Math Problem Statement

Use the method of substitution to solve the following system of equations. If the system is dependent, express the solution set in terms of one of the variables. Leave all fractional answers in fraction form. -x + 2y = -6 2x - 4y = 12

Solution

The system of equations displayed in your image is:

-x + 2y &= -6 \quad \text{(Equation 1)} \\ 2x - 4y &= 12 \quad \text{(Equation 2)} \end{aligned}$$ ### Step 1: Solve one of the equations for one variable. We can solve Equation 1 for $$x$$. Rewriting Equation 1: $$-x + 2y = -6$$ $$x = 2y + 6$$ ### Step 2: Substitute this into the second equation. Now substitute $$x = 2y + 6$$ into Equation 2: $$2(2y + 6) - 4y = 12$$ $$4y + 12 - 4y = 12$$ $$12 = 12$$ ### Step 3: Analyze the result. The result $$12 = 12$$ is a **true statement** for all values of $$y$$. This means that the system has **infinitely many solutions** and is **dependent**. ### Solution: Since the system is dependent, we express $$x$$ in terms of $$y$$. From Step 1, we already have: $$x = 2y + 6$$ Thus, the solution set is: $$(x, y) = (2y + 6, y) \quad \text{for all real values of } y$$ The answer you need to input for the dependent system should be: $$(2y + 6, y) \quad \text{where } y \in \mathbb{R}$$ Do you want more details or have any questions? ### Related questions: 1. How can we identify a dependent system from substitution? 2. What happens if substitution yields a false statement? 3. Can elimination be used instead of substitution for this system? 4. What is the graphical interpretation of a dependent system? 5. How does the result $$12 = 12$$ lead to infinitely many solutions? **Tip**: In a dependent system, both equations represent the same line, so the solutions will always be infinitely many.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations
Substitution Method

Formulas

-x + 2y = -6 (Rearranging to x = 2y + 6)
Substituting x into 2x - 4y = 12 and solving for y

Theorems

Consistency of Linear Systems
Dependent System Theorem

Suitable Grade Level

Grades 10-12

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