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Chapter 1: Problem 63

Solve each absolute value equation or indicate the equation has no solution. $$ |2 x-1|=5 $$

Short Answer

Expert verified
The solutions to the equation \(|2x - 1| = 5\) are \(x = 3\) and \(x = -2\).

Step by step solution

01

Set up the two possible solutions

Absolute value of a number can be either positive or negative, so set the equation \(2x - 1\) equal to 5 and -5.\nThis yields two equations: \(2x - 1 = 5\) and \(2x - 1 = -5\).
02

Solve the first equation

In the equation \(2x - 1 = 5\), add 1 to both sides to isolate \(2x\) on one side, resulting in \(2x = 6\). Then divide both sides by 2 to solve for \(x\) which gives \(x = 3\).
03

Solve the second equation

In the equation \(2x - 1 = -5\), add 1 to both sides to isolate \(2x\) on one side, which gives \(2x = -4\). Divide both sides of the equation by 2 to solve for \(x\) which gives \(x = -2\).

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Key Concepts

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

Solving Absolute Value Equations
Absolute value equations involve expressions where the value inside an absolute value sign can yield both positive and negative solutions. To solve equations like \(|2x - 1| = 5\), we need to consider two possible scenarios since the expression inside can be equal to both 5 and -5.
This means we set up two separate equations to tackle the absolute value:
  • First, ignore the absolute signs and solve as if it's positive: \(2x - 1 = 5\).
  • Second, solve it assuming the negative scenario: \(2x - 1 = -5\).
By solving each equation, we account for both possibilities that satisfy the absolute value condition.
Isolating the Variable
Isolating a variable means rearranging an equation so that the unknown variable is by itself on one side of the equation. This is a crucial step in solving equations, as it helps us find the value of the unknown.
To isolate \(x\) in our equation \(2x - 1 = 5\), we first need to eliminate any terms that are not attached to the variable. Similar steps follow for the second equation \(2x - 1 = -5\).
Here's how we isolate:
  • For \(2x - 1 = 5\), add 1 to both sides to get rid of the constant: \(2x = 6\).
  • For \(2x - 1 = -5\), the same process applies: add 1 to both sides to get \(2x = -4\).
Additive Inverse
The concept of an additive inverse is key in solving equations to maintain balance. When we have a number with a minus sign and want to eliminate it, we add its opposite, or additive inverse, to both sides of the equation.
For instance, in \(2x - 1 = 5\), the term \(-1\) is the part we need to cancel out. By adding 1 (the additive inverse of -1) to both sides, we maintain the balance of the equation while simplifying it.
This gives us:
  • \(2x = 6\) for the positive scenario.
  • \(2x = -4\) for the negative case.
Understanding additive inverses is crucial for rearranging and simplifying equations efficiently.
Division Method
Once the variable is isolated with a coefficient, you'll often need to use division to solve for the variable itself. This is what we call the division method.
After isolating \(2x\), the next goal is to decipher \(x\) by eliminating the coefficient attached to it. Here鈥檚 how we do it:
  • In \(2x = 6\), divide both sides by 2 to find \(x = 3\).
  • For \(2x = -4\), a similar division by 2 gives us \(x = -2\).
This straightforward division solves for \(x\) by ensuring that each side remains balanced and the solution is accurate.

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Most popular questions from this chapter

In Exercises \(29-44,\) perform the indicated operations and write the result in standard form. $$\frac{-6-\sqrt{-12}}{48}$$

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x+5}{x-2}>0 $$

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x^{2}-3 x+2}{x^{2}-2 x-3}>0 $$

Write a quadratic inequality whose solution set is \([-3,5]\)

In Exercises \(29-44,\) perform the indicated operations and write the result in standard form. $$5 \sqrt{-16}+3 \sqrt{-81}$$
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