Solve for y (7y-3)/(y-y^2)=1/(y-1)-5/(y(y-1))

$\frac{7 y - 3}{y - y^{2}} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$

Step 1

Factor $y$ out of $y - y^{2}$ .

Raise $y$ to the power of $1$ .

$\frac{7 y - 3}{y^{1} - y^{2}} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$

Factor $y$ out of $y^{1}$ .

$\frac{7 y - 3}{y \cdot 1 - y^{2}} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$

Factor $y$ out of $- y^{2}$ .

$\frac{7 y - 3}{y \cdot 1 + y (- y)} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$

Factor $y$ out of $y \cdot 1 + y (- y)$ .

$\frac{7 y - 3}{y \cdot (1 - y)} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$

Multiply $y$ by $1 - y$ .

$\frac{7 y - 3}{y (1 - y)} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$

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.

$y (1 - y), y - 1, y (y - 1)$

Since $y (1 - y), y - 1, y (y - 1)$ contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.

Steps to find the LCM for $y (1 - y), y - 1, y (y - 1)$ are:

1. Find the LCM for the numeric part $1, 1, 1$ .

2. Find the LCM for the variable part $y^{1}, y^{1}$ .

3. Find the LCM for the compound variable part $1 - y, y - 1, y - 1$ .

4. Multiply each LCM together.

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.

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

Not prime

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

$1$

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

$y^{1} = y$

$y$ occurs $1$ time.

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

$y$

The factor for $1 - y$ is $1 - y$ itself.

$(1 - y) = 1 - y$

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

The factor for $y - 1$ is $y - 1$ itself.

$(y - 1) = y - 1$

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

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

$(1 - y) (y - 1)$

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$y (1 - y) (y - 1)$

Step 3

Multiply each term in $\frac{7 y - 3}{y (1 - y)} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$ by $y (1 - y) (y - 1)$ to eliminate the fractions.

Multiply each term in $\frac{7 y - 3}{y (1 - y)} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$ by $y (1 - y) (y - 1)$ .

$\frac{7 y - 3}{y (1 - y)} (y (1 - y) (y - 1)) = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Simplify the left side.

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

Cancel the common factor.

$\frac{7 y - 3}{y (1 - y)} (y (1 - y) (y - 1)) = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Rewrite the expression.

$(7 y - 3) (y - 1) = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Expand $(7 y - 3) (y - 1)$ using the FOIL Method.

Apply the distributive property.

$7 y (y - 1) - 3 (y - 1) = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Apply the distributive property.

$7 y \cdot y + 7 y \cdot - 1 - 3 (y - 1) = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Apply the distributive property.

$7 y \cdot y + 7 y \cdot - 1 - 3 y - 3 \cdot - 1 = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Simplify and combine like terms.

Simplify each term.

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

Move $y$ .

$7 (y \cdot y) + 7 y \cdot - 1 - 3 y - 3 \cdot - 1 = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Multiply $y$ by $y$ .

$7 y^{2} + 7 y \cdot - 1 - 3 y - 3 \cdot - 1 = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Multiply $- 1$ by $7$ .

$7 y^{2} - 7 y - 3 y - 3 \cdot - 1 = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Multiply $- 3$ by $- 1$ .

$7 y^{2} - 7 y - 3 y + 3 = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Subtract $3 y$ from $- 7 y$ .

$7 y^{2} - 10 y + 3 = \frac{1}{y - 1} (y (1 - y) (y - 1)) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Simplify the right side.

Simplify each term.

Cancel the common factor of $y - 1$ .

Factor $y - 1$ out of $y (1 - y) (y - 1)$ .

$7 y^{2} - 10 y + 3 = \frac{1}{y - 1} ((y - 1) (y (1 - y))) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Cancel the common factor.

$7 y^{2} - 10 y + 3 = \frac{1}{y - 1} ((y - 1) (y (1 - y))) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Rewrite the expression.

$7 y^{2} - 10 y + 3 = y (1 - y) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Apply the distributive property.

$7 y^{2} - 10 y + 3 = y \cdot 1 + y (- y) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Multiply $y$ by $1$ .

$7 y^{2} - 10 y + 3 = y + y (- y) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Rewrite using the commutative property of multiplication.

$7 y^{2} - 10 y + 3 = y - y \cdot y - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

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

Move $y$ .

