/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Perform the addition or subtract... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Perform the addition or subtraction and simplify. $$\frac{1}{x^{2}}+\frac{1}{x^{2}+x}$$

Short Answer

Expert verified
\(\frac{2x + 1}{x^2(x + 1)}\)

Step by step solution

01

Identify a Common Denominator

To add the fractions \( \frac{1}{x^2} \) and \( \frac{1}{x^2 + x} \), we need a common denominator. The denominators are \( x^2 \) and \( x^2 + x \), which can be combined by multiplying them: \( x^2(x^2 + x) = x^4 + x^3 \).
02

Adjust Each Fraction to the Common Denominator

Rewrite each fraction with the common denominator \( x^4 + x^3 \):- For the first fraction \( \frac{1}{x^2} \), multiply both the numerator and the denominator by \( x^2 + x \): \[ \frac{(1 \cdot (x^2 + x))}{(x^2) \cdot (x^2 + x)} = \frac{x^2 + x}{x^4 + x^3} \]- For the second fraction \( \frac{1}{x^2 + x} \), multiply both the numerator and the denominator by \( x^2 \): \[ \frac{1 \cdot x^2}{(x^2 + x) \cdot x^2} = \frac{x^2}{x^4 + x^3} \]
03

Add the Fractions

With a common denominator, we can now add the fractions:\[\frac{x^2 + x}{x^4 + x^3} + \frac{x^2}{x^4 + x^3} = \frac{x^2 + x + x^2}{x^4 + x^3} = \frac{2x^2 + x}{x^4 + x^3}\]
04

Simplify the Result

Factor the numerator and denominator where possible:- Factor out an \( x \) from the numerator: \( 2x^2 + x = x(2x + 1) \).- Factor out an \( x^3 \) from the denominator: \( x^4 + x^3 = x^3(x + 1) \).The expression now is:\[\frac{x(2x + 1)}{x^3(x + 1)}\]Cancel a common \( x \) term from the numerator and denominator:\[\frac{2x + 1}{x^2(x + 1)}\]
05

Conclusion: Final Simplified Expression

The final simplified expression after performing the addition and cancellation is:\[\frac{2x + 1}{x^2(x + 1)}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Common Denominator
When dealing with algebraic fractions, finding a common denominator is essential for addition and subtraction. The common denominator allows us to combine the fractions into a single expression. A common denominator is essentially a shared multiple between the denominators of the fractions involved.

Let's take the original problem for instance: the denominators were \( x^2 \) and \( x^2 + x \). To combine them, we multiplied them together, resulting in a common denominator of \( x^4 + x^3 \). This common base lets us rewrite both fractions so that they can be added together properly.

Steps to find a common denominator:
  • Identify each fraction's denominator.
  • Multiply the denominators to find the least common multiple (though sometimes just the product is sufficient for algebraic fractions).
  • Rewrite each fraction with this newfound common denominator.
Once you have the common denominator, you can focus on the numerators to proceed with addition or subtraction.
Simplifying Expressions
Simplifying expressions in algebra involves several steps to reduce a mathematical expression to its simplest form. It often includes performing operations, like addition or subtraction, and then reducing complexity through factoring or canceling terms.

For example, after obtaining the common denominator and combining fractions, our expression was \( \frac{2x^2 + x}{x^4 + x^3} \). Simplifying begins with examining both numerator and denominator:
  • Factor out common elements.
  • Simplify by canceling common factors from the numerator and the denominator.
  • Ensure all expressions are in their simplest form at the end.
\(2x^2 + x\) simplifies through factoring to \(x(2x + 1)\), and \(x^4 + x^3\) factors to \(x^3(x + 1)\).

By canceling one \(x\) from the numerator and denominator, you get \(\frac{2x + 1}{x^2(x + 1)}\), which is the expression in its simplest form.
Factoring Polynomials
Factoring is a powerful tool in algebra used to simplify expressions and solve equations. It involves breaking down a polynomial into simpler, multiplicative components. Let's see this through the example provided:

The numerator, \( 2x^2 + x \), was factored into \( x(2x + 1) \), and the denominator \( x^4 + x^3 \) was factored into \( x^3(x + 1) \). This process allowed for canceling of the common term \( x \) between the numerator and denominator, simplifying the overall expression.

Steps to factor polynomials effectively:
  • Identify common factors first.
  • Look for patterns like the difference of squares or trinomials that fit factoring formulas.
  • Rewrite the expression in its factored form, simplifying any resulting expressions where possible.
A well-factored expression is much easier to work with and critical for solving algebraic problems. It reduces complexity and can reveal insights that might not be apparent in the expanded form of the equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Gas Mileage The gas mileage \(q\) (measured in milgal) for a particular vehicle, driven at \(v\) mi/h, is given by the formula \(g=10+0.9 v-0.01 v^{2},\) as long as \(v\) is between \(10 \mathrm{mi} / \mathrm{h}\) and \(75 \mathrm{mi} / \mathrm{h}\). For what range of speeds is the vehicle's mileage 30 milgal or better?

Solve the equation for the variable \(x\). The constants \(a\) and \(b\) represent positive real numbers. $$x^{4}-5 a x^{2}+4 a^{2}=0$$

Sketch the region given by the set. $$\left\\{(x, y) | x^{2}+y^{2}>4\right\\}$$

DISCOVER - PROVE: Relationship Between Solutions and Coefficients The Quadratic Formula gives us the solutions of a quadratic equation from its coefficients. We can also obtain the coefficients from the solutions. (a) Find the solutions of the equation \(x^{2}-9 x+20=0\) and show that the product of the solutions is the constant term 20 and the sum of the solutions is \(9,\) the negative of the coefficient of \(x\) (b) Show that the same relationship between solutions and coefficients holds for the following equations:$$ \begin{array}{l}x^{2}-2 x-8=0 \\\x^{2}+4 x+2=0\end{array}$$ (c) Use the Quadratic Formula to prove that in general, if the equation \(x^{2}+b x+c=0\) has solutions \(r_{1}\) and \(r_{2}\) then \(c=r_{1} r_{2}\) and \(b=-\left(r_{1}+r_{2}\right)\)

Shrinkage in Concrete Beams As concrete dries, it shrinks - the higher the water content, the greater the shrinkage. If a concrete beam has a water content of \(\bar{w} \mathrm{kg} / \mathrm{m}^{3},\) then it will shrink by a factor $$S=\frac{0.032 w-2.5}{10,000}$$ where \(S\) is the fraction of the original beam length that disappears due to shrinkage. (a) A beam \(12.025 \mathrm{m}\) long is cast in concrete that contains \(250 \mathrm{kg} / \mathrm{m}^{3}\) water. What is the shrinkage factor \(S ?\) How Iong will the beam be when it has dried? (b) A beam is \(10.014 \mathrm{m}\) long when wet. We want it to shrink to \(10.009 \mathrm{m},\) so the shrinkage factor should be \(S=0.00050 .\) What water content will provide this amount of shrinkage? PICTURE CANT COPY

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.