Math Problem Statement
Solve for x and y using this system of absolute value equations; |x+3| = |3x+1|
Solution
Let's solve the system of absolute value equations:
Step 1: Split the absolute values
We need to consider the possible cases for the absolute values. Since absolute values are defined as:
\begin{cases} A & \text{if } A \geq 0 \\ -A & \text{if } A < 0 \end{cases}$$ We will have different cases depending on the signs of the expressions inside the absolute values. We have two absolute value expressions: $$|x+3|$$ and $$|3x+1|$$. Thus, we need to consider four possible cases: ### Case 1: Both expressions are non-negative $$x + 3 \geq 0 \quad \text{and} \quad 3x + 1 \geq 0$$ This gives the conditions: $$x \geq -3 \quad \text{and} \quad x \geq -\frac{1}{3}$$ The solution for this case will be valid for $$x \geq -\frac{1}{3}$$. ### Case 2: $$x+3 \geq 0$$ and $$3x+1 < 0$$ $$x \geq -3 \quad \text{and} \quad x < -\frac{1}{3}$$ This gives the range $$-3 \leq x < -\frac{1}{3}$$. ### Case 3: $$x+3 < 0$$ and $$3x+1 \geq 0$$ $$x < -3 \quad \text{and} \quad x \geq -\frac{1}{3}$$ This case leads to no solution, as no $$x$$ can satisfy both inequalities. ### Case 4: Both expressions are negative $$x < -3 \quad \text{and} \quad 3x + 1 < 0$$ This gives: $$x < -3 \quad \text{and} \quad x < -\frac{1}{3}$$ This condition simplifies to $$x < -3$$, which is a valid range. --- ### Step 2: Solve each case #### Case 1: $$x \geq -\frac{1}{3}$$ For $$x \geq -\frac{1}{3}$$, the equation becomes: $$x + 3 = 3x + 1$$ Solve for $$x$$: $$x + 3 = 3x + 1 \quad \Rightarrow \quad 3 - 1 = 3x - x \quad \Rightarrow \quad 2 = 2x \quad \Rightarrow \quad x = 1$$ Since $$x = 1 \geq -\frac{1}{3}$$, this solution is valid. #### Case 2: $$-3 \leq x < -\frac{1}{3}$$ For this case, the equation becomes: $$x + 3 = -(3x + 1)$$ Solve for $$x$$: $$x + 3 = -3x - 1 \quad \Rightarrow \quad 3 + 1 = -3x - x \quad \Rightarrow \quad 4 = -4x \quad \Rightarrow \quad x = -1$$ Since $$-3 \leq -1 < -\frac{1}{3}$$, this solution is valid. #### Case 4: $$x < -3$$ For $$x < -3$$, the equation becomes: $$-(x + 3) = -(3x + 1)$$ Simplify and solve for $$x$$: $$-x - 3 = -3x - 1 \quad \Rightarrow \quad -3 + 1 = -3x + x \quad \Rightarrow \quad -2 = -2x \quad \Rightarrow \quad x = 1$$ However, $$x = 1$$ is not valid in this case because $$x < -3$$ was required. ### Final solutions The solutions for the system of equations are: $$x = 1 \quad \text{or} \quad x = -1$$ ### Answer: $$x = 1 \quad \text{or} \quad x = -1$$ --- Would you like a more detailed breakdown of any of the steps? Or do you have any further questions? Here are some related questions for you: 1. How do absolute value equations apply to geometric problems like distances? 2. What happens if we introduce a third absolute value equation in the system? 3. How can we graph the solutions of absolute value equations? 4. What are the key differences between solving linear and absolute value equations? 5. Can we solve a system involving more than two absolute values? Tip: Always split absolute value equations into cases based on the expressions inside them, considering when the expressions are positive or negative. This approach ensures that all possible solutions are found.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Absolute Value Equations
Piecewise Functions
Formulas
|A| = A if A >= 0, |A| = -A if A < 0
Theorems
Absolute Value Theorem
Piecewise Function Solving
Suitable Grade Level
Grades 9-11