Problem 22 - Algebra and Trigonometry

Question

\(5-22=\) Solve the equation. $$ |x+3|=|2 x+1| $$

Step-by-Step Solution

Answer

The solutions are \( x = 2 \) and \( x = -\frac{4}{3} \).

Step 1: Understand the Absolute Value Function

The equation involves absolute value expressions, which can be understood as the distance from zero. If \( |a| = |b| \), then either \( a = b \) or \( a = -b \). We will use this property to solve the equation.

Step 2: Set Up the Equations

Using the property of absolute values, we set up two cases. Case 1: \( x+3 = 2x+1 \) Case 2: \( x+3 = -(2x+1) \) These will be solved separately.

Step 3: Solve Case 1: \( x+3 = 2x+1 \)

Rearrange the equation to solve for \( x \): \[ x + 3 = 2x + 1 \Rightarrow 3 - 1 = 2x - x \Rightarrow 2 = x\] So, \( x = 2 \) is a solution for the first case.

Step 4: Solve Case 2: \( x+3 = -(2x+1) \)

Distribute the negative sign on the right side and solve for \( x \): \[ x + 3 = -2x - 1 \Rightarrow x + 2x = -1 - 3 \Rightarrow 3x = -4 \Rightarrow x = -\frac{4}{3}\]So, \( x = -\frac{4}{3} \) is a solution for the second case.

Step 5: Verify Solutions

Check both solutions in the original equation to ensure they are correct.For \( x = 2 \): \[|2 + 3| = |2(2) + 1| \Rightarrow |5| = |5| \Rightarrow 5 = 5\] This checks out.For \( x = -\frac{4}{3} \): \[\left| -\frac{4}{3} + 3 \right| = \left| 2 \left( -\frac{4}{3} \right) + 1 \right| \Rightarrow \left| \frac{5}{3} \right| = \left| -\frac{5}{3} \right| \Rightarrow \frac{5}{3} = \frac{5}{3}\]This checks out as well.

Key Concepts

Solving Equations

The process of solving equations is like finding a missing puzzle piece that makes everything fit together perfectly. When dealing with absolute value equations, it's essential to understand the nature of absolute values as representing the distance from zero. This means the number inside the absolute value can either be positive or negative, but its magnitude remains the same.

For the given equation \(|x+3|=|2x+1|\), the property of absolute values tells us that \(|a| = |b|\) can lead to two scenarios: either \(a = b\) or \(a = -b\). We need to consider both possibilities to find all potential solutions.

• In Case 1, we assume the expressions inside the absolute values are equal: \(x+3 = 2x+1\).
• In Case 2, we consider them to be negatives of each other: \(x+3 = -(2x+1)\).

By analyzing these two cases separately, we ensure that we explore all possible values of \(x\) that satisfy the original equation.

Algebraic Techniques

Once we've set up our equations in each case, it's time to bring some algebraic techniques into play. Solving each equation involves rearranging terms to isolate the variable \(x\) on one side of the equation.

In Case 1, from \(x+3 = 2x+1\), we move the \(x\) and constants around:

• Subtract \(x\) from both sides to get \(3 = x + 1\).
• Subtract 1 from both sides to isolate \(x\), leaving you with \(x = 2\).

In Case 2, for the equation \(x+3 = -(2x+1)\):

• First, distribute the negative sign to get \(x + 3 = -2x - 1\).
• Add \(2x\) to both sides resulting in \(3x + 3 = -1\).
• Subtract 3 from both sides to get \(3x = -4\).
• Finally, divide by 3 to find \(x = -\frac{4}{3}\).

Using these steps ensures that each potential solution is thoroughly explored, using structural changes in the equation to isolate and solve for \(x\).

Verification of Solutions

After calculating potential solutions, it's vital to verify these to avoid any sneaky missteps. By substituting each solution back into the original equation \(|x+3|=|2x+1|\), you can ensure both sides of the equation remain equal.

For \(x = 2\):

• Substitute into the absolute values: \(|2+3|\) and \(|2(2)+1|\).
• This gives \(|5| = |5|\) resulting in a true statement, confirming \(x = 2\) is correct.

For \(x = -\frac{4}{3}\):

• Insert into the equation: \(|-\frac{4}{3} + 3|\) and \(|2(-\frac{4}{3})+1|\).
• This results in \(\left|\frac{5}{3}\right| = \left| -\frac{5}{3} \right|\), which is also a true statement.

Verification ensures that our calculated solutions are indeed solutions to the original equation. It solidifies our understanding and confidence in solving absolute value equations correctly.