Which Statement Is True Solving Absolute Value Equations

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In the realm of mathematics, absolute value equations often present a unique challenge, requiring a careful understanding of the absolute value concept and its implications. This article delves into the intricacies of solving absolute value equations, dissecting several statements to determine which one holds true. We'll explore the fundamental principles behind absolute value, examine the techniques for solving such equations, and ultimately, unveil the correct statement through a rigorous analysis of each option.

Understanding Absolute Value Equations

Absolute value equations, at their core, involve finding the values of a variable that satisfy an equation containing an absolute value expression. The absolute value of a number represents its distance from zero on the number line, irrespective of direction. This means that the absolute value of a number is always non-negative. For instance, |5| = 5 and |-5| = 5. This concept is crucial when dealing with equations like the ones presented.

To solve absolute value equations, we need to consider two possibilities: the expression inside the absolute value can be either positive or negative. This leads to two separate equations that need to be solved independently. For example, if we have the equation |x| = 3, we need to consider both x = 3 and x = -3 as potential solutions.

Now, let's analyze the given statements, applying these principles to determine their validity.

Dissecting the Statements: A Step-by-Step Analysis

We are presented with four statements, each involving an absolute value equation. Our task is to identify the one statement that is undeniably true. To accomplish this, we will systematically examine each equation, employing algebraic techniques to find potential solutions and assess their validity.

Statement 1: The Equation โˆ’3โˆฃ2x+1.2โˆฃ=โˆ’1-3|2x + 1.2| = -1 has no solution.

To determine if this statement is true, we must attempt to solve the equation. The initial step involves isolating the absolute value expression. We can achieve this by dividing both sides of the equation by -3:

โˆฃ2x+1.2โˆฃ=โˆ’1โˆ’3|2x + 1.2| = \frac{-1}{-3}

โˆฃ2x+1.2โˆฃ=13|2x + 1.2| = \frac{1}{3}

Now, we have an absolute value expression equal to a positive number. This means there are potentially two solutions. We need to consider both cases:

Case 1: 2x+1.2=132x + 1.2 = \frac{1}{3}

2x=13โˆ’1.22x = \frac{1}{3} - 1.2

To subtract 1.2 from 13\frac{1}{3}, we need to convert 1.2 into a fraction. 1.2 is equal to 1210\frac{12}{10}, which simplifies to 65\frac{6}{5}. Therefore:

2x=13โˆ’652x = \frac{1}{3} - \frac{6}{5}

To subtract these fractions, we need a common denominator, which is 15:

2x=515โˆ’18152x = \frac{5}{15} - \frac{18}{15}

2x=โˆ’13152x = \frac{-13}{15}

Now, divide both sides by 2:

x=โˆ’1315รท2x = \frac{-13}{15} \div 2

x=โˆ’1315โ‹…12x = \frac{-13}{15} \cdot \frac{1}{2}

x=โˆ’1330x = \frac{-13}{30}

Case 2: 2x+1.2=โˆ’132x + 1.2 = -\frac{1}{3}

2x=โˆ’13โˆ’1.22x = -\frac{1}{3} - 1.2

Again, convert 1.2 to a fraction:

2x=โˆ’13โˆ’652x = -\frac{1}{3} - \frac{6}{5}

Find a common denominator:

2x=โˆ’515โˆ’18152x = -\frac{5}{15} - \frac{18}{15}

2x=โˆ’23152x = \frac{-23}{15}

Divide both sides by 2:

x=โˆ’2315รท2x = \frac{-23}{15} \div 2

x=โˆ’2315โ‹…12x = \frac{-23}{15} \cdot \frac{1}{2}

x=โˆ’2330x = \frac{-23}{30}

We have found two distinct solutions for this equation: x=โˆ’1330x = \frac{-13}{30} and x=โˆ’2330x = \frac{-23}{30}. Therefore, Statement 1 is false. The equation does have solutions.

Statement 2: The Equation 3.5โˆฃ6xโˆ’2โˆฃ=3.53.5|6x - 2| = 3.5 has one solution.

Let's analyze this statement. First, isolate the absolute value expression by dividing both sides by 3.5:

โˆฃ6xโˆ’2โˆฃ=3.53.5|6x - 2| = \frac{3.5}{3.5}

โˆฃ6xโˆ’2โˆฃ=1|6x - 2| = 1

Now, we consider the two cases:

Case 1: 6xโˆ’2=16x - 2 = 1

6x=1+26x = 1 + 2

6x=36x = 3

x=36x = \frac{3}{6}

x=12x = \frac{1}{2}

Case 2: 6xโˆ’2=โˆ’16x - 2 = -1

6x=โˆ’1+26x = -1 + 2

6x=16x = 1

x=16x = \frac{1}{6}

We have found two distinct solutions: x=12x = \frac{1}{2} and x=16x = \frac{1}{6}. Therefore, Statement 2 is false. The equation has two solutions, not one.

Statement 3: The Equation 5โˆฃโˆ’3.1x+6.9โˆฃ=โˆ’3.55|-3.1x + 6.9| = -3.5 has two solutions.

To assess this statement, isolate the absolute value expression by dividing both sides by 5:

โˆฃโˆ’3.1x+6.9โˆฃ=โˆ’3.55|-3.1x + 6.9| = \frac{-3.5}{5}

โˆฃโˆ’3.1x+6.9โˆฃ=โˆ’0.7|-3.1x + 6.9| = -0.7

Here, we encounter a crucial point. The absolute value of any expression can never be negative. Since we have the absolute value of an expression equaling -0.7, which is a negative number, there is no solution to this equation. Therefore, Statement 3 is false. The equation has no solutions.

Statement 4: The Equation โˆ’0.3โˆฃ3+8xโˆฃ=0.9-0.3|3 + 8x| = 0.9 has no solution.

Let's examine this statement. Divide both sides by -0.3 to isolate the absolute value expression:

โˆฃ3+8xโˆฃ=0.9โˆ’0.3|3 + 8x| = \frac{0.9}{-0.3}

โˆฃ3+8xโˆฃ=โˆ’3|3 + 8x| = -3

Similar to Statement 3, we have the absolute value of an expression equaling a negative number (-3). This is impossible. The absolute value of any expression cannot be negative. Therefore, there is no solution to this equation, and Statement 4 is true.

Conclusion: The Verdict

After a meticulous analysis of each statement, we have determined that Statement 4 is the only true statement. The equation โˆ’0.3โˆฃ3+8xโˆฃ=0.9-0.3|3 + 8x| = 0.9 indeed has no solution because the absolute value of an expression cannot be negative.

This exploration highlights the importance of understanding the fundamental properties of absolute value when solving equations. By carefully considering the implications of the absolute value, we can accurately determine the number of solutions and identify statements that hold true in the world of mathematics.

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