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Chapter 7: Problem 53

Perform each indicated operation. Simplify if possible. \(\frac{13}{x^{2}-5 x+6}-\frac{5}{x-3}\)

Short Answer

Expert verified
The simplified expression is \( \frac{-5x + 23}{(x-2)(x-3)} \).

Step by step solution

01

Factor the Denominators

First, factor the quadratic expression in the denominator of the first fraction, \(x^2 - 5x + 6\). Look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. Thus, \(x^2 - 5x + 6\) factors to \((x-2)(x-3)\).
02

Write as Single Fraction

Rewrite each fraction with a common denominator. The common denominator for \((x-2)(x-3)\) and \(x-3\) is \((x-2)(x-3)\). For the first term, the denominator is already the common denominator. For the second term, multiply the numerator and denominator by \(x-2\): \[ \frac{5}{x-3} \times \frac{x-2}{x-2} = \frac{5(x-2)}{(x-3)(x-2)} \] Now, write the expression as a single fraction:\[ \frac{13 - 5(x-2)}{(x-2)(x-3)} \]
03

Simplify the Numerator

Simplify the numerator by distributing and combining like terms:\[ 13 - 5(x-2) = 13 - 5x + 10 = -5x + 23 \]So the expression becomes:\[ \frac{-5x + 23}{(x-2)(x-3)} \]
04

Final Simplification

The numerator \(-5x + 23\) does not have common factors with the denominator \((x-2)(x-3)\), so it is already in its simplest form. Therefore, the simplified expression is:\[ \frac{-5x + 23}{(x-2)(x-3)} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factoring Polynomials
Polynomials can often be broken down into simpler, easier-to-manage components through a process known as factoring. This is the process of finding which smaller expressions multiply together to produce the polynomial. In our exercise, we deal with the expression \(x^2 - 5x + 6\).

To factor this quadratic, we are looking for two numbers that multiply to +6 (the constant term) and add to -5 (the coefficient of the linear term, x). The numbers -2 and -3 fit these requirements perfectly, because
  • -2 multiplied by -3 equals 6
  • -2 plus -3 equals -5
This leads to the factorization of the quadratic as \((x-2)(x-3)\). Factoring is an essential skill in algebra, highly valuable in simplifying complex expressions and solving equations.
Common Denominators
Finding a common denominator is crucial when performing operations like addition and subtraction on fractions. By finding the same denominator for every term, you can easily combine the fractions into a single expression.

In the exercise, our fractions are \(\frac{13}{(x-2)(x-3)}\) and \(\frac{5}{x-3}\). The first denominator is already \((x-2)(x-3)\).

To develop a common denominator, multiply the numerator and denominator of the second fraction, \(\frac{5}{x-3}\), by \(x-2\):
  • This gives: \(\frac{5(x-2)}{(x-3)(x-2)}\)
  • Now, both fractions share the same common denominator: \((x-2)(x-3)\)
Having a common denominator allows straightforward algebraic operations across both fractions.
Simplifying Fractions
Simplifying fractions is about reducing them to their simplest form, making them easier to work with. To simplify a fraction, any common factors in the numerator and denominator should be canceled out.

After addressing the common denominator, the expression becomes a single fraction: \(\frac{13 - 5(x-2)}{(x-2)(x-3)}\). Simplifying involves first dealing with the numerator:
  • Distribute the -5 across the \((x-2)\)
  • This results in: \(13 - 5x + 10\)
  • Combining like terms gives \(-5x + 23\)
Since \(-5x + 23\) has no factors in common with \((x-2)(x-3)\), this fraction is already simplified.
Algebraic Operations
Algebraic operations enable us to manipulate and solve equations or expressions using various mathematical rules and processes. These operations typically include:
  • Addition and subtraction, as seen in combining terms \(13 - 5(x-2)\)
  • Multiplication, for example adjusting denominators and numerators \(5(x-2)\)
  • Distribution, used here to expand \(-5(x-2)\) into \(-5x + 10\)
Understanding these basic operations is vital to performing more complex algebraic tasks. They allow us to combine and simplify expressions, engaging in logical steps to derive answers efficiently.

The operations showcased here highlight how algebra harmoniously combines skills like factoring, distributing, and simplifying fractions to reach a solution.

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Most popular questions from this chapter

Then list four equivalent forms for each rational expression. $$ -\frac{8 y-1}{y-15} $$

Write two rational expressions with the same denominator whose sum is \(\frac{5}{3 x-1}\).

Find the \(L C D\) for each list of rational expressions. $$ \frac{9 x^{2}}{7 x-14}, \frac{6 x}{(x-2)^{2}} $$

Rewrite each rational expression as an equivalent rational expression with the given denominator. $$ \frac{5}{4 y^{2} x}=\frac{ }{32 y^{3} x^{2}} $$

Find the \(L C D\) for each list of rational expressions. $$ \frac{19}{2 x}, \frac{5}{4 x^{3}} $$
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