Problem 88
Question
For the following problems, convert the given rational expressions to rational expressions having the same denominators. $$ \frac{x+7}{x^{2}-2 x-3}, \frac{x+3}{x^{2}-6 x-7} $$
Step-by-Step Solution
Verified Answer
Question: Given the rational expressions \(\frac{x+7}{x^2 - 2x -3}\) and \(\frac{x+3}{x^2 - 6x - 7}\), rewrite them with a common denominator.
Answer: The rational expressions with the same denominators are \(\frac{(x+7)(x-7)}{(x-3)(x+1)(x-7)}\) and \(\frac{(x+3)(x-3)}{(x-3)(x+1)(x-7)}\).
1Step 1: Factor the denominators
First, let's factor the denominators of the rational expressions:
$$
x^2 - 2x - 3 = (x - 3)(x+1) \\
x^2 - 6x - 7 = (x - 7)(x+1)
$$
2Step 2: Find the least common multiple (LCM) of the denominators
The least common multiple (LCM) of the two factored denominators is the product of the highest powers of all factors present in the denominators:
$$
LCM((x-3)(x+1), (x-7)(x+1)) = (x-3)(x+1)(x-7)
$$
3Step 3: Rewrite the given rational expressions with the same denominator
Now, let's rewrite each of the original rational expressions with the common denominator (LCM) and adjust their numerators accordingly:
For the first rational expression, we need to multiply by \(\frac{(x-7)}{(x-7)}\) to reach the common denominator:
$$
\frac{x+7}{(x-3)(x+1)} \cdot \frac{(x-7)}{(x-7)} = \frac{(x+7)(x-7)}{(x-3)(x+1)(x-7)}
$$
For the second rational expression, we need to multiply by \(\frac{(x-3)}{(x-3)}\) to reach the common denominator:
$$
\frac{x+3}{(x-7)(x+1)} \cdot \frac{(x-3)}{(x-3)} = \frac{(x+3)(x-3)}{(x-3)(x+1)(x-7)}
$$
So, the rational expressions with the same denominators are:
$$
\frac{(x+7)(x-7)}{(x-3)(x+1)(x-7)} , \frac{(x+3)(x-3)}{(x-3)(x+1)(x-7)}
$$
Key Concepts
Factoring PolynomialsLeast Common MultipleEquivalent Rational Expressions
Factoring Polynomials
When you encounter a polynomial, factoring is like breaking it down into simpler pieces that, when multiplied together, give you the original polynomial. It's a bit like finding out what ingredients went into making a cake so that you can understand it better and work with each component individually.
For example, the expression \(x^2 - 2x - 3\) can be factored into \(x - 3\) and \(x + 1\). How do we get there? We look for two numbers that multiply to give us the constant term (-3), and add to give us the coefficient of the x term (-2). In this case, -3 and +1 do the trick. So, \(x^2 - 2x - 3\) becomes \(x - 3)(x + 1\).
Factoring is a fundamental skill because it's used in simplifying expressions, solving equations, and finding least common multiples, which are all part of working with rational expressions.
For example, the expression \(x^2 - 2x - 3\) can be factored into \(x - 3\) and \(x + 1\). How do we get there? We look for two numbers that multiply to give us the constant term (-3), and add to give us the coefficient of the x term (-2). In this case, -3 and +1 do the trick. So, \(x^2 - 2x - 3\) becomes \(x - 3)(x + 1\).
Factoring is a fundamental skill because it's used in simplifying expressions, solving equations, and finding least common multiples, which are all part of working with rational expressions.
Least Common Multiple
The least common multiple (LCM) of two or more polynomials is the simplest expression that all of the polynomials can divide into without leaving any remainder. Think of it as a common ground that all the expressions can share. It's like scheduling a meeting that fits everyone's available times.
In our example, we have the factored denominators \(x - 3)(x + 1)\) and \(x - 7)(x + 1)\). To find the LCM, we take each different factor at its highest power that appears in any of the polynomials. Since \(x + 1\) is common to both, we only count it once. The LCM of our denominators will then be \(x - 3)(x + 1)(x - 7)\).
Understanding how to find the LCM is very useful in adding, subtracting, and comparing fractions or rational expressions with different denominators, as it allows for a common basis for these operations.
In our example, we have the factored denominators \(x - 3)(x + 1)\) and \(x - 7)(x + 1)\). To find the LCM, we take each different factor at its highest power that appears in any of the polynomials. Since \(x + 1\) is common to both, we only count it once. The LCM of our denominators will then be \(x - 3)(x + 1)(x - 7)\).
Understanding how to find the LCM is very useful in adding, subtracting, and comparing fractions or rational expressions with different denominators, as it allows for a common basis for these operations.
Equivalent Rational Expressions
Creating equivalent rational expressions with the same denominators is similar to balancing scales. You adjust the expressions so they have a common weight without changing their intrinsic value. This process is essential when adding, subtracting, or comparing fractions.
As we did in our example, to find equivalent expressions with the same denominator, multiply the numerator and denominator of each expression by whatever factors are needed to obtain the LCM as the new denominator. When we did this for \(\frac{x+7}{x^{2}-2 x-3}\), we multiplied by \(\frac{x-7}{x-7}\) to make the denominator \(x - 3)(x + 1)(x - 7)\). Similarly, for \(\frac{x+3}{x^{2}-6 x-7}\), we multiplied by \(\frac{x-3}{x-3}\).
This ensures that the original value of the expressions remains unchanged, while allowing us to perform operations between them seamlessly. It's critical to remember that the goal is to balance the expressions, not to change their inherent values.
As we did in our example, to find equivalent expressions with the same denominator, multiply the numerator and denominator of each expression by whatever factors are needed to obtain the LCM as the new denominator. When we did this for \(\frac{x+7}{x^{2}-2 x-3}\), we multiplied by \(\frac{x-7}{x-7}\) to make the denominator \(x - 3)(x + 1)(x - 7)\). Similarly, for \(\frac{x+3}{x^{2}-6 x-7}\), we multiplied by \(\frac{x-3}{x-3}\).
This ensures that the original value of the expressions remains unchanged, while allowing us to perform operations between them seamlessly. It's critical to remember that the goal is to balance the expressions, not to change their inherent values.
Other exercises in this chapter
Problem 87
Classify the polynomial \(4 x y+2 y\) as a monomial, binomial, or trinomial. State its degree and write the numerical coefficient of each term.
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For the following problems, solve the rational equations. Solve \(\frac{1}{R}=\frac{1}{E}+\frac{1}{F}\) for \(E\).
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Find the product: \(y^{2}(2 y-1)(2 y+1)\).
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For the following problems, solve the rational equations. Solve \(Q=\frac{2 m n}{s+t}\) for \(t\).
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