Solve Using the Quadratic Formula 3/(x-1)-(2x+10)/(x^2+2x-3)=1/3

$\frac{3}{x - 1} - \frac{2 x + 10}{x^{2} + 2 x - 3} = \frac{1}{3}$

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify each term.

Factor $2$ out of $2 x + 10$ .

Factor $2$ out of $2 x$ .

$\frac{3}{x - 1} - \frac{2 (x) + 10}{x^{2} + 2 x - 3} = \frac{1}{3}$

Factor $2$ out of $10$ .

$\frac{3}{x - 1} - \frac{2 x + 2 \cdot 5}{x^{2} + 2 x - 3} = \frac{1}{3}$

Factor $2$ out of $2 x + 2 \cdot 5$ .

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

Factor $x^{2} + 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$\frac{3}{x - 1} - \frac{2 (x + 5)}{(x - 1) (x + 3)} = \frac{1}{3}$

Subtract $\frac{1}{3}$ from both sides of the equation.

$\frac{3}{x - 1} - \frac{2 (x + 5)}{(x - 1) (x + 3)} - \frac{1}{3} = 0$

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.

$x - 1, (x - 1) (x + 3), 3, 1$

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.

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

Not prime

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

The LCM of $1, 1, 3, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$3$

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

$(x - 1) = x - 1$

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

The factor for $x + 3$ is $x + 3$ itself.

$(x + 3) = x + 3$

$(x + 3)$ occurs $1$ time.

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

$(x - 1) (x + 3)$

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

$3 (x - 1) (x + 3)$

Multiply each term in $\frac{3}{x - 1} - \frac{2 (x + 5)}{(x - 1) (x + 3)} - \frac{1}{3} = 0$ by $3 (x - 1) (x + 3)$ to eliminate the fractions.

Multiply each term in $\frac{3}{x - 1} - \frac{2 (x + 5)}{(x - 1) (x + 3)} - \frac{1}{3} = 0$ by $3 (x - 1) (x + 3)$ .

$\frac{3}{x - 1} (3 (x - 1) (x + 3)) - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 \frac{3}{x - 1} ((x - 1) (x + 3)) - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Multiply $3 \frac{3}{x - 1}$ .

Combine $3$ and $\frac{3}{x - 1}$ .

$\frac{3 \cdot 3}{x - 1} ((x - 1) (x + 3)) - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Multiply $3$ by $3$ .

$\frac{9}{x - 1} ((x - 1) (x + 3)) - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{9}{x - 1} ((x - 1) (x + 3)) - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Rewrite the expression.

$9 (x + 3) - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Apply the distributive property.

$9 x + 9 \cdot 3 - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Multiply $9$ by $3$ .

$9 x + 27 - \frac{2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

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

Move the leading negative in $- \frac{2 (x + 5)}{(x - 1) (x + 3)}$ into the numerator.

$9 x + 27 + \frac{- 2 (x + 5)}{(x - 1) (x + 3)} (3 (x - 1) (x + 3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Factor $(x - 1) (x + 3)$ out of $3 (x - 1) (x + 3)$ .

$9 x + 27 + \frac{- 2 (x + 5)}{(x - 1) (x + 3)} ((x - 1) (x + 3) (3)) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Cancel the common factor.

$9 x + 27 + \frac{- 2 (x + 5)}{(x - 1) (x + 3)} ((x - 1) (x + 3) \cdot 3) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Rewrite the expression.

$9 x + 27 - 2 (x + 5) \cdot 3 - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Multiply $3$ by $- 2$ .

$9 x + 27 - 6 (x + 5) - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Apply the distributive property.

$9 x + 27 - 6 x - 6 \cdot 5 - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Multiply $- 6$ by $5$ .

$9 x + 27 - 6 x - 30 - \frac{1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$9 x + 27 - 6 x - 30 + \frac{- 1}{3} (3 (x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Factor $3$ out of $3 (x - 1) (x + 3)$ .

$9 x + 27 - 6 x - 30 + \frac{- 1}{3} (3 ((x - 1) (x + 3))) = 0 (3 (x - 1) (x + 3))$

Cancel the common factor.

$9 x + 27 - 6 x - 30 + \frac{- 1}{3} (3 ((x - 1) (x + 3))) = 0 (3 (x - 1) (x + 3))$

Rewrite the expression.

$9 x + 27 - 6 x - 30 - ((x - 1) (x + 3)) = 0 (3 (x - 1) (x + 3))$

Expand $(x - 1) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$9 x + 27 - 6 x - 30 - (x (x + 3) - 1 (x + 3)) = 0 (3 (x - 1) (x + 3))$

Apply the distributive property.

