Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$\frac{7}{y^{2}-4}-\frac{4}{y+2}=\frac{5}{y-2}$$

Step-by-Step Solution

Answer

The solution is \( y = \frac{5}{9} \).

Step 1: Identify the Restrictions

Before solving the equation, note the values that make the denominators zero to identify the restrictions. The denominators are \( y^2 - 4 \), \( y + 2 \), and \( y - 2 \). 1. \( y^2 - 4 = (y+2)(y-2) \), so \( y eq 2 \) and \( y eq -2 \).2. \( y + 2 eq 0 \Rightarrow y eq -2 \).3. \( y - 2 eq 0 \Rightarrow y eq 2 \).Therefore, the restrictions are \( y eq 2 \) and \( y eq -2 \).

Step 2: Rewrite the Equation with a Common Denominator

The least common denominator (LCD) for the equation is \((y+2)(y-2)\), which matches \(y^2-4\). Rewrite all fractions with this common denominator:\[\frac{7}{(y+2)(y-2)} - \frac{4(y-2)}{(y+2)(y-2)} = \frac{5(y+2)}{(y+2)(y-2)}\]

Step 3: Combine Fractions

Now that all terms have a common denominator, combine the numerators over this denominator:\[\frac{7 - 4(y-2) - 5(y+2)}{(y+2)(y-2)}\] Simplify the numerator:

Step 4: Simplify the Numerator

Distribute the constants and simplify:\[7 - 4(y-2) - 5(y+2) = 7 - 4y + 8 - 5y - 10 \]Combine like terms:\[15 - 4y - 5y - 10 = -9y + 15 - 10 = -9y + 5\] So, the numerator simplifies to \(-9y + 5\).

Step 5: Solve the Equation

Now the equation is:\[\frac{-9y + 5}{(y+2)(y-2)} = 0\] A fraction is zero when its numerator is zero, so solve the equation: \(-9y + 5 = 0\).Solve for \(y\):\[-9y + 5 = 0 \-9y = -5 \y = \frac{5}{9}\]

Step 6: Check the Solution Against Restrictions

The solution \( y = \frac{5}{9} \) does not violate the restrictions \( y eq 2 \) and \( y eq -2 \). Thus, it is valid.

Key Concepts

Common Denominator

When you're dealing with rational equations, finding a common denominator is crucial. Much like adding or subtracting fractions, each fraction in the equation needs to share the same denominator. This simplifies the equation and allows you to combine the numerators effectively.

In our exercise, the original denominators were: \( y^2 - 4 \, \, y + 2 \, \, \text{and} \, y - 2 \). To find the least common denominator (LCD), it's beneficial to factorize them if possible. Here, \( y^2 - 4 \) factorizes to \( (y+2)(y-2) \). Thus, the LCD is \( (y+2)(y-2) \, \,\text{matching}\, y^2-4 \). This approach ensures that all terms will line up efficiently when combined, paving the way for simplifying the equation further.

By rewriting the fractions with this common denominator, they become easier to handle and combine.

Numerator Simplification

Once the common denominator is established, the next step is to simplify the numerators. This involves combining multiple expressions into one, essentially by putting everything over the same denominator.

For the exercise, the equation after finding a common denominator was: \[\frac{7 - 4(y-2) - 5(y+2)}{(y+2)(y-2)}\] The goal here is to simplify the top part of the fraction. This requires:

• Distributing constants correctly
• Combining like terms

Distributing gives us: \[7 - 4(y-2) - 5(y+2) = 7 - 4y + 8 - 5y - 10\] By grouping similar terms, we get a cleaner form: \[-9y + 5\]This simplification is essential to move on to solving the rational equation effectively.

Equation Restrictions

Before diving into solving the equation, it's important to define its restrictions. Equation restrictions are values that make any denominator zero, which would make the fraction undefined. Identifying these ensures you won’t end up with an invalid solution.

For the exercise, the restrictions we recognized included:

• From \( y^2 - 4 = (y+2)(y-2)\), avoid \( y=2 \) and \( y=-2\)
• From the denominator \( y + 2 \, \,y eq -2\)
• From the denominator \( y - 2 \, \,y eq 2\)

Thus, the valid \( y\) values must exclude \( 2 \) and \( -2 \. \) This crucial step safeguards against dividing by zero in mathematical solutions.

Solving Rational Equations

After setting up the equation with a common denominator and simplifying the numerators, solving the equation becomes more straightforward. The heart of solving rational equations often lies in setting the numerator equal to zero. This is because a rational expression equals zero when only the numerator is zero.

In the exercise, the simplified equation was: \[\frac{-9y + 5}{(y+2)(y-2)} = 0\] To solve, follow these steps:

• Set the numerator, \(-9y + 5\), equal to zero.
• Solve for \( y \): \(-9y + 5 = 0\)
• Rearrange to \(-9y = -5\)
• Conclude with \( y = \frac{5}{9}\)

Finally, verify that \( y = \frac{5}{9} \) doesn’t violate the equation's restrictions \( y eq 2 \) and \( y eq -2\). \This solution is valid!\ \Understanding rational equations means not only finding solutions but ensuring they are mathematically valid, making it an invaluable skill in algebra.