Solve for x (7x)/(3x+3)-5/(4x-4)=(3x)/(2x+2)

$\frac{7 x}{3 x + 3} - \frac{5}{4 x - 4} = \frac{3 x}{2 x + 2}$

Factor each term.

Factor $3$ out of $3 x + 3$ .

Factor $3$ out of $3 x$ .

$\frac{7 x}{3 (x) + 3} - \frac{5}{4 x - 4} = \frac{3 x}{2 x + 2}$

Factor $3$ out of $3$ .

$\frac{7 x}{3 (x) + 3 (1)} - \frac{5}{4 x - 4} = \frac{3 x}{2 x + 2}$

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

$\frac{7 x}{3 (x + 1)} - \frac{5}{4 x - 4} = \frac{3 x}{2 x + 2}$

Factor $4$ out of $4 x - 4$ .

Factor $4$ out of $4 x$ .

$\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x) - 4} = \frac{3 x}{2 x + 2}$

Factor $4$ out of $- 4$ .

$\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x) + 4 (- 1)} = \frac{3 x}{2 x + 2}$

Factor $4$ out of $4 (x) + 4 (- 1)$ .

$\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x - 1)} = \frac{3 x}{2 x + 2}$

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

Factor $2$ out of $2 x$ .

$\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x - 1)} = \frac{3 x}{2 (x) + 2}$

Factor $2$ out of $2$ .

$\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x - 1)} = \frac{3 x}{2 (x) + 2 (1)}$

Factor $2$ out of $2 (x) + 2 (1)$ .

$\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x - 1)} = \frac{3 x}{2 (x + 1)}$

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.

$3 (x + 1), 4 (x - 1), 2 (x + 1)$

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

$3$ is a prime number

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

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

$2$ is a prime number

The LCM of $3, 4, 2$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$2 \cdot 2 \cdot 3$

The LCM of $3, 4, 2$ is $2 \cdot 2 \cdot 3 = 12$ .

Multiply $2$ by $2$ .

$4 \cdot 3$

Multiply $4$ by $3$ .

$12$

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

$(x + 1) = x + 1$

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

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

$(x - 1) = x - 1$

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

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

$(x + 1) = x + 1$

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

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

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

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

$12 (x + 1) (x - 1)$

Multiply each term in $\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x - 1)} = \frac{3 x}{2 (x + 1)}$ by $12 (x + 1) (x - 1)$ to eliminate the fractions.

Multiply each term in $\frac{7 x}{3 (x + 1)} - \frac{5}{4 (x - 1)} = \frac{3 x}{2 (x + 1)}$ by $12 (x + 1) (x - 1)$ .

$\frac{7 x}{3 (x + 1)} (12 (x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Simplify the left side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$12 \frac{7 x}{3 (x + 1)} ((x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Cancel the common factor of $3$ .

Factor $3$ out of $12$ .

$3 (4) \frac{7 x}{3 (x + 1)} ((x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Cancel the common factor.

$3 \cdot 4 \frac{7 x}{3 (x + 1)} ((x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Rewrite the expression.

$4 \frac{7 x}{x + 1} ((x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Combine $4$ and $\frac{7 x}{x + 1}$ .

$\frac{4 (7 x)}{x + 1} ((x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Multiply $7$ by $4$ .

$\frac{28 x}{x + 1} ((x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{28 x}{x + 1} ((x + 1) (x - 1)) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Rewrite the expression.

$28 x (x - 1) - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Apply the distributive property.

$28 x \cdot x + 28 x \cdot - 1 - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

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

Move $x$ .

$28 (x \cdot x) + 28 x \cdot - 1 - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Multiply $x$ by $x$ .

$28 x^{2} + 28 x \cdot - 1 - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Multiply $- 1$ by $28$ .

$28 x^{2} - 28 x - \frac{5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Cancel the common factor of $4 (x - 1)$ .

Move the leading negative in $- \frac{5}{4 (x - 1)}$ into the numerator.

$28 x^{2} - 28 x + \frac{- 5}{4 (x - 1)} (12 (x + 1) (x - 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Factor $4 (x - 1)$ out of $12 (x + 1) (x - 1)$ .

$28 x^{2} - 28 x + \frac{- 5}{4 (x - 1)} (4 (x - 1) (3 (x + 1))) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Cancel the common factor.

$28 x^{2} - 28 x + \frac{- 5}{4 (x - 1)} (4 (x - 1) (3 (x + 1))) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Rewrite the expression.

$28 x^{2} - 28 x - 5 (3 (x + 1)) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Multiply $3$ by $- 5$ .

$28 x^{2} - 28 x - 15 (x + 1) = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Apply the distributive property.

