Problem 20 - Elementary Algebra

Question

$$ \text { For Problems 1-32, solve each equation. (Objective 1) } $$ $$ 2+\frac{4}{y-1}=\frac{4}{y^{2}-y} $$

Step-by-Step Solution

Answer

The solution is \( y = -2 \).

Step 1: Understand the Equation

The given equation is \( 2 + \frac{4}{y-1} = \frac{4}{y^2-y} \). Notice that the denominator on the right-hand side, \( y^2 - y \), can be factored.

Step 2: Factor the Denominator

Rewrite \( y^2 - y \) as \( y(y-1) \). The equation becomes \( 2 + \frac{4}{y-1} = \frac{4}{y(y-1)} \).

Step 3: Find a Common Denominator

The least common denominator (LCD) of \( y-1 \) and \( y(y-1) \) is \( y(y-1) \). Multiply every term by \( y(y-1) \) to eliminate denominators.

Step 4: Clear the Fractions

Multiply the entire equation by \( y(y-1) \): \( 2y(y-1) + 4y = 4 \). This simplifies to \( 2y^2 - 2y + 4y = 4 \).

Step 5: Simplify the Equation

Combine like terms: \( 2y^2 + 2y = 4 \).

Step 6: Move All Terms to One Side

Subtract 4 from both sides to set the equation to zero: \( 2y^2 + 2y - 4 = 0 \).

Step 7: Solve the Quadratic Equation

Factor the quadratic equation: \( 2(y^2 + y - 2) = 0 \). Divide by 2 to simplify to \( y^2 + y - 2 = 0 \). Factor it: \( (y + 2)(y - 1) = 0 \).

Step 8: Find the Solutions

Set each factor to zero: 1) \( y + 2 = 0 \Rightarrow y = -2 \) 2) \( y - 1 = 0 \Rightarrow y = 1 \). Check for extraneous solutions from the original equation's denominators.

Step 9: Verify the Solutions

\( y = 1 \) makes the denominators zero, hence it's not valid. Thus, the valid solution is \( y = -2 \).

Key Concepts

Factoring Expressions

Factoring is a crucial skill in algebra as it simplifies expressions and makes solving equations more manageable. When you factor an expression, you find two or more expressions that multiply together to form the original expression. In this problem, we encounter the expression \(y^2 - y\).To factor such an expression, you look for common factors and use algebraic identities. Here, the expression can be rewritten as \(y \cdot (y - 1)\). This indicates that each term shares a common factor of \(y\), leaving us with the factored form. Remember:

• Look for common terms or factors first.
• Use identities like difference of squares or trinomials if applicable.

By factoring, we simplify the complexity of expressions, which is critical in solving more challenging equations later on.

Least Common Denominator

When dealing with equations that have fractions, aligning them to a common denominator simplifies the process of eliminating the fractions. This is especially useful for rational equations, such as the one we have here. The least common denominator (LCD) is the smallest expression that all denominators can divide into without leaving a remainder. For the given equation, the denominators are \(y-1\) and \(y(y-1)\). The LCD is \(y(y-1)\).

• Identify all denominators in the equation.
• Determine the LCD by finding the least common multiple of these denominators.

After you find the LCD, use it to eliminate the fractions by multiplying through the entire equation, thus simplifying your expressions significantly.

Quadratic Equations

A quadratic equation is a polynomial equation of the second degree, typically expressed in the form \(ax^2 + bx + c = 0\). In our exercise, we arrived at the quadratic equation \(y^2 + y - 2 = 0\). Solving such equations usually involves factoring, completing the square, or using the quadratic formula.In our step-by-step solution, we solved it by factoring. The equation factors to \((y + 2)(y - 1) = 0\). From here:

• Set each factor to zero: \(y + 2 = 0\) and \(y - 1 = 0\).
• Solve each equation to find potential solutions for \(y\).

Always check your solutions in the original equation, as some may not be valid for the given context.

Extraneous Solutions

Extraneous solutions are results from algebraic manipulation that may satisfy transformed equations but are not solutions to the original equation. This can happen when dealing with rational expressions.In this problem, upon solving \((y + 2)(y - 1) = 0\), we found potential solutions \(y = -2\) and \(y = 1\). However, we need to verify if these solutions fit the original equation. The solution \(y = 1\) makes the original denominators zero, rendering it invalid or extraneous.

• Always substitute solutions back into the original equation.
• Check for any restrictions caused by denominator terms becoming zero.

The only valid solution in the context of this problem is \(y = -2\). It's essential to complete this verification step to ensure accuracy in your results.