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Chapter 0: Problem 56

Find the value of each expression if \(x=3\) and \(y=-2\) \(|3 x+2 y|\)

Short Answer

Expert verified
5

Step by step solution

01

Substitute the values

Start by substituting the given values of 3 for x and -2 for y into the expression The expression is |3x + 2y| Substitute x = 3 and y = -2 into the expression: |3(3) + 2(-2)|
02

Simplify inside the absolute value

Simplify the expression inside the absolute value sign: |3(3) + 2(-2)| = |9 + (-4)| = |9 - 4|
03

Calculate the result

Finish simplifying the expression inside the absolute value and compute the absolute value: |9 - 4| = |5| = 5

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Key Concepts

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

Substitution in Algebra
Substitution in algebra is a critical concept. We often need to replace variables with their given values to solve expressions. For instance, in the exercise, we substitute the values of \(x\) and \(y\) into the expression \(|3x + 2y|\).
Here is how it is done:
1. Replace \(x\) with 3 and \(y\) with -2.
2. The expression \(|3x + 2y|\) then becomes \(|3(3) + 2(-2)|\).
This step helps to transform the algebraic problem into a numerical one, making it easier to solve.
Simplifying Expressions
Simplifying expressions is a cornerstone of algebra. It involves performing arithmetic operations to reduce an expression to its simplest form.
For example, in the substituted expression \(|3(3) + 2(-2)|\), we first carry out the multiplication:
1. \(3 \times 3 = 9\)
2. \(2 \times -2 = -4\)
Now, we add these results together: \(9 + (-4)\) which simplifies to \(9 - 4\).
At this point, the expression inside the absolute value is simplified to \(5\). Breaking down each operation is crucial for clarity and accurate results.
Absolute Value Calculations
The absolute value of a number is its distance from zero on a number line, regardless of direction. In simpler terms, it is always a non-negative value.
In our problem, after simplifying the expression inside the absolute value, we get \(|9 - 4| = |5|\).
The absolute value of 5 is simply 5, because 5 units from zero is 5.
The final step: After calculating the inside expression, wrap up by determining the absolute value, giving us the result 5.
Absolute value ensures our final answer is always non-negative, simplifying and clarifying the result.

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Most popular questions from this chapter

Simplify each expression. Assume that all variables are positive when they appear. $$\sqrt[4]{162 x^{9} y^{12}}$$

Expressions that occur in calculus are given. Write each expression as a single quotient in which only positive exponents and radicals appear. $$\frac{\frac{x^{2}}{\left(x^{2}-1\right)^{1 / 2}}-\left(x^{2}-1\right)^{1 / 2}}{x^{2}} \quad x<-1 \text { or } x>1$$

Expressions that occur in calculus are given. Write each expression as a single quotient in which only positive exponents and radicals appear. $$2 x\left(x^{2}+1\right)^{1 / 2}+x^{2} \cdot \frac{1}{2}\left(x^{2}+1\right)^{-1 / 2} \cdot 2 x$$

Simplify each expression. Express your answer so that only positive exponents occur. Assume that the variables are positive. $$\frac{\left(x^{2} y\right)^{1 / 3}\left(x y^{2}\right)^{2 / 3}}{x^{2 / 3} y^{2 / 3}}$$

Use a calculator to approximate each radical. Round your answer to two decimal places. $$\frac{3 \sqrt[3]{5}-\sqrt{2}}{\sqrt{3}}$$
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