Problem 63
Question
Simplify each expression. $$ \frac{3 y+1}{2 y-10}+\frac{1}{y^{2}-2 y-15} $$
Step-by-Step Solution
Verified Answer
The simplified expression is \(\frac{3y^2 + 10y + 5}{2(y-5)(y+3)}\).
1Step 1: Factor the Denominators
Start by factoring the denominators of the fractions. For the first fraction, the denominator is \(2y-10\) which factors to \(2(y-5)\). For the second fraction, the denominator is \(y^2 - 2y - 15\), which factors to \((y-5)(y+3)\).
2Step 2: Find a Common Denominator
Next, find a common denominator for the two fractions. The denominators are \(2(y-5)\) and \((y-5)(y+3)\). The least common denominator (LCD) is \(2(y-5)(y+3)\).
3Step 3: Adjust Each Fraction to the LCD
Adjust the first fraction \(\frac{3y+1}{2(y-5)}\) to have the common denominator \(2(y-5)(y+3)\). Multiply the numerator and the denominator by \((y+3)\), resulting in \(\frac{(3y+1)(y+3)}{2(y-5)(y+3)}\). For the second fraction \(\frac{1}{(y-5)(y+3)}\), multiply the numerator and denominator by 2 to get \(\frac{2}{2(y-5)(y+3)}\).
4Step 4: Combine the Fractions
Combine the fractions since they now have a common denominator. Add the numerators: \((3y+1)(y+3) + 2\), resulting in \[\frac{(3y+1)(y+3) + 2}{2(y-5)(y+3)}\].
5Step 5: Simplify the Numerator
Expand and simplify the numerator. \((3y+1)(y+3)\) expands to \(3y^2 + 9y + y + 3\), which simplifies to \(3y^2 + 10y + 3\). Add 2 to get \(3y^2 + 10y + 5\).
6Step 6: Final Simplified Expression
The final simplified expression is \[\frac{3y^2 + 10y + 5}{2(y-5)(y+3)}\].
Key Concepts
Factoring PolynomialsRational ExpressionsLeast Common DenominatorCombine Fractions
Factoring Polynomials
Factoring polynomials is a crucial step when simplifying algebraic expressions, particularly rational expressions. It involves expressing a polynomial as a product of its factors to simplify complex expressions.
For example, consider the polynomial in the second fraction’s denominator:
For example, consider the polynomial in the second fraction’s denominator:
- Original polynomial: \(y^2 - 2y - 15\)
- This can be factored into: \((y-5)(y+3)\)
Rational Expressions
Rational expressions are fractions where the numerator and/or the denominator are polynomials. To work with them effectively, we must understand how to manipulate these expressions.
- Look for opportunities to factor both the numerator and the denominator.
- Simplify them much like you would numerical fractions, by reducing common factors.
Least Common Denominator
When adding or subtracting rational expressions, finding the least common denominator (LCD) is essential. This allows you to combine the fractions efficiently.
In our example, rational expressions have the following denominators:- \(2(y-5)\) and- \((y-5)(y+3)\)The LCD is determined by taking each distinct factor with the highest power found in any denominator:
In our example, rational expressions have the following denominators:- \(2(y-5)\) and- \((y-5)(y+3)\)The LCD is determined by taking each distinct factor with the highest power found in any denominator:
- Combine unique factors: \(2\), \(y-5\), and \(y+3\).
- Resulting LCD: \(2(y-5)(y+3)\).
Combine Fractions
Once we’ve found the least common denominator, combining fractions becomes straightforward. All we have to do is adjust each fraction so it has this common denominator.In the given problem, after we've modified:- First fraction: \( \frac{(3y+1)(y+3)}{2(y-5)(y+3)}\)- Second fraction: \( \frac{2}{2(y-5)(y+3)}\)The fractions now share the same denominator \(2(y-5)(y+3)\). We can add their numerators directly:
- Add numerators: \((3y+1)(y+3) + 2\)
- Keep the common denominator: \(2(y-5)(y+3)\)
Other exercises in this chapter
Problem 62
Simplify each expression. $$ \frac{w+12}{4 w-16}-\frac{w+4}{2 w-8} $$
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PREREQUISITE SKILL Solve each equation. $$ (x-1)(x+1)=0 $$
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PREREQUISITE SKILL Solve each equation. $$ x(x+2)=0 $$
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Find the value of each function. $$ \cos 150^{\circ} $$
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