Solve for x x+(3x+5)/(x+5)=(2x)/(x+5)

Solution:

Step:1

Identify the Least Common Denominator (LCD) for the equation's terms.

Step:1.1

To find the LCD, determine the Least Common Multiple (LCM) of the denominators $1$, $x + 5$, and $x + 5$.

Step:1.2

The LCM is the smallest number that each of the given numbers can divide into without a remainder. To find it: 1. Prime factorize each number. 2. Take the highest power of each prime factor found in any of the numbers.

Step:1.3

Since the number $1$ has only itself as a factor, it is not considered a prime number.

Step:1.4

The LCM of $1$, $1$, and $1$ is obtained by multiplying the highest power of all prime factors present in any of the numbers, which is simply $1$.

Step:1.5

The expression $x + 5$ is already in its simplest form and appears once.

Step:1.6

The LCM of $x + 5$ and $x + 5$ is simply $x + 5$, as it appears only once.

Step:2

Clear the fractions by multiplying every term in the equation $x + \frac{3x + 5}{x + 5} = \frac{2x}{x + 5}$ by the LCD $x + 5$.

Step:2.1

Multiply each term by $x + 5$: $x(x + 5) + \frac{3x + 5}{x + 5}(x + 5) = \frac{2x}{x + 5}(x + 5)$.

Step:2.2

Simplify the left-hand side of the equation.

Step:2.2.1

Apply simplification to each term.

Step:2.2.1.1

Use the distributive property: $x \cdot x + x \cdot 5 + \frac{3x + 5}{x + 5}(x + 5) = \frac{2x}{x + 5}(x + 5)$.

Step:2.2.1.2

Multiply $x$ by $x$: $x^2 + x \cdot 5 + \frac{3x + 5}{x + 5}(x + 5) = \frac{2x}{x + 5}(x + 5)$.

Step:2.2.1.3

Reorder the multiplication: $x^2 + 5 \cdot x + \frac{3x + 5}{x + 5}(x + 5) = \frac{2x}{x + 5}(x + 5)$.

Step:2.2.1.4

Eliminate the common factors of $x + 5$.

Step:2.2.1.4.1

Remove the common factor: $x^2 + 5x + \frac{3x + 5}{\cancel{x + 5}}(\cancel{x + 5}) = \frac{2x}{x + 5}(x + 5)$.

Step:2.2.1.4.2

Rewrite without the fraction: $x^2 + 5x + 3x + 5 = 2x$.

Step:2.2.2

Combine like terms: $x^2 + 8x + 5 = 2x$.

Step:2.3

Simplify the right-hand side.

Step:2.3.1

Remove the common factor of $x + 5$.

Step:2.3.1.1

Eliminate the common factor: $x^2 + 8x + 5 = \frac{2x}{\cancel{x + 5}}(\cancel{x + 5})$.

Step:2.3.1.2

Rewrite without the fraction: $x^2 + 8x + 5 = 2x$.

Step:3

Solve the resulting quadratic equation.

Step:3.1

Move all terms with $x$ to one side: $x^2 + 8x + 5 - 2x = 0$.

Step:3.1.1

Subtract $2x$ from both sides: $x^2 + 6x + 5 = 0$.

Step:3.2

Factor the quadratic expression $x^2 + 6x + 5$.

Step:3.2.1

Find two numbers that multiply to $5$ and add up to $6$: $1$ and $5$.

Step:3.2.2

Write the factors: $(x + 1)(x + 5) = 0$.

Step:3.3

Set each factor equal to zero: $x + 1 = 0$ and $x + 5 = 0$.

Step:3.4

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

Step:3.4.1

Set $x + 1$ to zero: $x + 1 = 0$.

Step:3.4.2

Subtract $1$ from both sides to find $x$: $x = -1$.

Step:3.5

Solve for $x$ when $x + 5 = 0$.

Step:3.5.1

Set $x + 5$ to zero: $x + 5 = 0$.

Step:3.5.2

Subtract $5$ from both sides to find $x$: $x = -5$.

Step:3.6

Combine the solutions to get the final answer: $x = -1$ or $x = -5$.

Step:4

Check which solutions satisfy the original equation: $x = -1$.

Knowledge Notes:

1. Least Common Denominator (LCD): The LCD is the smallest multiple that is common to the denominators of a set of fractions. It is used to combine fractions into a single fraction or to eliminate fractions from an equation.

2. Least Common Multiple (LCM): The LCM of a set of numbers is the smallest number that is a multiple of each of the numbers in the set. It is found by prime factorization and taking the highest power of each prime factor that appears in any of the numbers.

3. Prime Factorization: Breaking down a number into its prime factors, which are prime numbers that multiply together to give the original number.

4. Distributive Property: A property of multiplication over addition or subtraction, stating that $a(b + c) = ab + ac$.

5. Factoring Quadratics: The process of breaking down a quadratic expression into a product of two binomials. The AC method involves finding two numbers that multiply to give the product of the coefficient of $x^2$ and the constant term (AC) and add up to the coefficient of the middle term (B).

6. Zero Product Property: If the product of two factors is zero, then at least one of the factors must be zero. This property is used to solve quadratic equations by setting each factor equal to zero and solving for the variable.

7. Checking Solutions: After solving an equation, it is important to substitute the solutions back into the original equation to verify that they satisfy the equation. This step ensures that no extraneous solutions are included in the final answer.