$28 x^{2} - 28 x - 15 x - 15 \cdot 1 = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Multiply $- 15$ by $1$ .

$28 x^{2} - 28 x - 15 x - 15 = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Subtract $15 x$ from $- 28 x$ .

$28 x^{2} - 43 x - 15 = \frac{3 x}{2 (x + 1)} (12 (x + 1) (x - 1))$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$28 x^{2} - 43 x - 15 = 12 \frac{3 x}{2 (x + 1)} ((x + 1) (x - 1))$

Cancel the common factor of $2$ .

Factor $2$ out of $12$ .

$28 x^{2} - 43 x - 15 = 2 (6) \frac{3 x}{2 (x + 1)} ((x + 1) (x - 1))$

Cancel the common factor.

$28 x^{2} - 43 x - 15 = 2 \cdot 6 \frac{3 x}{2 (x + 1)} ((x + 1) (x - 1))$

Rewrite the expression.

$28 x^{2} - 43 x - 15 = 6 \frac{3 x}{x + 1} ((x + 1) (x - 1))$

Combine $6$ and $\frac{3 x}{x + 1}$ .

$28 x^{2} - 43 x - 15 = \frac{6 (3 x)}{x + 1} ((x + 1) (x - 1))$

Multiply $3$ by $6$ .

$28 x^{2} - 43 x - 15 = \frac{18 x}{x + 1} ((x + 1) (x - 1))$

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$28 x^{2} - 43 x - 15 = \frac{18 x}{x + 1} ((x + 1) (x - 1))$

Rewrite the expression.

$28 x^{2} - 43 x - 15 = 18 x (x - 1)$

Apply the distributive property.

$28 x^{2} - 43 x - 15 = 18 x \cdot x + 18 x \cdot - 1$

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

Move $x$ .

$28 x^{2} - 43 x - 15 = 18 (x \cdot x) + 18 x \cdot - 1$

Multiply $x$ by $x$ .

$28 x^{2} - 43 x - 15 = 18 x^{2} + 18 x \cdot - 1$

Multiply $- 1$ by $18$ .

$28 x^{2} - 43 x - 15 = 18 x^{2} - 18 x$

Solve the equation.

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

Subtract $18 x^{2}$ from both sides of the equation.

$28 x^{2} - 43 x - 15 - 18 x^{2} = - 18 x$

Add $18 x$ to both sides of the equation.

$28 x^{2} - 43 x - 15 - 18 x^{2} + 18 x = 0$

Subtract $18 x^{2}$ from $28 x^{2}$ .

$10 x^{2} - 43 x - 15 + 18 x = 0$

Add $- 43 x$ and $18 x$ .

$10 x^{2} - 25 x - 15 = 0$

Factor the left side of the equation.

Factor $5$ out of $10 x^{2} - 25 x - 15$ .

Factor $5$ out of $10 x^{2}$ .

$5 (2 x^{2}) - 25 x - 15 = 0$

Factor $5$ out of $- 25 x$ .

$5 (2 x^{2}) + 5 (- 5 x) - 15 = 0$

Factor $5$ out of $- 15$ .

$5 (2 x^{2}) + 5 (- 5 x) + 5 (- 3) = 0$

Factor $5$ out of $5 (2 x^{2}) + 5 (- 5 x)$ .

$5 (2 x^{2} - 5 x) + 5 (- 3) = 0$

Factor $5$ out of $5 (2 x^{2} - 5 x) + 5 (- 3)$ .

$5 (2 x^{2} - 5 x - 3) = 0$

Factor.

Factor by grouping.

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 $b = - 5$ .

Factor $- 5$ out of $- 5 x$ .

$5 (2 x^{2} - 5 x - 3) = 0$

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

$5 (2 x^{2} + (1 - 6) x - 3) = 0$

Apply the distributive property.

$5 (2 x^{2} + 1 x - 6 x - 3) = 0$

Multiply $x$ by $1$ .

$5 (2 x^{2} + x - 6 x - 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$5 ((2 x^{2} + x) - 6 x - 3) = 0$

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

$5 (x (2 x + 1) - 3 (2 x + 1)) = 0$

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

$5 ((2 x + 1) (x - 3)) = 0$

Remove unnecessary parentheses.

$5 (2 x + 1) (x - 3) = 0$

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

$2 x + 1 = 0$

$x - 3 = 0$

Set $2 x + 1$ equal to $0$ and solve for $x$ .

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $5 (2 x + 1) (x - 3) = 0$ true.

$x = - \frac{1}{2}, 3$

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1}{2}, 3$

Decimal Form:

$x = - 0.5, 3$

Solve for x (7x)/(3x+3)-5/(4x-4)=(3x)/(2x+2)

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