Problem 47 - Elementary and Intermediate Algebra

Question

Solve each equation. \(|5 x-12|=|4 x-16|\)

Step-by-Step Solution

Answer

The solutions are \(x = -4\) and \(x = \frac{28}{9}\).

Step 1: Set Up Equations from the Absolute Values

Because both sides of the equation have absolute values, we need to set up two cases. The first case considers the expressions within the absolute values as equal: \[5x - 12 = 4x - 16\] The second case considers one expression as the negative of the other:\[5x - 12 = -(4x - 16)\]

Step 2: Solve the First Case

For the equation from the first case, simplify and solve for \(x\):1. Subtract \(4x\) from both sides: \[5x - 4x - 12 = -16\] \[x - 12 = -16\]2. Add 12 to both sides: \[x = -16 + 12\] \[x = -4\]

Step 3: Solve the Second Case

For the equation from the second case, first distribute the negative sign, then simplify and solve for \(x\):\[5x - 12 = -4x + 16\]1. Add \(4x\) to both sides: \[5x + 4x - 12 = 16\] \[9x - 12 = 16\]2. Add 12 to both sides: \[9x = 16 + 12\] \[9x = 28\]3. Divide both sides by 9: \[x = \frac{28}{9}\]

Step 4: Verify Solutions

We need to ensure both solutions satisfy the original equation:1. For \(x = -4\): - Left side: \(|5(-4) - 12| = |-20 - 12| = |-32| = 32\) - Right side: \(|4(-4) - 16| = |-16 - 16| = |-32| = 32\) - Both sides equal, so \(x = -4\) is a solution.2. For \(x = \frac{28}{9}\): - Calculate separately for precision: - Left side: \(|5(\frac{28}{9}) - 12| = |\frac{140}{9} - \frac{108}{9}| = |\frac{32}{9}|\) - Right side: \(|4(\frac{28}{9}) - 16| = |\frac{112}{9} - \frac{144}{9}| = |-\frac{32}{9}| = \frac{32}{9}|\) - Both sides equal, so \(x = \frac{28}{9}\) is also a solution.

Key Concepts

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols to solve problems. Think of it as a generalized arithmetic where numbers are represented by letters (like \( x \) in this example). By using algebra, we can solve equations and understand how different variables relate to one another.

In algebra, you'll see expressions, which combine numbers and variables. We use these expressions to form equations which can be solved.

• Variables like \( x \) stand in for unknown numbers or quantities.
• Equations express equality between two expressions.

In this problem, both sides of the equation have variables within absolute values. Our goal is to determine which values of \( x \) make the equation true by solving algebraically.

Equation Solving

Solving an equation means finding the value(s) of the variable that make the equation true. This involves manipulating the equation to isolate the variable. Let's follow the step-by-step solution provided to solve these specific types of equations.

1. **Set Up Two Cases**: In this particular problem, because both sides include absolute values, we must consider two possibilities:

• The expressions inside the absolute values are equal.
• One expression is the negative of the other.

2. **Solving Steps**: For each case, we simplify and solve:
- For the first case, subtract \(4x\) from both sides to isolate \(x\), then solve as you would a typical algebraic equation.
- For the second case, distribute the negative sign to simplify before solving.

3. **Checking Solutions**: After finding possible solutions for \(x\), substitute them back into the original equation to verify that they satisfy it. This helps to confirm that you didn't make any mistakes along the way.

Absolute Values

Absolute value refers to how far a number is from zero on the number line, regardless of direction. It is always non-negative.

The absolute value of a number \(a\) is denoted as \(|a|\).

• If \(a \geq 0\), then \(|a| = a\).
• If \(a < 0\), then \(|a| = -a\).

When solving absolute value equations like the one in this exercise, where \(|5x-12|=|4x-16|\), it's crucial to remember that:

• The expressions inside the absolute value bars can either be equal (positive case) or opposite signs (negative case).
• This leads us to consider two distinct equation cases, as solving both will provide potential solutions.

Understanding this concept simplifies problem-solving because it turns an absolute value equation into simpler linear equations that you can solve using familiar algebraic techniques.