Math Problem Statement
Solve the following system of equations. Verify your answers algebraically and graphically.
Solution
The system of equations provided is:
1. & \quad x + y = 4 \\ 2. & \quad x^2 - 2xy - 3 = 0 \end{aligned}$$ ### Step 1: Solve the first equation for one variable We can solve the first equation for $$y$$: $$y = 4 - x$$ ### Step 2: Substitute into the second equation Substitute $$y = 4 - x$$ into the second equation: $$x^2 - 2x(4 - x) - 3 = 0$$ Simplify the equation: $$x^2 - 8x + 2x^2 - 3 = 0$$ Combine like terms: $$3x^2 - 8x - 3 = 0$$ ### Step 3: Solve the quadratic equation To solve $$3x^2 - 8x - 3 = 0$$, we can use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Where $$a = 3$$, $$b = -8$$, and $$c = -3$$: $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-3)}}{2(3)}$$ $$x = \frac{8 \pm \sqrt{64 + 36}}{6}$$ $$x = \frac{8 \pm \sqrt{100}}{6}$$ $$x = \frac{8 \pm 10}{6}$$ This gives us two solutions for $$x$$: 1. $$x = \frac{18}{6} = 3$$ 2. $$x = \frac{-2}{6} = -\frac{1}{3}$$ ### Step 4: Find corresponding $$y$$ values Using $$y = 4 - x$$, calculate the corresponding $$y$$ values: - For $$x = 3$$: $$y = 4 - 3 = 1$$ - For $$x = -\frac{1}{3}$$: $$y = 4 - \left(-\frac{1}{3}\right) = 4 + \frac{1}{3} = \frac{13}{3}$$ ### Solutions: The solutions are: - $$(3, 1)$$ - $$\left(-\frac{1}{3}, \frac{13}{3}\right)$$ ### Step 5: Verification #### Algebraic Verification Substitute the solutions back into the second equation to verify: 1. For $$(3, 1)$$: $$3^2 - 2(3)(1) - 3 = 9 - 6 - 3 = 0$$ Verified. 2. For $$\left(-\frac{1}{3}, \frac{13}{3}\right)$$: $$\left(-\frac{1}{3}\right)^2 - 2\left(-\frac{1}{3}\right)\left(\frac{13}{3}\right) - 3 = \frac{1}{9} + \frac{26}{9} - 3 = \frac{27}{9} - 3 = 0$$ Verified. #### Graphical Verification Graphing these equations will show the points of intersection corresponding to the solutions $$(3, 1)$$ and $$\left(-\frac{1}{3}, \frac{13}{3}\right)$$. Would you like a graphical representation of the solution? Let me know if you need further details or explanations. ### Related Questions: 1. How do you derive the quadratic formula? 2. What are the different methods to solve a system of equations? 3. How can you verify solutions graphically? 4. What is the significance of the discriminant in a quadratic equation? 5. Can you explain the geometric interpretation of these solutions? 6. How does substitution help in solving systems of equations? 7. What happens if the discriminant is zero or negative in a quadratic equation? 8. How do you plot equations like these on a graph? ### Tip: Always check your solutions by substituting them back into the original equations. This helps confirm the correctness of your results.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
System of Equations
Quadratic Equations
Algebra
Formulas
Quadratic Formula
Theorems
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Suitable Grade Level
Grades 9-12