1. Lial/Hungerford/Holcomb: Mathematics with Applications 11e Finite Mathematics with Applications 11e
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Chapter 1
Algebra and Equations
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Section 1.1
The Real Numbers
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13. Graph all real numbers x such that .
Example:
Graph all real numbers x such that
Solution:
This graph includes all the real numbers between 1 and 5, not just the integers. Graph these numbers by drawing a heavy line from 1 to 5 on the number line.
Parentheses at 1 and 5 show that neither of these points belongs to the graph.
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Section 1.2
Polynomials
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Example:
Subtract:
Solution:
Eliminate parentheses.
Group like terms.
Combine like terms.
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Section 1.3
Factoring
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Example:
Factor:
Solution:
We must find integers b and d such that
Since the constant coefficients on each side of the equation must be equal, we must have that is, b and d are factors of 18. Similarly, the coefficients of x must be the same, so that
The possibilities are summarized in this table:
There is no need to list negative factors, such as because their sum is negative. The table suggests that 6 and 3 will work. Verify that
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Section 1.4
Rational Expressions
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Example:
Divide
Solution:
Invert the second expression and multiply (division rule):
Invert and multiply.
Multiply.
Lowest terms
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35. Exponents and Radicals
Section 1.5
Exponents and Radicals
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41. Use the inversion property to compute each of the following. (a) (b)
Example:
Use the inversion property to compute each of the following. (a) (b)
Solution:
(a)
(b)
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57. First-Degree Equations
Section 1.6
First-Degree Equations
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Example:
Solve
Solution:
First, simplify the equation by using the distributive property on the left-side term and the right-side term
One way to proceed is to add to both sides:
The solution is 2. Check this result by substituting 2 for k in the original equation.
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Section 1.7
Quadratic Equations
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Example:
Solve
Solution:
Rewrite the equation as
Now factor to get
By the zero-factor property, the product can equal 0 only if
Solving each of these equations separately gives the solutions of the original equation:
Verify that both 1/3 and −3/2 are solutions by substituting them into the original equation.
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