Problem 12 - Elementary Algebra Within Reach

Question

Determine whether each value of \(x\) is a solution of the inequality. \(2 x+1<3\) (a) \(x=0\) (b) \(x=4\) (c) \(x=-\frac{2}{5}\) (d) \(x=\frac{1}{2}\)

Step-by-Step Solution

Answer

The values of \(x = 0\), \(x = -\frac{2}{5}\), and \(x = \frac{1}{2}\) are solutions to the inequality \(2x + 1 < 3\), while \(x = 4\) is not a solution.

Step 1: Check \(x = 0\)

Substitute \(x = 0\) into the inequality \(2x + 1 < 3\). It becomes \(2*0+1=1\), which is less than 3, so \(x=0\) is a solution to the inequality.

Step 2: Check \(x = 4\)

Substitute \(x = 4\) into the inequality \(2x + 1 < 3\). It becomes \(2*4+1=9\), which is not less than 3, so \(x=4\) is not a solution to the inequality.

Step 3: Check \(x = -\frac{2}{5}\)

Substitute \(x = -\frac{2}{5}\) into the inequality \(2x + 1 < 3\). It becomes \(2*(-\frac{2}{5})+1= -\frac{4}{5} +1 = 0.2\), which is less than 3, so \(x=-\frac{2}{5}\) is a solution to the inequality.

Step 4: Check \(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into the inequality \(2x + 1 < 3\). It becomes \(2*(\frac{1}{2})+1= 2\), which is less than 3, so \(x=\frac{1}{2}\) is a solution to the inequality.

Key Concepts

Solutions

In algebra, a solution refers to the value of a variable that satisfies an equation or inequality. When we talk about solutions for inequalities, these are the values that make the inequality statement true. Understanding how to find and verify solutions is crucial because it determines which values can be considered valid answers for the problem at hand.

• In the exercise, the inequality is given as \(2x + 1 < 3\).
• The goal is to determine which values of \(x\) satisfy this inequality.

To verify whether a specific value of \(x\) is a solution, we substitute it into the inequality. If the resulting expression is true, then the value is indeed a solution. For example, substituting \(x = 0\):

• Plug it into the inequality resulting in \(2(0) + 1 = 1\), which is less than 3.
• Thus, \(x = 0\) is a solution.

Developing the skill to determine solutions helps ensure you can tackle various algebraic problems effectively.

Algebra

Algebra is a major branch of mathematics that involves symbols and the rules for manipulating these symbols. It forms the basis for advanced mathematics and many real-world applications. In the context of our exercise, we're dealing with linear inequalities, which are an essential part of algebra.

• An inequality such as \(2x + 1 < 3\) describes a relationship where one side is not equal but rather less than the other.
• This inequality tells us about the range of values that \(x\) can take rather than a precise value.

The beauty of algebra lies in its ability to generalize mathematical concepts and enable us to solve problems systematically.

• By engaging with algebra, you improve your problem-solving skills and your ability to think logically.
• It also equips you with the tools needed to understand and work with more complex mathematical approaches later.

Whether dealing with equations or inequalities, mastering the basics of algebra is vital for your overall progress in mathematics.

Substitution Method

The substitution method is a powerful algebraic tool used to determine if specific values satisfy an equation or inequality. This process involves taking a proposed value and "substituting" it into the given equation or inequality to see if it holds true.

• For inequalities, like in our exercise, the same concept applies by testing different values of \(x\).
• We plug each proposed value into the inequality \(2x + 1 < 3\) to check its validity.

This method is simple but effective:

• Start with the inequality or equation and the value you want to test.
• Replace the variable with the given number to check if the resulting statement is true.

Through this exercise:

• When \(x = \frac{1}{2}\), substituting it in gives \(2(\frac{1}{2}) + 1 = 2\), which is less than 3.
• Hence, it is confirmed as a valid solution to the inequality.

Employing the substitution method allows one to systematically verify solutions and strengthen their understanding of algebraic manipulation.