$7 y^{2} - 10 y + 3 = y - (y \cdot y) - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

Multiply $y$ by $y$ .

$7 y^{2} - 10 y + 3 = y - y^{2} - \frac{5}{y (y - 1)} (y (1 - y) (y - 1))$

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

Move the leading negative in $- \frac{5}{y (y - 1)}$ into the numerator.

$7 y^{2} - 10 y + 3 = y - y^{2} + \frac{- 5}{y (y - 1)} (y (1 - y) (y - 1))$

Factor $y (y - 1)$ out of $y (1 - y) (y - 1)$ .

$7 y^{2} - 10 y + 3 = y - y^{2} + \frac{- 5}{y (y - 1)} (y (y - 1) (1 - y))$

Cancel the common factor.

$7 y^{2} - 10 y + 3 = y - y^{2} + \frac{- 5}{y (y - 1)} (y (y - 1) (1 - y))$

Rewrite the expression.

$7 y^{2} - 10 y + 3 = y - y^{2} - 5 (1 - y)$

Apply the distributive property.

$7 y^{2} - 10 y + 3 = y - y^{2} - 5 \cdot 1 - 5 (- y)$

Multiply $- 5$ by $1$ .

$7 y^{2} - 10 y + 3 = y - y^{2} - 5 - 5 (- y)$

Multiply $- 1$ by $- 5$ .

$7 y^{2} - 10 y + 3 = y - y^{2} - 5 + 5 y$

Add $y$ and $5 y$ .

$7 y^{2} - 10 y + 3 = 6 y - y^{2} - 5$

Step 4

Solve the equation.

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

Subtract $6 y$ from both sides of the equation.

$7 y^{2} - 10 y + 3 - 6 y = - y^{2} - 5$

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

$7 y^{2} - 10 y + 3 - 6 y + y^{2} = - 5$

Add $7 y^{2}$ and $y^{2}$ .

$8 y^{2} - 10 y + 3 - 6 y = - 5$

Subtract $6 y$ from $- 10 y$ .

$8 y^{2} - 16 y + 3 = - 5$

Add $5$ to both sides of the equation.

$8 y^{2} - 16 y + 3 + 5 = 0$

Add $3$ and $5$ .

$8 y^{2} - 16 y + 8 = 0$

Factor the left side of the equation.

Factor $8$ out of $8 y^{2} - 16 y + 8$ .

Factor $8$ out of $8 y^{2}$ .

$8 (y^{2}) - 16 y + 8 = 0$

Factor $8$ out of $- 16 y$ .

$8 (y^{2}) + 8 (- 2 y) + 8 = 0$

Factor $8$ out of $8$ .

$8 (y^{2}) + 8 (- 2 y) + 8 (1) = 0$

Factor $8$ out of $8 (y^{2}) + 8 (- 2 y)$ .

$8 (y^{2} - 2 y) + 8 (1) = 0$

Factor $8$ out of $8 (y^{2} - 2 y) + 8 (1)$ .

$8 (y^{2} - 2 y + 1) = 0$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$8 (y^{2} - 2 y + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 y = 2 \cdot y \cdot 1$

Rewrite the polynomial.

$8 (y^{2} - 2 \cdot y \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = y$ and $b = 1$ .

$8 {(y - 1)}^{2} = 0$

Divide each term in $8 {(y - 1)}^{2} = 0$ by $8$ and simplify.

Divide each term in $8 {(y - 1)}^{2} = 0$ by $8$ .

$\frac{8 {(y - 1)}^{2}}{8} = \frac{0}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 {(y - 1)}^{2}}{8} = \frac{0}{8}$

Divide ${(y - 1)}^{2}$ by $1$ .

${(y - 1)}^{2} = \frac{0}{8}$

Simplify the right side.

Divide $0$ by $8$ .

${(y - 1)}^{2} = 0$

Set the $y - 1$ equal to $0$ .

$y - 1 = 0$

Add $1$ to both sides of the equation.

$y = 1$

Step 5

Exclude the solutions that do not make $\frac{7 y - 3}{y - y^{2}} = \frac{1}{y - 1} - \frac{5}{y (y - 1)}$ true.

$No solution$

Solve for y (7y-3)/(y-y^2)=1/(y-1)-5/(y(y-1))

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