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Chapter 1: Problem 26

Find the real solutions, if any, of each equation. $$ 5-\left|\frac{1}{2} x\right|=3 $$

Short Answer

Expert verified
The real solutions are x = 4 and x = -4.

Step by step solution

01

- Isolate the absolute value expression

Subtract 5 from both sides of the equation to isolate the absolute value expression: $$5 - \frac{1}{2}|x| - 5 = 3 - 5$$ $$-\frac{1}{2} |x|= -2$$
02

- Remove the coefficient of absolute value

Divide both sides by -1/2 to solve for the absolute value expression: $$|x| = \frac{2}{\frac{1}{2}}$$ $$|x| = 4$$
03

- Solve the absolute value equation

The absolute value equation $$|x| = 4$$ has two solutions: $$x = 4$$ and $$x = -4$$
04

- Verify the solutions

Substitute each solution back into the original equation to verify: For $$x = 4$$: $$5 - |\frac{1}{2} \times 4| = 5 - 2 = 3$$ (True) For $$x = -4$$: $$5 - |\frac{1}{2} \times (-4)| = 5 - 2 = 3$$ (True) Since both solutions satisfy the original equation, they are correct.

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Key Concepts

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

absolute value isolation
When dealing with absolute value equations, the first step is to isolate the absolute value expression. This means getting \(|x|\) on one side of the equation by itself.
In our example, the equation is $$5 - \left| \frac{1}{2} x \right| = 3$$.
To isolate the absolute value, we subtract 5 from both sides:
$$5 - \frac{1}{2}|x| - 5 = 3 - 5$$. Simplifying this, we get:
$$-\frac{1}{2} |x| = -2$$.
The absolute value expression is now isolated.
equation solving steps
Once the absolute value expression is isolated, the next step is to solve the equation. First, remove any coefficient from \(|x|\).
In our example, we have the equation: $$-\frac{1}{2} |x| = -2$$.
Divide both sides by -1/2:
$$|x| = \frac{2}{\frac{1}{2}}$$.
Simplifying this, we get:
$$|x| = 4$$.
Next, solve the absolute value equation. Since the absolute value of a number \(|x|\) can be either positive or negative, we get two solutions:
- $$x = 4$$
- $$x = -4$$.
We reached the solutions for the equation.
verification of solutions
Verifying solutions is an essential step in solving absolute value equations.
To ensure our solutions are correct, substitute them back into the original equation.
For $$x = 4$$:
$$5 - |\frac{1}{2} \times 4| = 5 - 2 = 3$$.
Since this is true, $$x = 4$$ is a correct solution.
For $$x = -4$$:
$$5 - |\frac{1}{2} \times (-4)| = 5 - 2 = 3$$.
This is also true, so $$x = -4$$ is another correct solution.
Verifying helps you confirm the accuracy of your results.
algebraic manipulation
Algebraic manipulation involves using algebraic techniques to simplify and solve equations.
In the given exercise, we initially had:\( 5 - \left| \frac{1}{2} x \right| = 3 \).
To isolate \(|x|\), subtract 5 from both sides:\br> $$5 - \frac{1}{2} |x| - 5 = 3 - 5$$,
which simplifies to: $$-\frac{1}{2} |x| = -2$$.
Then, to remove the coefficient of \(|x|\), divide both sides by -1/2:
$$|x| = \frac{2}{\frac{1}{2}}$$, simplifying to $$|x| = 4$$.
This process of algebraic manipulation helps isolate variables and solve equations systematically.

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Most popular questions from this chapter

Constructing a Border around a Garden A landscaper, who just completed a rectangular flower garden measuring 6 feet by 10 feet, orders 1 cubic yard of premixed cement, all of which is to be used to create a border of uniform width around the garden. If the border is to have a depth of 3 inches, how wide will the border be? (Hint: 1 cubic yard \(=27\) cubic feet)

Solve: \(\sqrt{3 x+5}-\sqrt{x-2}=\sqrt{x+3}\)

Discuss the step in the solving process for radical equations that leads to the possibility of extraneous solutions. Why is there no such possibility for linear and quadratic equations?

Find the real solutions, if any, of each equation. $$ \left(\frac{v}{v+1}\right)^{2}+\frac{2 v}{v+1}=8 $$

Challenge Problem If \(\frac{2}{3} \leq \frac{5-x}{3} \leq 3,\) find the largest possible value of \(2 x^{2}-3\)
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