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Chapter 1: Problem 15

Solve the equations in Exercises \(13-18\) $$ |2 t+5|=4 $$

Short Answer

Expert verified
The solutions are \(t = -\frac{1}{2}\) and \(t = -\frac{9}{2}\).

Step by step solution

01

Understand Absolute Value Equation

The equation given is \(|2t + 5| = 4\). An absolute value \(|x|\) represents the distance of \(x\) from zero on a number line. Thus, it can be either \(x = 4\) or \(x = -4\).
02

Set Up Two Separate Equations

Since \(|2t + 5| = 4\), this implies two possibilities: 1. \(2t + 5 = 4\) 2. \(2t + 5 = -4\).
03

Solve the First Equation

Start with the first possibility, \(2t + 5 = 4\). Subtract 5 from both sides: \[2t = 4 - 5\]\[2t = -1\]Divide both sides by 2:\[t = -\frac{1}{2}\].
04

Solve the Second Equation

Now solve the second possibility, \(2t + 5 = -4\).Subtract 5 from both sides:\[2t = -4 - 5\]\[2t = -9\]Divide both sides by 2:\[t = -\frac{9}{2}\].
05

Write General Solution

The solutions for \(|2t + 5| = 4\) are \(t = -\frac{1}{2}\) and \(t = -\frac{9}{2}\).

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Key Concepts

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

Solving Absolute Value Equations
When we encounter an absolute value equation like \(|2t + 5| = 4\), it signifies the distance on the number line. The absolute value takes into account both the positive and negative scenarios of the number. This is why we end up with two separate equations to solve - one for the positive case and one for the negative case. To break it down further, if we have \(|x| = a\), then the solutions can be expressed as \(x = a\) or \(x = -a\). This is because the absolute value represents a measure without direction, so both scenarios are valid solutions. In our example:
  • Setting the inside of the absolute value to the positive, \(2t + 5 = 4\).
  • Setting to the negative, \(2t + 5 = -4\).
Using this method, we can solve and find all possible values for the variable in an absolute value equation.
Absolute Value Properties
Absolute value properties are helpful when tackling equations that incorporate \(|x|\). The absolute value of any number \(x\) denoted as \(|x|\) is always non-negative. Certain notable properties include:
  • Non-Negativity: \(|x| \geq 0\) for any \(x\). That means the result of an absolute value is never negative.
  • Identity Property: If \(|x| = 0\), then \(x = 0\). This makes sense because zero has no distance from itself on the number line.
  • Symmetry: \(|-x| = |x|\). Negative values become positive, thus symmetrical around zero.
  • Triangle Inequality: \(|x + y| \leq |x| + |y|\), which helps us understand how sums of absolute values work compared to the absolute value of a sum.
These properties facilitate an understanding of why absolute value equations require two solutions and help in comprehending their application in real-world scenarios.
Linear Equations Solutions
Solving the linear equations is the next step after setting up the equation through absolute value understanding. In our case, once we have the equations \(2t + 5 = 4\) and \(2t + 5 = -4\), they need to be solved individually to find the values of \(t\).Let's consider the first equation:
  • Subtract 5 from both sides which gives \(2t = -1\).
  • Divide by 2 which simplifies to \(t = -\frac{1}{2}\).
For the second equation, repeat the steps:
  • Subtract 5 to end up with \(2t = -9\).
  • Again, divide by 2 arriving at \(t = -\frac{9}{2}\).
Therefore, for an absolute value equation, after transforming it into linear equations, performing basic algebraic isolation gives all possible solutions for the variable in question. It鈥檚 a straightforward process of isolating the variable through addition, subtraction, multiplication, or division.

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Most popular questions from this chapter

A magic trick You may have heard of a magic trick that goes like this: Take any number. Add \(5 .\) Double the result. Subtract \(6 .\) Divide by \(2 .\) Subtract \(2 .\) Now tell me your answer, and I'll tell you what you started with. Pick a number and try it. You can see what is going on if you let \(x\) be your original number and follow the steps to make a formula \(f(x)\) for the number you end up with.

Express the side length of a square as a function of the length \(d\) of the square's diagonal. Then express the area as a function of the diagonal length.

In Exercises \(67-70,\) you will explore graphically the general sine function $$f(x)=37 \sin \left(\frac{2 \pi}{365}(x-101)\right)+25$$ as you change the values of the constants \(A, B, C,\) and \(D .\) Use a CAS or computer grapher to perform the steps in the exercises. The vertical shift \(D \quad\) Set the constants \(A=3, B=6, C=0\) a. Plot \(f(x)\) for the values \(D=0,1,\) and 3 over the interval \(-4 \pi \leq x \leq 4 \pi\) . Describe what happens to the graph of the general sine function as \(D\) increases through positive values. b. What happens to the graph for negative values of \(D ?\)

Graph the following equations and explain why they are not graphs of functions of \(x .\) a. \(|x|+|y|=1 \quad\) b. \(|x+y|=1\)

Exercises \(19-28\) tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation. $$ y=1 / x^{2} \quad \text { Left } 2, \text { down } 1 $$
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