Textbook Question
Add or subtract, as indicated. (7x + 8)/(3x + 2) - (x + 4)/(3x + 2)
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0. Review of Algebra
Algebraic Expressions
9:30 minutesProblem 88a
Textbook Question
Simplify each complex fraction. [ 2/[(x+h)2 + 16] - 2/(x2+16)] / h
Verified step by step guidance1
Identify the complex fraction: the numerator is the difference of two fractions, and the denominator is a single variable \(h\). The expression is: \(\frac{\frac{2}{(x+h)^2 + 16} - \frac{2}{x^2 + 16}}{h}\).
Focus on simplifying the numerator first, which is \(\frac{2}{(x+h)^2 + 16} - \frac{2}{x^2 + 16}\). To combine these two fractions, find a common denominator, which is \(\left((x+h)^2 + 16\right) \left(x^2 + 16\right)\).
Rewrite each fraction with the common denominator: \(\frac{2(x^2 + 16)}{\left((x+h)^2 + 16\right) \left(x^2 + 16\right)} - \frac{2((x+h)^2 + 16)}{\left((x+h)^2 + 16\right) \left(x^2 + 16\right)}\).
Combine the numerators over the common denominator: \(\frac{2(x^2 + 16) - 2((x+h)^2 + 16)}{\left((x+h)^2 + 16\right) \left(x^2 + 16\right)}\).
Now, rewrite the entire original expression as a single fraction by dividing this result by \(h\), which is equivalent to multiplying by \(\frac{1}{h}\). So the expression becomes: \(\frac{2(x^2 + 16) - 2((x+h)^2 + 16)}{h \cdot \left((x+h)^2 + 16\right) \left(x^2 + 16\right)}\).
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Video duration:
9mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Complex Fractions
A complex fraction is a fraction where the numerator, denominator, or both contain fractions themselves. Simplifying involves rewriting the expression as a single fraction by finding common denominators or multiplying by the least common denominator to eliminate the smaller fractions.
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Common Denominator
To combine or subtract fractions, you must express them with a common denominator. This involves finding the least common multiple of the denominators and rewriting each fraction accordingly, allowing the numerators to be combined directly.
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Difference Quotient and Simplification
The given expression resembles a difference quotient, often used in calculus. Simplifying such expressions requires careful algebraic manipulation, including expanding binomials, combining like terms, and factoring to reduce the expression, especially when dividing by a variable like h.
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