Math Problem Statement
Solve the system of equations: \( \frac{x}{2} + \frac{3y - 6}{2} = y \) and \( \frac{2y - 6}{3} - \frac{y - 2}{2} = 2x \), and verify if the point \((0,6)\) is a solution.
Solution
The image shows a system of two equations:
\frac{x}{2} + \frac{3y - 6}{2} = y \\ \frac{2y - 6}{3} - \frac{y - 2}{2} = 2x \end{cases}$$ The goal is typically to solve for $$x$$ and $$y$$ or verify if a given point, such as $$(0,6)$$, is a solution to the system. Let's solve this step-by-step. ### Step 1: Simplify the First Equation The first equation is: $$\frac{x}{2} + \frac{3y - 6}{2} = y$$ Combine the terms on the left side: $$\frac{x + 3y - 6}{2} = y$$ Multiply both sides by 2 to clear the fraction: $$x + 3y - 6 = 2y$$ Move terms involving $$y$$ to one side: $$x + y = 6$$ Thus, the first equation simplifies to: $$x + y = 6$$ ### Step 2: Simplify the Second Equation The second equation is: $$\frac{2y - 6}{3} - \frac{y - 2}{2} = 2x$$ To eliminate the fractions, find the least common multiple of 3 and 2, which is 6. Multiply every term by 6: $$6 \cdot \frac{2y - 6}{3} - 6 \cdot \frac{y - 2}{2} = 6 \cdot 2x$$ Simplify each term: $$2 \cdot (2y - 6) - 3 \cdot (y - 2) = 12x$$ Expanding each term: $$4y - 12 - 3y + 6 = 12x$$ Combine like terms: $$y - 6 = 12x$$ Add 6 to both sides: $$y = 12x + 6$$ ### Step 3: Substitute and Solve We now have two equations: 1. $$x + y = 6$$ 2. $$y = 12x + 6$$ Substitute $$y = 12x + 6$$ into the first equation: $$x + (12x + 6) = 6$$ Combine like terms: $$13x + 6 = 6$$ Subtract 6 from both sides: $$13x = 0$$ Divide by 13: $$x = 0$$ ### Step 4: Find $$y$$ Substitute $$x = 0$$ into $$y = 12x + 6$$: $$y = 12(0) + 6 = 6$$ ### Solution The solution to the system is: $$(x, y) = (0, 6)$$ This matches the point $$(0,6)$$ given in the image, confirming that it is indeed a solution to the system. Would you like further details or have any other questions? ### Additional Questions 1. How would the solution change if the constant term in the first equation were different? 2. Can we solve the system using matrix methods? If so, what would the matrix look like? 3. How do we verify if the solution is unique in this type of linear system? 4. How would the solution method differ if the equations were nonlinear? 5. How can substitution and elimination methods be combined in solving systems of equations? **Tip:** Always simplify each equation before attempting to solve the system, as it can make the solution process clearer and quicker.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Systems of Linear Equations
Substitution Method
Elimination Method
Formulas
Combining like terms
Solving linear equations by substitution or elimination
Theorems
Linear equation properties
Suitable Grade Level
Grades 10-12