Problem 54
Question
For exercises \(25-68\), evaluate or simplify. $$ \frac{\frac{x}{x-2}}{\frac{x+4}{x+3}} $$
Step-by-Step Solution
Verified Answer
\( \frac{x(x+3)}{(x-2)(x+4)} \)
1Step 1 - Rewrite the fraction
Rewrite the given complex fraction as a division of two fractions: \ \( \frac{x}{x-2} \ \frac{x+4}{x+3} \)
2Step 2 - Multiply by the reciprocal
Convert the division into a multiplication by taking the reciprocal of the divisor: \ \( \frac{x}{x-2} \times \frac{x+3}{x+4} \)
3Step 3 - Simplify the expression
Now, multiply the numerators together and the denominators together: \ \( \frac{x \times (x+3)}{(x-2) \times (x+4)} \)
4Step 4 - Final simplified form
The simplified expression is: \ \( \frac{x(x+3)}{(x-2)(x+4)} \)
Key Concepts
multiplying fractionsreciprocal of a fractionalgebraic expressions
multiplying fractions
Multiplying fractions is a straightforward process that makes use of the numerators (the top parts) and denominators (the bottom parts) of the given fractions.
To multiply fractions, simply multiply the numerators together and the denominators together.
For example, given two fractions, \(\frac{a}{b} \) and \(\frac{c}{d} \), their product is \(\frac{a \times c}{b \times d} \).
This concept is crucial when simplifying complex fractions.
In our exercise, when converting the division problem into multiplication by the reciprocal, we multiplied: \(\frac{x}{x-2} \times \frac{x+3}{x+4} \).
Here’s how to do it step-by-step:
To multiply fractions, simply multiply the numerators together and the denominators together.
For example, given two fractions, \(\frac{a}{b} \) and \(\frac{c}{d} \), their product is \(\frac{a \times c}{b \times d} \).
This concept is crucial when simplifying complex fractions.
In our exercise, when converting the division problem into multiplication by the reciprocal, we multiplied: \(\frac{x}{x-2} \times \frac{x+3}{x+4} \).
Here’s how to do it step-by-step:
- Multiply the numerators: \(x \times (x + 3) \)
- Multiply the denominators: \((x - 2) \times (x + 4) \)
reciprocal of a fraction
The reciprocal of a fraction is simply flipping the fraction.
This means that the numerator becomes the denominator and the denominator becomes the numerator.
For example, the reciprocal of \(\frac{a}{b} \) is \(\frac{b}{a} \).
This is important because when dividing by a fraction, you actually multiply by its reciprocal.
In our exercise, we were given \(\frac{\frac{x}{x-2}}{\frac{x+4}{x+3}} \).
To simplify this, we took the reciprocal of \(\frac{x+4}{x+3} \), which is \(\frac{x+3}{x+4} \), and then multiplied:
This means that the numerator becomes the denominator and the denominator becomes the numerator.
For example, the reciprocal of \(\frac{a}{b} \) is \(\frac{b}{a} \).
This is important because when dividing by a fraction, you actually multiply by its reciprocal.
In our exercise, we were given \(\frac{\frac{x}{x-2}}{\frac{x+4}{x+3}} \).
To simplify this, we took the reciprocal of \(\frac{x+4}{x+3} \), which is \(\frac{x+3}{x+4} \), and then multiplied:
- Reciprocal of \(\frac{x+4}{x+3} \) is \(\frac{x+3}{x+4} \)
- Multiply \(\frac{x}{x-2} \) by the reciprocal: \(\frac{x}{x-2} \times \frac{x+3}{x+4} \)
algebraic expressions
Algebraic expressions are mathematical phrases that can contain numbers, variables, and operators.
In our exercise, we worked with expressions like \(\frac{x}{x-2} \) and \(\frac{x+4}{x+3} \).
Understanding and simplifying these expressions is vital for solving complex fractions.
Here are a few things to remember:
Finally, combining these simplified forms results in \(\frac{x(x+3)}{(x-2)(x+4)} \), which is much easier to work with.
In our exercise, we worked with expressions like \(\frac{x}{x-2} \) and \(\frac{x+4}{x+3} \).
Understanding and simplifying these expressions is vital for solving complex fractions.
Here are a few things to remember:
- Variables represent unknown values
- Expressions can be fractions, sums, differences, products, or quotients of numbers and variables
- You can simplify algebraic expressions by combining like terms or using algebraic rules
Finally, combining these simplified forms results in \(\frac{x(x+3)}{(x-2)(x+4)} \), which is much easier to work with.
Other exercises in this chapter
Problem 53
For exercises 1-66, simplify. $$ \frac{9 n^{2}-48 n+64}{9 n^{2}-64} $$
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For exercises 53-56, the formula \(F=\frac{100 S_{u} C_{p}}{S_{p} C_{u}}\) describes the fractional excretion of sodium, \(F\). Is the relationship of the given
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For exercises \(35-86\), simplify. $$ \frac{4}{z^{2}+3 z}+\frac{5}{z^{2}+9 z} $$
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A student is simplifying \(\frac{x+3}{x+4}\). He thinks that the \(x\) in the numerator and the \(x\) in the denominator are common factors and that the express
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