Problem 39 - College Algebra with Modeling and Visualization

Question

Solve the absolute value equation. $$|15 x-5|=|35-5 x|$$

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = -3\).

Step 1: Set Up the Equations from the Absolute Values

Since both sides of the equation are absolute values, set up two different equations to solve:1) Solve when both expressions are equal without the absolute values: \\[15x - 5 = 35 - 5x\]2) Solve when one expression equals the negation of the other (considering the negative scenario on one side): \\[15x - 5 = -(35 - 5x)\]

Step 2: Solve the First Equation

Solve the equation \(15x - 5 = 35 - 5x\):1) Add \(5x\) to both sides: \\[15x + 5x - 5 = 35\]2) Simplify: \\[20x - 5 = 35\]3) Add 5 to both sides: \\[20x = 40\]4) Divide both sides by 20: \\[x = 2\]

Step 3: Solve the Second Equation

Solve the equation \(15x - 5 = -(35 - 5x)\):1) Distribute the negative on the right-hand side: \\[15x - 5 = -35 + 5x\]2) Subtract \(5x\) from both sides: \\[10x - 5 = -35\]3) Add 5 to both sides: \\[10x = -30\]4) Divide both sides by 10: \\[x = -3\]

Step 4: Verify the Solutions

Substitute \(x = 2\) and \(x = -3\) back into the original equation to verify:1) For \(x = 2\):- LHS: \(15(2) - 5 = 30 - 5 = 25\)- RHS: \(35 - 5(2) = 35 - 10 = 25\)- Both sides equal 25.2) For \(x = -3\):- LHS: \(15(-3) - 5 = -45 - 5 = -50\)- RHS: \(35 - 5(-3) = 35 + 15 = 50\)- Both sides equal "| -50 | = | 50 |" which are equal.Both solutions satisfy the original equation, confirming they are correct.

Key Concepts

Solving Equations

When we deal with absolute value equations, we're looking to find the value of the variable that satisfies the equation. Absolute value refers to the non-negative reflection of a number or expression on the number line. Understanding how to solve these requires setting up different potential scenarios.

To solve the equation \[|15x - 5| = |35 - 5x|,\]we need to consider two separate situations:

• When the expressions inside the absolute values are equal, \[15x - 5 = 35 - 5x\]
• When one expression is the negation of the other, \[15x - 5 = -(35 - 5x).\]

By tackling each situation individually, we ensure we're considering the full range of possible answers.

Verification of Solutions

Verification is a crucial part of any equation-solving process. It's about checking whether the solutions we found actually make the original equation true. For our equation, we found two potential solutions: \( x = 2 \) and \( x = -3 \).

To verify these solutions, we substitute them back into the original equation and see if both sides of the equation are equal:

• For \( x = 2 \), substitute to get \[| 15(2) - 5 | = | 35 - 5(2) |\]which simplifies to \[| 25 | = | 25 |.\]Both sides equal 25, confirming it's a valid solution.
• For \( x = -3 \), substitute to find \[| 15(-3) - 5 | = | 35 - 5(-3) |\]which leads to \[| -50 | = | 50 |\]showing both sides equal when considering absolute values.

Both solutions satisfy the equation, verifying their correctness.

Algebraic Manipulation

Algebraic manipulation allows us to simplify and solve equations systematically by performing the same mathematical operation on both sides of the equation.

With the equation \[15x - 5 = 35 - 5x,\]we follow a series of steps to isolate \( x \):

• Add \( 5x \) to both sides to consolidate terms involving \( x \).
• Combine like terms: \[15x + 5x - 5 = 35\]becomes \[20x - 5 = 35.\]
• Add 5 to both sides to eliminate constants on the side with \( x \): \[20x = 40.\]
• Divide by 20 to solve for \( x \), yielding \[x = 2.\]

Algebraic manipulation empowers us to systematically isolate and solve for variables, highlighting the importance of careful step-by-step operations.