$9 x + 27 - 6 x - 30 - (x \cdot x + x \cdot 3 - 1 (x + 3)) = 0 (3 (x - 1) (x + 3))$

Apply the distributive property.

$9 x + 27 - 6 x - 30 - (x \cdot x + x \cdot 3 - 1 x - 1 \cdot 3) = 0 (3 (x - 1) (x + 3))$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$9 x + 27 - 6 x - 30 - (x^{2} + x \cdot 3 - 1 x - 1 \cdot 3) = 0 (3 (x - 1) (x + 3))$

Move $3$ to the left of $x$ .

$9 x + 27 - 6 x - 30 - (x^{2} + 3 \cdot x - 1 x - 1 \cdot 3) = 0 (3 (x - 1) (x + 3))$

Rewrite $- 1 x$ as $- x$ .

$9 x + 27 - 6 x - 30 - (x^{2} + 3 x - x - 1 \cdot 3) = 0 (3 (x - 1) (x + 3))$

Multiply $- 1$ by $3$ .

$9 x + 27 - 6 x - 30 - (x^{2} + 3 x - x - 3) = 0 (3 (x - 1) (x + 3))$

Subtract $x$ from $3 x$ .

$9 x + 27 - 6 x - 30 - (x^{2} + 2 x - 3) = 0 (3 (x - 1) (x + 3))$

Apply the distributive property.

$9 x + 27 - 6 x - 30 - x^{2} - (2 x) - - 3 = 0 (3 (x - 1) (x + 3))$

Simplify.

Multiply $2$ by $- 1$ .

$9 x + 27 - 6 x - 30 - x^{2} - 2 x - - 3 = 0 (3 (x - 1) (x + 3))$

Multiply $- 1$ by $- 3$ .

$9 x + 27 - 6 x - 30 - x^{2} - 2 x + 3 = 0 (3 (x - 1) (x + 3))$

Simplify by adding terms.

Subtract $6 x$ from $9 x$ .

$3 x + 27 - 30 - x^{2} - 2 x + 3 = 0 (3 (x - 1) (x + 3))$

Subtract $2 x$ from $3 x$ .

$x + 27 - 30 - x^{2} + 3 = 0 (3 (x - 1) (x + 3))$

Subtract $30$ from $27$ .

$x - 3 - x^{2} + 3 = 0 (3 (x - 1) (x + 3))$

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

Add $- 3$ and $3$ .

$x - x^{2} + 0 = 0 (3 (x - 1) (x + 3))$

Add $x - x^{2}$ and $0$ .

$x - x^{2} = 0 (3 (x - 1) (x + 3))$

Simplify the right side.

Simplify by multiplying through.

Apply the distributive property.

$x - x^{2} = 0 ((3 x + 3 \cdot - 1) (x + 3))$

Multiply $3$ by $- 1$ .

$x - x^{2} = 0 ((3 x - 3) (x + 3))$

Expand $(3 x - 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$x - x^{2} = 0 (3 x (x + 3) - 3 (x + 3))$

Apply the distributive property.

$x - x^{2} = 0 (3 x \cdot x + 3 x \cdot 3 - 3 (x + 3))$

Apply the distributive property.

$x - x^{2} = 0 (3 x \cdot x + 3 x \cdot 3 - 3 x - 3 \cdot 3)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x - x^{2} = 0 (3 (x \cdot x) + 3 x \cdot 3 - 3 x - 3 \cdot 3)$

Multiply $x$ by $x$ .

$x - x^{2} = 0 (3 x^{2} + 3 x \cdot 3 - 3 x - 3 \cdot 3)$

Multiply $3$ by $3$ .

$x - x^{2} = 0 (3 x^{2} + 9 x - 3 x - 3 \cdot 3)$

Multiply $- 3$ by $3$ .

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

Subtract $3 x$ from $9 x$ .

$x - x^{2} = 0 (3 x^{2} + 6 x - 9)$

Multiply $0$ by $3 x^{2} + 6 x - 9$ .

$x - x^{2} = 0$

Solve the equation.

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$u - u^{2} = 0$

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

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

$u - u^{2} = 0$

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

$u \cdot 1 - u^{2} = 0$

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

$u \cdot 1 + u (- u) = 0$

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

$u (1 - u) = 0$

Replace all occurrences of $u$ with $x$ .

$x (1 - x) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$1 - x = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $1 - x$ equal to $0$ and solve for $x$ .

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

The final solution is all the values that make $x (1 - x) = 0$ true.

$x = 0, 1$

Exclude the solutions that do not make $\frac{3}{x - 1} - \frac{2 (x + 5)}{(x - 1) (x + 3)} - \frac{1}{3} = 0$ true.

$x = 0$

Solve Using the Quadratic Formula 3/(x-1)-(2x+10)/(x^2+2x-3)=1/3

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