}$$

Reduce the fraction \frac{-14}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-28±14}{6} when ± is minus. Subtract 14 from -28.

$$x=-\frac{42}{6}$$

Divide -42 by 6.

$$x=-7$$

The equation is now solved.

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

Variable x cannot be equal to -\frac{7}{3}.

$$x=-7$$

Steps for Completing the Square

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x+1}{x+1}.

$$\frac{\frac{3\left(x+1\right)}{x+1}+\frac{4}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

Since \frac{3\left(x+1\right)}{x+1} and \frac{4}{x+1} have the same denominator, add them by adding their numerators.

$$\frac{\frac{3\left(x+1\right)+4}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

Do the multiplications in 3\left(x+1\right)+4.

$$\frac{\frac{3x+3+4}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

Combine like terms in 3x+3+4.

$$\frac{\frac{3x+7}{x+1}}{3+\frac{1}{x+2}}=\frac{5}{6}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x+2}{x+2}.

$$\frac{\frac{3x+7}{x+1}}{\frac{3\left(x+2\right)}{x+2}+\frac{1}{x+2}}=\frac{5}{6}$$

Since \frac{3\left(x+2\right)}{x+2} and \frac{1}{x+2} have the same denominator, add them by adding their numerators.

$$\frac{\frac{3x+7}{x+1}}{\frac{3\left(x+2\right)+1}{x+2}}=\frac{5}{6}$$

Do the multiplications in 3\left(x+2\right)+1.

$$\frac{\frac{3x+7}{x+1}}{\frac{3x+6+1}{x+2}}=\frac{5}{6}$$

Combine like terms in 3x+6+1.

$$\frac{\frac{3x+7}{x+1}}{\frac{3x+7}{x+2}}=\frac{5}{6}$$

Variable x cannot be equal to -2 since division by zero is not defined. Divide \frac{3x+7}{x+1} by \frac{3x+7}{x+2} by multiplying \frac{3x+7}{x+1} by the reciprocal of \frac{3x+7}{x+2}.

$$\frac{\left(3x+7\right)\left(x+2\right)}{\left(x+1\right)\left(3x+7\right)}=\frac{5}{6}$$

Use the distributive property to multiply 3x+7 by x+2 and combine like terms.

$$\frac{3x^{2}+13x+14}{\left(x+1\right)\left(3x+7\right)}=\frac{5}{6}$$

Use the distributive property to multiply x+1 by 3x+7 and combine like terms.

$$\frac{3x^{2}+13x+14}{3x^{2}+10x+7}=\frac{5}{6}$$

Variable x cannot be equal to any of the values -\frac{7}{3},-1 since division by zero is not defined. Multiply both sides of the equation by 6\left(x+1\right)\left(3x+7\right), the least common multiple of 3x^{2}+10x+7,6.

$$6\left(3x^{2}+13x+14\right)=5\left(x+1\right)\left(3x+7\right)$$

Use the distributive property to multiply 6 by 3x^{2}+13x+14.

$$18x^{2}+78x+84=5\left(x+1\right)\left(3x+7\right)$$

Use the distributive property to multiply 5 by x+1.

$$18x^{2}+78x+84=\left(5x+5\right)\left(3x+7\right)$$

Use the distributive property to multiply 5x+5 by 3x+7 and combine like terms.

$$18x^{2}+78x+84=15x^{2}+50x+35$$

Subtract 15x^{2} from both sides.

$$18x^{2}+78x+84-15x^{2}=50x+35$$

Combine 18x^{2} and -15x^{2} to get 3x^{2}.

$$3x^{2}+78x+84=50x+35$$

Subtract 50x from both sides.

$$3x^{2}+78x+84-50x=35$$

Combine 78x and -50x to get 28x.

$$3x^{2}+28x+84=35$$

Subtract 84 from both sides.

$$3x^{2}+28x=35-84$$

Subtract 84 from 35 to get -49.

$$3x^{2}+28x=-49$$

Divide both sides by 3.

$$\frac{3x^{2}+28x}{3}=-\frac{49}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{28}{3}x=-\frac{49}{3}$$

Divide \frac{28}{3}, the coefficient of the x term, by 2 to get \frac{14}{3}. Then add the square of \frac{14}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{28}{3}x+\left(\frac{14}{3}\right)^{2}=-\frac{49}{3}+\left(\frac{14}{3}\right)^{2}$$

Square \frac{14}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{28}{3}x+\frac{196}{9}=-\frac{49}{3}+\frac{196}{9}$$

Add -\frac{49}{3} to \frac{196}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{28}{3}x+\frac{196}{9}=\frac{49}{9}$$

Factor x^{2}+\frac{28}{3}x+\frac{196}{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}.

$$\left(x+\frac{14}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{14}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x+\frac{14}{3}=\frac{7}{3} x+\frac{14}{3}=-\frac{7}{3}$$

Subtract \frac{14}{3} from both sides of the equation.

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

Variable x cannot be equal to -\frac{7}{3}.

$$x=-7$$