Solve for b b/2-31/6=b/3+5/b

$\frac{b}{2} - \frac{31}{6} = \frac{b}{3} + \frac{5}{b}$

Step 1

Add $\frac{31}{6}$ to both sides of the equation.

$\frac{b}{2} = \frac{b}{3} + \frac{5}{b} + \frac{31}{6}$

Step 2

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$2, 3, b, 6$

Since $2, 3, b, 6$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $2, 3, 1, 6$ then find LCM for the variable part $b^{1}$ .

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Since $2$ has no factors besides $1$ and $2$ .

$2$ is a prime number

Since $3$ has no factors besides $1$ and $3$ .

$3$ is a prime number

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

$6$ has factors of $2$ and $3$ .

$2 \cdot 3$

Multiply $2$ by $3$ .

$6$

The factor for $b^{1}$ is $b$ itself.

$b^{1} = b$

$b$ occurs $1$ time.

The LCM of $b^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$b$

The LCM for $2, 3, b, 6$ is the numeric part $6$ multiplied by the variable part.

$6 b$

Step 3

Multiply each term in $\frac{b}{2} = \frac{b}{3} + \frac{5}{b} + \frac{31}{6}$ by $6 b$ to eliminate the fractions.

Multiply each term in $\frac{b}{2} = \frac{b}{3} + \frac{5}{b} + \frac{31}{6}$ by $6 b$ .

$\frac{b}{2} (6 b) = \frac{b}{3} (6 b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Simplify the left side.

Rewrite using the commutative property of multiplication.

$6 \frac{b}{2} b = \frac{b}{3} (6 b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$2 (3) \frac{b}{2} b = \frac{b}{3} (6 b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Cancel the common factor.

$2 \cdot 3 \frac{b}{2} b = \frac{b}{3} (6 b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Rewrite the expression.

$3 b \cdot b = \frac{b}{3} (6 b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Multiply $b$ by $b$ by adding the exponents.

Move $b$ .

$3 (b \cdot b) = \frac{b}{3} (6 b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Multiply $b$ by $b$ .

$3 b^{2} = \frac{b}{3} (6 b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Simplify the right side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 b^{2} = 6 \frac{b}{3} b + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$3 b^{2} = 3 (2) \frac{b}{3} b + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Cancel the common factor.

$3 b^{2} = 3 \cdot 2 \frac{b}{3} b + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Rewrite the expression.

$3 b^{2} = 2 b \cdot b + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Multiply $b$ by $b$ by adding the exponents.

Move $b$ .

$3 b^{2} = 2 (b \cdot b) + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Multiply $b$ by $b$ .

$3 b^{2} = 2 b^{2} + \frac{5}{b} (6 b) + \frac{31}{6} (6 b)$

Rewrite using the commutative property of multiplication.

$3 b^{2} = 2 b^{2} + 6 \frac{5}{b} b + \frac{31}{6} (6 b)$

Multiply $6 \frac{5}{b}$ .

Combine $6$ and $\frac{5}{b}$ .

$3 b^{2} = 2 b^{2} + \frac{6 \cdot 5}{b} b + \frac{31}{6} (6 b)$

Multiply $6$ by $5$ .

$3 b^{2} = 2 b^{2} + \frac{30}{b} b + \frac{31}{6} (6 b)$

Cancel the common factor of $b$ .

Cancel the common factor.

$3 b^{2} = 2 b^{2} + \frac{30}{b} b + \frac{31}{6} (6 b)$

Rewrite the expression.

$3 b^{2} = 2 b^{2} + 30 + \frac{31}{6} (6 b)$

Cancel the common factor of $6$ .

Factor $6$ out of $6 b$ .

$3 b^{2} = 2 b^{2} + 30 + \frac{31}{6} (6 (b))$

Cancel the common factor.

$3 b^{2} = 2 b^{2} + 30 + \frac{31}{6} (6 b)$

Rewrite the expression.

$3 b^{2} = 2 b^{2} + 30 + 31 b$

Step 4

Solve the equation.

Since $b$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 b^{2} + 30 + 31 b = 3 b^{2}$

Move all terms containing $b$ to the left side of the equation.

Subtract $3 b^{2}$ from both sides of the equation.

$2 b^{2} + 30 + 31 b - 3 b^{2} = 0$

Subtract $3 b^{2}$ from $2 b^{2}$ .

$- b^{2} + 30 + 31 b = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - 1$ , $b = 31$ , and $c = 30$ into the quadratic formula and solve for $b$ .

$\frac{- 31 \pm \sqrt{31^{2} - 4 \cdot (- 1 \cdot 30)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

Raise $31$ to the power of $2$ .

$b = \frac{- 31 \pm \sqrt{961 - 4 \cdot - 1 \cdot 30}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 30$ .

Multiply $- 4$ by $- 1$ .

$b = \frac{- 31 \pm \sqrt{961 + 4 \cdot 30}}{2 \cdot - 1}$

Multiply $4$ by $30$ .

$b = \frac{- 31 \pm \sqrt{961 + 120}}{2 \cdot - 1}$

Add $961$ and $120$ .

$b = \frac{- 31 \pm \sqrt{1081}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$b = \frac{- 31 \pm \sqrt{1081}}{- 2}$

Simplify $\frac{- 31 \pm \sqrt{1081}}{- 2}$ .

$b = \frac{31 \pm \sqrt{1081}}{2}$

The final answer is the combination of both solutions.

$b = \frac{31 + \sqrt{1081}}{2}, \frac{31 - \sqrt{1081}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$b = \frac{31 + \sqrt{1081}}{2}, \frac{31 - \sqrt{1081}}{2}$

Decimal Form:

$b = 31.93928222 \dots, - 0.93928222 \dots$

Solve for b b/2-31/6=b/3+5/b

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