Problem 25 - Elementary Algebra

Question

$$ \text { For Problems 1-32, solve each equation. (Objective 1) } $$ $$ x-\frac{5 x}{x-2}=\frac{-10}{x-2} $$

Step-by-Step Solution

Answer

The solution is \( x = 5 \).

Step 1: Find a Common Denominator

The given equation is \( x - \frac{5x}{x-2} = \frac{-10}{x-2} \). The terms \( \frac{5x}{x-2} \) and \( \frac{-10}{x-2} \) have the common denominator \( x-2 \). We'll multiply all terms by \( x-2 \) to eliminate the denominators. This gives: \[ x(x-2) - 5x = -10 \].

Step 2: Distribute and Simplify

Distribute \( x \) to \( x-2 \) in the expression \( x(x-2) - 5x = -10 \), resulting in: \[ x^2 - 2x - 5x = -10 \]. Combine like terms: \[ x^2 - 7x = -10 \].

Step 3: Rearrange into a Standard Quadratic Form

Add 10 to both sides to rearrange the equation into the standard quadratic form: \( x^2 - 7x + 10 = 0 \).

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \( x^2 - 7x + 10 = 0 \). The factors of 10 that add to -7 are -2 and -5. So, we rewrite the quadratic as: \((x - 2)(x - 5) = 0 \).

Step 5: Solve for x

Set each factor equal to zero: \( x - 2 = 0 \) and \( x - 5 = 0 \). Solving these gives \( x = 2 \) and \( x = 5 \).

Step 6: Check for Extraneous Solutions

Since the equation originally had the term \( \frac{5x}{x-2} \), \( x = 2 \) would make the denominator zero, which is undefined. Hence, \( x = 2 \) is an extraneous solution. The only valid solution is \( x = 5 \).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is two. The general form of a quadratic equation is given by \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In our exercise, we encounter the quadratic equation \( x^2 - 7x + 10 = 0 \) after simplifying the algebraic expression.Understanding quadratic equations is fundamental for solving them effectively. These equations can model various real-world situations like the trajectory of a projectile or the area of a rectangle after adjusting dimensions.To solve a quadratic equation, we have several methods at our disposal:- **Factoring:** This involves writing the quadratic as the product of two binomial expressions.- **Completing the square:** A technique that rewrites the quadratic to reveal the vertex form.- **Quadratic formula:** Given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), it provides a solution directly.The selection of the method depends on the equation's specific form and the numbers involved. Mastery of these methods will enable you to tackle a wide range of quadratic equations confidently.

Factoring Quadratic Expressions

Factoring is the process of breaking down an expression into simpler components, which when multiplied together give the original expression. In the context of quadratic equations like \( x^2 - 7x + 10 = 0 \), factoring involves expressing it as \( (x - 2)(x - 5) = 0 \).This technique relies on finding two numbers that multiply to the constant term (here, 10) and add up to the middle coefficient (here, -7). In our example:

• The two numbers that multiply to 10 are -2 and -5.
• These numbers add up to -7, helping us break down the quadratic into two binomials.

The factored form is particularly useful because it allows us to apply the zero-product property. This property states that if the product of two factors equals zero, at least one of the factors must be zero. By setting each factor equal to zero, we solve the quadratic equation.Factoring is often the quickest and most straightforward way to solve quadratics when applicable. However, always ensure the factorization is correct by expanding back to check.

Extraneous Solutions

Extraneous solutions are solutions that arise from the process of solving an equation but do not satisfy the original equation. These often occur when manipulating equations algebraically, particularly when dealing with rational expressions.During our exercise, we derived the potential solutions \( x = 2 \) and \( x = 5 \) after factoring. However, the original equation contained the term \( \frac{5x}{x-2} \), which means \( x = 2 \) would cause a division by zero—an undefined operation.To avoid accepting extraneous solutions, always:

• Substitute potential solutions back into the original equation.
• Verify that they don't result in undefined expressions or contradictions.

In this case, \( x = 2 \) was identified as extraneous, leaving \( x = 5 \) as the valid solution. Recognizing and discarding extraneous solutions ensures that the answers you provide are both correct and meaningful in the context of the problem.