Problem 30 Give four examples of pairs of r... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 0: Problem 30

Give four examples of pairs of real numbers $a$ and $b$ such that $|a+b|=3$ and $|a|+|b|=11$.

Short Answer

Expert verified

Our four pairs of real numbers (a, b) that satisfy both equations are: 1) $(7, -4)$ 2) $(-4, 7)$ 3) $(-7, 4)$ 4) $(4, -7)$

Step by step solution

Understand Absolute Value Properties

Absolute value is a function that converts any number into its non-negative equivalent. It is defined as: $ |x| = \begin{cases} x & \text{for } x \geq 0 \\ -x & \text{for } x < 0 \end{cases} $ Given this definition, we can derive a few properties of absolute values that are helpful when solving absolute value equations: 1) $|x|\geq 0$ 2) $|xy|=|x|\cdot |y|$ 3) $|x+y|\leq |x|+|y|$ Keep these properties in mind as we solve the equations.

Analyze Equation 1

Given the first equation, $|a+b|=3$, we can say that either $a+b=3$ or $a+b=-3$. We will need to consider both cases when solving the system. Case 1: $a+b=3$ Case 2: $a+b=-3$

Analyze Equation 2

We should also analyze the second equation, $|a|+|b|=11$, in the context of the given cases. For Case 1, if $a+b=3$, we can rewrite Equation 2 as $|a|+|3-a|=11$. For Case 2, if $a+b=-3$, we rewrite Equation 2 as $|a|+|-3-a|=11$.

Solve for Case 1

Now let's solve these equations. First, for Case 1: $3-a \leq 0 \Rightarrow a \geq 3 \Rightarrow \text{we have } |3-a|\ = \ a-3 \ \text{and} \ -a+3\ ) So our equation becomes: Case 1A: \(|a|+a-3=11 \Rightarrow 2a=14 \Rightarrow a=7 \ \text{and} \ b=-4 $ (This is a valid solution) Case 1B: $|a|-a+3 = 11 \Rightarrow -a+3=11 \Rightarrow a=-8 $ (This is invalid, as it contradicts the initial assumption that $a+b=3$)

Solve for Case 2

Now let's examine Case 2: $-3-a \leq 0 \Rightarrow a \geq -3 \Rightarrow \text{we have } |-3-a|\ = \ -3-a \ \text{and} \ a+3\ ) So our equation becomes: Case 2A: \(|a|-a-3=11 \Rightarrow -2a=14 \Rightarrow a=-7 \ \text{and} \ b=4$ (This is a valid solution) Case 2B: $|a|+a+3=11 \Rightarrow 2a=8 \Rightarrow a=4$ (This is invalid, as it contradicts the initial assumption that $a+b=-3$)

Combine the Valid Solutions

To meet the requirement of providing four examples, we need to find two more solutions. Note that the equations are symmetric, meaning that if $(a, b)$ is a solution, then $(b, a)$ is also a solution. Thus, we have the following four pairs of real numbers (a, b): 1) $(7, -4)$ 2) $(-4, 7)$ 3) $(-7, 4)$ 4) $(4, -7)$

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91影视

Give four examples of pairs of real numbers \(a\) and \(b\) such that \(|a+b|=3\) and \(|a|+|b|=11\).

Short Answer

Step by step solution

Understand Absolute Value Properties

Analyze Equation 1

Analyze Equation 2

Solve for Case 1

Solve for Case 2

Combine the Valid Solutions

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