Math Problem Statement

Part 1 Solve the nonlinear system of equations for real solutions. left brace Start 2 By 1 Matrix 1st Row 1st Column y equals x squared minus 6 2nd Row 1st Column y equals x squared minus 6 x EndMatrix

Solution

The given system of nonlinear equations is:

y = x^2 - 6 \\ y = x^2 - 6x \end{cases}$$ ### Step 1: Set the two expressions for $$y$$ equal Since both equations are equal to $$y$$, we can set them equal to each other: $$x^2 - 6 = x^2 - 6x$$ ### Step 2: Simplify the equation Cancel out the $$x^2$$ terms on both sides: $$-6 = -6x$$ ### Step 3: Solve for $$x$$ Solve for $$x$$ by dividing both sides by -6: $$x = 1$$ ### Step 4: Substitute $$x = 1$$ into one of the original equations Substitute $$x = 1$$ into the first equation, $$y = x^2 - 6$$: $$y = 1^2 - 6 = 1 - 6 = -5$$ ### Final Solution: The solution to the system is: $$x = 1, \quad y = -5$$ Thus, the only real solution is $$(1, -5)$$. --- Would you like to see a more detailed explanation or have additional questions on this topic? Here are some related questions: 1. How can you verify this solution by substituting it into both original equations? 2. What happens if we solve the system using the second equation first? 3. How would this solution change if we had an additional constraint? 4. What is the general approach for solving nonlinear systems of equations? 5. Can there be more than one solution to a system like this, and why? **Tip:** Always check the consistency of your solution by plugging it back into the original equations!

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Math Problem Analysis

Mathematical Concepts

Algebra
Nonlinear Systems
Quadratic Equations

Formulas

Equality of two functions y = f(x)
Linear equation: ax + b = 0

Theorems

Basic principles of solving systems of equations
Properties of quadratic functions

Suitable Grade Level

Grades 9-11

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