Basic Math Examples

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{2 y^{2} - 5 y - 3}$

Step 1

Factor by grouping.

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Step 1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 5$ .

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Step 1.1.1

Factor $- 5$ out of $- 5 y$ .

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{2 y^{2} - 5 (y) - 3}$

Step 1.1.2

Rewrite $- 5$ as $1$ plus $- 6$

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{2 y^{2} + (1 - 6) y - 3}$

Step 1.1.3

Apply the distributive property.

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{2 y^{2} + 1 y - 6 y - 3}$

Step 1.1.4

Multiply $y$ by $1$ .

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{2 y^{2} + y - 6 y - 3}$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{(2 y^{2} + y) - 6 y - 3}$

Step 1.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{y (2 y + 1) - 3 (2 y + 1)}$

Step 1.3

Factor the polynomial by factoring out the greatest common factor, $2 y + 1$ .

$\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)}$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

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

$2 y + 1, y - 3, (2 y + 1) (y - 3)$

Step 2.2

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.

Step 2.3

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

Not prime

Step 2.4

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 2.5

The factor for $2 y + 1$ is $2 y + 1$ itself.

$(2 y + 1) = 2 y + 1$

$(2 y + 1)$ occurs $1$ time.

Step 2.6

The factor for $y - 3$ is $y - 3$ itself.

$(y - 3) = y - 3$

$(y - 3)$ occurs $1$ time.

Step 2.7

The factor for $2 y + 1$ is $2 y + 1$ itself.

$(2 y + 1) = 2 y + 1$

$(2 y + 1)$ occurs $1$ time.

Step 2.8

The factor for $y - 3$ is $y - 3$ itself.

$(y - 3) = y - 3$

$(y - 3)$ occurs $1$ time.

Step 2.9

The LCM of $2 y + 1, y - 3, 2 y + 1, y - 3$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(2 y + 1) (y - 3)$

Step 3

Multiply each term in $\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)}$ by $(2 y + 1) (y - 3)$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{4}{2 y + 1} - \frac{y}{y - 3} = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)}$ by $(2 y + 1) (y - 3)$ .

$\frac{4}{2 y + 1} ((2 y + 1) (y - 3)) - \frac{y}{y - 3} ((2 y + 1) (y - 3)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $2 y + 1$ .

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Step 3.2.1.1.1

Cancel the common factor.

$\frac{4}{2 y + 1} ((2 y + 1) (y - 3)) - \frac{y}{y - 3} ((2 y + 1) (y - 3)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.1.2

Rewrite the expression.

$4 (y - 3) - \frac{y}{y - 3} ((2 y + 1) (y - 3)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.2

Apply the distributive property.

$4 y + 4 \cdot - 3 - \frac{y}{y - 3} ((2 y + 1) (y - 3)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.3

Multiply $4$ by $- 3$ .

$4 y - 12 - \frac{y}{y - 3} ((2 y + 1) (y - 3)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.4

Cancel the common factor of $y - 3$ .

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Step 3.2.1.4.1

Move the leading negative in $- \frac{y}{y - 3}$ into the numerator.

$4 y - 12 + \frac{- y}{y - 3} ((2 y + 1) (y - 3)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.4.2

Factor $y - 3$ out of $(2 y + 1) (y - 3)$ .

$4 y - 12 + \frac{- y}{y - 3} ((y - 3) (2 y + 1)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.4.3

Cancel the common factor.

$4 y - 12 + \frac{- y}{y - 3} ((y - 3) (2 y + 1)) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.4.4

Rewrite the expression.

$4 y - 12 - y (2 y + 1) = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.5

Apply the distributive property.

$4 y - 12 - y (2 y) - y \cdot 1 = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.6

Rewrite using the commutative property of multiplication.

$4 y - 12 - 1 \cdot 2 y \cdot y - y \cdot 1 = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.7

Multiply $- 1$ by $1$ .

$4 y - 12 - 1 \cdot 2 y \cdot y - y = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.8

Simplify each term.

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Step 3.2.1.8.1

Multiply $y$ by $y$ by adding the exponents.

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Step 3.2.1.8.1.1

Move $y$ .

$4 y - 12 - 1 \cdot 2 (y \cdot y) - y = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.8.1.2

Multiply $y$ by $y$ .

$4 y - 12 - 1 \cdot 2 y^{2} - y = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.1.8.2

Multiply $- 1$ by $2$ .

$4 y - 12 - 2 y^{2} - y = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.2.2

Subtract $y$ from $4 y$ .

$3 y - 12 - 2 y^{2} = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.3

Simplify the right side.

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Step 3.3.1

Cancel the common factor of $(2 y + 1) (y - 3)$ .

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Step 3.3.1.1

Cancel the common factor.

$3 y - 12 - 2 y^{2} = \frac{- 3 y^{2} + 3 y + 37}{(2 y + 1) (y - 3)} ((2 y + 1) (y - 3))$

Step 3.3.1.2

Rewrite the expression.

$3 y - 12 - 2 y^{2} = - 3 y^{2} + 3 y + 37$

Step 4

Solve the equation.

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Step 4.1

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

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Step 4.1.1

Add $3 y^{2}$ to both sides of the equation.

$3 y - 12 - 2 y^{2} + 3 y^{2} = 3 y + 37$

Step 4.1.2

Subtract $3 y$ from both sides of the equation.

$3 y - 12 - 2 y^{2} + 3 y^{2} - 3 y = 37$

Step 4.1.3

Combine the opposite terms in $3 y - 12 - 2 y^{2} + 3 y^{2} - 3 y$ .

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Step 4.1.3.1

Subtract $3 y$ from $3 y$ .

$- 12 - 2 y^{2} + 3 y^{2} + 0 = 37$

Step 4.1.3.2

Add $- 12 - 2 y^{2} + 3 y^{2}$ and $0$ .

$- 12 - 2 y^{2} + 3 y^{2} = 37$

Step 4.1.4

Add $- 2 y^{2}$ and $3 y^{2}$ .

$- 12 + y^{2} = 37$

Step 4.2

Move all terms not containing $y$ to the right side of the equation.

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Step 4.2.1

Add $12$ to both sides of the equation.

$y^{2} = 37 + 12$

Step 4.2.2

Add $37$ and $12$ .

$y^{2} = 49$

Step 4.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{49}$

Step 4.4

Simplify $\pm \sqrt{49}$ .

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Step 4.4.1

Rewrite $49$ as $7^{2}$ .

$y = \pm \sqrt{7^{2}}$

Step 4.4.2

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 7$

Step 4.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.5.1

First, use the positive value of the $\pm$ to find the first solution.

$y = 7$

Step 4.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 7$

Step 4.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = 7, - 7$