Problem 2
Question
Write each rational expression in simplest form and list the values of the variables for which the fraction is undefined. If \(\frac{a}{b}=\frac{c}{d},\) then is \(\frac{a+b}{b}=\frac{c+d}{d} ?\) Justify your answer.
Step-by-Step Solution
Verified Answer
Yes, \(\frac{a+b}{b} = \frac{c+d}{d}\) is true if \(\frac{a}{b} = \frac{c}{d}\), except when \(b = 0\) or \(d = 0\), as these cases are undefined.
1Step 1: Recall the Property of Equivalent Fractions
Given the two equivalent fractions \(\frac{a}{b} = \frac{c}{d}\), this implies that when we cross-multiply, we have \(a \cdot d = b \cdot c\). This relationship is crucial in understanding the properties of these fractions.
2Step 2: Analyze the Given Assertion
We are asked if \(\frac{a+b}{b} = \frac{c+d}{d}\) is true given \(\frac{a}{b} = \frac{c}{d}\). To assess this, understand that \(\frac{a+b}{b}\) can be rewritten as \(\frac{a}{b} + 1\) and \(\frac{c+d}{d}\) can be rewritten as \(\frac{c}{d} + 1\).
3Step 3: Simplify and Compare Both Sides
Since \(\frac{a}{b} = \frac{c}{d}\), it follows that \(\frac{a}{b} + 1 = \frac{c}{d} + 1\). This implies that \(\frac{a+b}{b} = \frac{c+d}{d}\) is indeed true. This is because adding 1 to both fractions that were originally equal results in equivalency between \(\frac{a+b}{b}\) and \(\frac{c+d}{d}\).
4Step 4: Identify Undefined Values
For the fractions \(\frac{a+b}{b}\) and \(\frac{c+d}{d}\), we need to ensure that the denominators are not zero. Therefore, the expression is undefined for \(b = 0\) and \(d = 0\). It is critical to note that any fraction is undefined if its denominator equals zero.
Key Concepts
Rational ExpressionsUndefined Values in FractionsCross MultiplicationSimplifying Fractions
Rational Expressions
Rational expressions are essentially fractions, where the numerator and the denominator are both polynomials. Understanding rational expressions is key to solving many algebraic problems. Consider a rational expression to be similar to any fraction that you are familiar with, like \( \frac{1}{2} \).
In mathematics, we often use rational expressions to represent divisions between two polynomials. These expressions are not limited to just numbers; they can include variables as well. For instance, \( \frac{x^2 + 2x + 1}{x+1} \) is a rational expression.
In mathematics, we often use rational expressions to represent divisions between two polynomials. These expressions are not limited to just numbers; they can include variables as well. For instance, \( \frac{x^2 + 2x + 1}{x+1} \) is a rational expression.
- Rational expressions can be simplified by factoring the numerator and denominator.
- It's crucial to understand that the value of the variable can affect the expression. Making the denominator zero, for example, can cause issues.
- Equivalent rational expressions have the same value, even if they look different.
Undefined Values in Fractions
Fractions become undefined when their denominator equals zero. This is a vital principle when dealing with rational expressions. No matter what number is in the numerator, if the denominator becomes zero, we are left with something undefined.
The same rule applies in algebra when working with variables.
Let's take \( \frac{2}{x-3} \): this fraction is undefined if \( x-3=0 \), meaning \( x=3 \). If \( x \) were to be 3, the expression would become \( \frac{2}{0} \), which is undefined.
The same rule applies in algebra when working with variables.
Let's take \( \frac{2}{x-3} \): this fraction is undefined if \( x-3=0 \), meaning \( x=3 \). If \( x \) were to be 3, the expression would become \( \frac{2}{0} \), which is undefined.
- Always check the denominator and find the values that make it zero.
- Avoid these values as they will make the rational expression undefined.
Cross Multiplication
Cross multiplication is a valuable tool when working with equations involving fractions. It provides an effective method to find equivalences between two fractions.
When fractions are said to be equal, like \( \frac{a}{b} = \frac{c}{d} \), you can cross-multiply to verify or clear the equation of fractions. It involves multiplying across the equals sign in a criss-cross manner.
What this means is:
When fractions are said to be equal, like \( \frac{a}{b} = \frac{c}{d} \), you can cross-multiply to verify or clear the equation of fractions. It involves multiplying across the equals sign in a criss-cross manner.
What this means is:
- The product of the numerator of the first fraction and the denominator of the second \( (a \cdot d) \) should equal the product of the numerator of the second and the denominator of the first \( (b \cdot c) \).
- Cross multiplication can simplify solving and checking rational expressions, verifying the equality without maintaining the fractions' structure.
Simplifying Fractions
Simplifying fractions involves reducing them to their simplest form. In both numerical and algebraic fractions, simplification is crucial for clarity and ease of use.
To simplify a fraction like \( \frac{6x^2}{3x} \), begin by factoring out common factors from the numerator and the denominator. In this case, both can be divided by 3 and \( x \), simplifying to \( 2x \).
To simplify a fraction like \( \frac{6x^2}{3x} \), begin by factoring out common factors from the numerator and the denominator. In this case, both can be divided by 3 and \( x \), simplifying to \( 2x \).
- Identifying common factors helps reduce fractions, making them much simpler.
- While simplifying, ensure not to cancel out terms incorrectly. Only common factors can be canceled.
- Simplified fractions often make further algebraic manipulation easier and less error-prone.
Other exercises in this chapter
Problem 2
Gabriel wrote \(\frac{12 x}{5 x+10} \div \frac{4}{5}=\frac{12 x \div 4}{(5 x+10) \div 5}=\frac{3 x}{x+2} .\) Is Gabriel's solution correct? Justify your answer.
View solution Problem 2
Bebe said that since each of the denominators in the complex fraction \(\frac{\frac{d}{4}+\frac{3}{5}}{2-\frac{d^{2}}{2}}\) is a non-zero constant, the fraction
View solution Problem 2
Explain the difference between the additive inverse and the multiplicative inverse.
View solution Problem 3
In \(3-14,\) solve and check each inequality. $$ \frac{a}{4}>\frac{a}{2}+6 $$
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