Problem 77 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{5}{y-1}+\frac{3}{y-3}=\frac{8}{y-2}\)

Step-by-Step Solution

Answer

The solution is \(y = 6\), and it is not extraneous.

Step 1: Identify the least common denominator (LCD)

The denominators are \(y-1\), \(y-3\), and \(y-2\). To eliminate the fractions, find the least common denominator, which is \((y-1)(y-3)(y-2)\).

Step 2: Clear the fractions

Multiply every term in the equation by the least common denominator \((y-1)(y-3)(y-2)\) to clear the fractions:\[ (y-1)(y-3)(y-2) \left( \frac{5}{y-1} + \frac{3}{y-3} - \frac{8}{y-2} \right) = 0 \]Simplifying, we get:\[ 5(y-3)(y-2) + 3(y-1)(y-2) = 8(y-1)(y-3) \]

Step 3: Expand the expressions

Expand each term in the equation:\[ 5(y^2 - 5y + 6) + 3(y^2 - 3y + 2) = 8(y^2 - 4y + 3) \]This simplifies to:\[ 5y^2 - 25y + 30 + 3y^2 - 9y + 6 = 8y^2 - 32y + 24 \]

Step 4: Combine like terms

Combine the like terms on each side:\[ 8y^2 - 34y + 36 = 8y^2 - 32y + 24 \]

Step 5: Simplify the equation

Subtract \(8y^2\) from both sides to reduce the equation:\[ -34y + 36 = -32y + 24 \]

Step 6: Solve for \(y\)

Rearrange to solve for \(y\):\[ -34y + 32y = 24 - 36 \]This simplifies to:\[ -2y = -12 \]Divide each side by -2:\[ y = 6 \]

Step 7: Check for extraneous solutions

Determine if \(y = 6\) makes any original denominators zero:- **Denominator 1**: \(y-1 = 6-1 = 5 eq 0\)- **Denominator 2**: \(y-3 = 6-3 = 3 eq 0\)- **Denominator 3**: \(y-2 = 6-2 = 4 eq 0\)Since none of the denominators are zero, \(y = 6\) is not an extraneous solution.

Key Concepts

Least Common Denominator

When solving rational equations, like \( \frac{5}{y-1}+\frac{3}{y-3}=\frac{8}{y-2} \), it's crucial to work with a least common denominator (LCD). The LCD is the smallest expression that all the denominators can divide into without leaving a remainder. This makes calculations simpler and helps clear the fractions in the equation.

To find the LCD, look at each denominator in the equation. In this specific problem, we have denominators \( y-1 \), \( y-3 \), and \( y-2 \). To determine the LCD, take each distinct factor found in the denominators and multiply them:

• \( y-1 \)
• \( y-3 \)
• \( y-2 \)

Thus, the LCD here is \((y-1)(y-3)(y-2)\). By working with the LCD, you can efficiently eliminate all fractions, making it easier to solve the equation.

Extraneous Solutions

Extraneous solutions are values that appear as solutions when solving an equation but do not satisfy the original equation. These types of solutions often arise when we perform certain operations, like multiplying both sides by an expression that could potentially be zero, which introduces new solutions that weren't there initially.

In the given equation \(\frac{5}{y-1}+\frac{3}{y-3}=\frac{8}{y-2}\), we solved for \(y = 6\). To check if it is an extraneous solution, substitute \(y = 6\) back into each original denominator:

• \( y-1 = 6-1 = 5 \), which is not zero.
• \( y-3 = 6-3 = 3 \), which is not zero.
• \( y-2 = 6-2 = 4 \), which is not zero.

Since none of these expressions are zero, this means \( y = 6 \) is not an extraneous solution. Hence, it is a valid solution to the original equation. Always remember to check for these, as skipping this step might lead you to incorrect conclusions.

Clearing Fractions

Clearing fractions is a strategic approach when solving rational equations as it simplifies the calculation significantly. The idea is to transform the equation into one without fractions, making it easier to handle. Here is how it's done:

Start with the equation \( \frac{5}{y-1} + \frac{3}{y-3} = \frac{8}{y-2} \). Previously, we determined the least common denominator (LCD) is \((y-1)(y-3)(y-2)\). We multiply every term in the equation by this LCD.

By doing this, each denominator cancels out:

• The term \( \frac{5}{y-1} \) becomes \( 5(y-3)(y-2) \)
• The term \( \frac{3}{y-3} \) becomes \( 3(y-1)(y-2) \)
• The term \( \frac{8}{y-2} \) becomes \( 8(y-1)(y-3) \)

As a result, you have a neat and manageable equation without fractions to solve: 5(y-3)(y-2) + 3(y-1)(y-2) = 8(y-1)(y-3). This makes the solving process straightforward and less prone to errors inherent in handling multiple fractions.