Math Problem Statement

Solve the following system of equations: \(\frac{x - 2}{x + 2} = \frac{y - 7}{y - 5}\) \(\frac{x + 1}{x - 1} = \frac{y - 3}{y - 5}\)

Solution

The system of equations provided in the image is as follows:

\frac{x - 2}{x + 2} = \frac{y - 7}{y - 5} \\ \frac{x + 1}{x - 1} = \frac{y - 3}{y - 5} \end{cases}$$ ### Steps to solve the system of equations: #### 1. Simplify the first equation: $$\frac{x - 2}{x + 2} = \frac{y - 7}{y - 5}$$ Cross multiply: $$(x - 2)(y - 5) = (x + 2)(y - 7)$$ Expanding both sides: $$xy - 5x - 2y + 10 = xy - 7x - 2y + 14$$ Cancel terms and simplify: $$-5x + 10 = -7x + 14$$ Solving for $$x$$: $$2x = 4 \implies x = 2$$ #### 2. Substitute $$x = 2$$ into the second equation: $$\frac{x + 1}{x - 1} = \frac{y - 3}{y - 5}$$ Substitute $$x = 2$$: $$\frac{2 + 1}{2 - 1} = \frac{y - 3}{y - 5} \implies \frac{3}{1} = \frac{y - 3}{y - 5}$$ Cross multiply: $$3(y - 5) = y - 3$$ Expand and simplify: $$3y - 15 = y - 3$$ Solving for $$y$$: $$2y = 12 \implies y = 6$$ Thus, the solution to the system is: $$x = 2, \quad y = 6$$ Would you like a detailed explanation of any specific step or concept? Here are five related questions you might consider: 1. How do you approach solving systems of nonlinear equations? 2. What happens if a system of equations has no solution or infinitely many solutions? 3. Can you graphically interpret the solution to this system of equations? 4. What types of systems of equations require different methods (e.g., substitution, elimination)? 5. How would the solution change if a parameter were introduced into one of the equations? **Tip:** When solving nonlinear systems, always check if simplifying or factoring expressions can reduce the complexity of the equations.

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Math Problem Analysis

Mathematical Concepts

Algebra
Rational Equations
Systems of Equations

Formulas

Cross-multiplication of rational equations: \(\frac{a}{b} = \frac{c}{d} \implies ad = bc\)

Theorems

Solving systems of equations by substitution and cross-multiplication

Suitable Grade Level

Grades 10-12

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