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Solve each equation. See Examples I through 6. \(\frac{y}{2 y+2}+\frac{2 y-16}{4 y+4}=\frac{2 y-3}{y+1}\)

Short Answer

Expert verified
The equation has no valid solutions (extraneous).

Step by step solution

01

Identify the terms and the common denominator

From the equation \( \frac{y}{2y+2} + \frac{2y-16}{4y+4} = \frac{2y-3}{y+1} \), note the denominators are \(2y+2\), \(4y+4\), and \(y+1\). Simplify each denominator where possible. The common denominator should be the least common multiple of these simplified denominators.
02

Simplify the denominators

Factor each denominator: \(2y+2 = 2(y+1)\), \(4y+4 = 4(y+1)\), and \(y+1\) remains \(y+1\). These factors help us to find the least common denominator (LCD), which is \(4(y+1)\).
03

Rewrite each fraction with the common denominator

Convert \(\frac{y}{2(y+1)}\) to have the LCD by multiplying both numerator and denominator by 2, resulting in \(\frac{2y}{4(y+1)}\). The second fraction \(\frac{2y-16}{4(y+1)}\) already has the common denominator. The right-hand side \(\frac{2y-3}{y+1}\) is converted by multiplying both numerator and denominator by 4: \(\frac{4(2y-3)}{4(y+1)}\).
04

Combine all fractions into a single equation

Combine the left side fractions: \(\frac{2y + (2y-16)}{4(y+1)}\) to get \(\frac{4y-16}{4(y+1)}\). The right side remains \(\frac{8y-12}{4(y+1)}\).
05

Solve the equation by equating numerators

Since the denominators are equal, set the numerators equal to each other: \(4y - 16 = 8y - 12\). Solve for \(y\) by subtracting \(4y\) from both sides: \(-16 = 4y - 12\).
06

Isolate the variable

Add 12 to both sides to simplify: \(-4 = 4y\). Finally, divide both sides by 4: \(y = -1\).
07

Check for extraneous solutions

Substitute \(y = -1\) back into the original equation to verify both sides are undefined due to zero denominators, confirming \(y = -1\) is an extraneous solution. Consequently, there are no valid solutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation Solving
When solving rational equations, the goal is to find the variable that satisfies the equation. These equations often involve algebraic fractions, which can make them appear complicated.
To simplify the process, follow these general steps:
  • Identify all denominators in the equation.
  • Determine the least common denominator (LCD) to eliminate fractions.
  • Reconstruct the equation using the LCD as a common base.
  • Combine fractions effectively into a single equation.
  • Solve for the variable by focusing on the numerators.
  • Check your solution, ensuring it doesn't create undefined terms, such as division by zero.
This structured approach helps manage the complexity of rational equations and leads to the correct solution if executed correctly.
Least Common Denominator
The least common denominator (LCD) is a fundamental concept when working with rational equations. It allows you to combine fractions effectively by aligning their bases. For our equation, we first factor the given denominators:
  • \(2y+2 = 2(y+1)\)
  • \(4y+4 = 4(y+1)\)
  • \(y+1\) remains unchanged.
After factoring, you can determine the LCD by identifying the largest expression that can be formed from these factors, which is \(4(y+1)\).
Rewriting each fraction with this common denominator simplifies the solving process and allows you to easily combine and manipulate the equation's terms. Using the LCD in rational equations eliminates the complexity of different denominators, leading you faster to the solution.
Extraneous Solutions
Extraneous solutions occur when a solution makes any denominator in the original equation equal to zero. This is critical because division by zero is undefined. In our exercise, after finding \(y = -1\), we find it's extraneous by checking its validity against the original equation.
Substituting \(y = -1\) into the denominators \(2(y+1)\), \(4(y+1)\), and \(y+1\) all yield 0, making the fractions undefined.
To avoid extraneous solutions, always return to the original equation to verify that your solution does not result in undefined terms. This step ensures that the solution is valid within the context of the original problem.
Algebraic Fractions
Algebraic fractions behave similarly to regular fractions, but they contain polynomials in either their numerators or denominators. In rational equations, these fractions can be manipulated using the same rules as numerical fractions, plus additional considerations for variables.
Following our exercise, we dealt with algebraic fractions such as \(\frac{y}{2(y+1)}\) and \(\frac{2y-16}{4(y+1)}\). Using common methods like factorizing and finding the least common denominator helps simplify these fractions.
Essentially, understanding how to handle algebraic fractions enables you to convert a complex equation into something more manageable. This foundational knowledge is vital for solving not only rational equations but a broad range of algebraic problems.

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Most popular questions from this chapter

Solve the following linear and quadratic equations. See Sections 2.3 and 6.6 $$ 4(x+6)+3=-3 $$

Simplify. $$ \frac{(x+2)^{-1}+(x-2)^{-1}}{\left(x^{2}-4\right)^{-1}} $$

Simplify each complex fraction. See Examples 1 and 2. $$ \frac{\frac{x+3}{12}}{\frac{4 x-5}{15}} $$

One of the great algebraists of ancient times a man named Diophantus. Litle is known of his life other than the lived and worked in Alexandria. Some historians believe he lived during the first century of the Christian era, about the time of Nero. The only che to his personal life is the following epigram found in a collection called the Palatine A nthology. God granted him youth for a sixth of his life and added a twelfth part to this. He clothed his cheeks in down. He lit him the light of wedlock after a seventh part and five years after his marriage. He granted him a son. Alas, lateborn wretched child. After attaining the measure of half his father's life, cruel fate overtook him, thus leaving Diophantus during the last four years of his life only such consolation as the science of numbers. How old was Diophantus at his death?" We are looking for Diophantus' age when he died, so let x represent that age. If we sum the parts of his life, we should get the total age. Parts of his life \(\left\\{\begin{array}{l}{\frac{1}{6} x+\frac{1}{12} x \text { is the time of his youth. }} \\ {\frac{1}{7} x \text { is the time between his youth and when }} \\ {\text { he married. }} \\ {5 \text { years is the time between his marriage }} \\ {\text { and the birth of his son. }} \\\ {\frac{1}{2} x \text { is the time Diophantus had with his son. }} \\ {4 \text { years is the time between his son's death }} \\ {\text { and his own. }}\end{array}\right.\) The sum of these parts should equal Diophantus' age when he died. $$ \frac{1}{6} \cdot x+\frac{1}{12} \cdot x+\frac{1}{7} \cdot x+5+\frac{1}{2} \cdot x+4=x $$ Solve the epigram.

Perform each indicated operation. In your own words, explain how to subtract two rational expressions with different denominators.

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