Math Problem Statement
Solve the following system of equations. Verify your answers algebraically and graphically.
x + y = 4
x^2 - 2xy - 3 = 0
Solution
To solve this system of equations:
x + y = 4 \\ x^2 - 2xy - 3 = 0 \end{cases}$$ we'll proceed algebraically first, and then sketch the graphs to find the points of intersection. ### Step 1: Substitute $$y = 4 - x$$ into the second equation From the first equation, we have: $$y = 4 - x$$ Substitute $$y = 4 - x$$ into the second equation: $$x^2 - 2x(4 - x) - 3 = 0$$ ### Step 2: Expand and simplify Expanding $$-2x(4 - x)$$: $$x^2 - 8x + 2x^2 - 3 = 0$$ Combine like terms: $$3x^2 - 8x - 3 = 0$$ ### Step 3: Solve the quadratic equation Now we solve $$3x^2 - 8x - 3 = 0$$ using the quadratic formula: $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 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$$ values For each $$x$$-value, substitute back into $$y = 4 - x$$: 1. When $$x = 3$$: $$y = 4 - 3 = 1$$ So, one solution is $$(3, 1)$$. 2. When $$x = -\frac{1}{3}$$: $$y = 4 - \left(-\frac{1}{3}\right) = 4 + \frac{1}{3} = \frac{12 + 1}{3} = \frac{13}{3}$$ So, the other solution is $$\left(-\frac{1}{3}, \frac{13}{3}\right)$$. ### Final Answer The solutions to the system are: $$(3, 1) \quad \text{and} \quad \left(-\frac{1}{3}, \frac{13}{3}\right)$$ ### Verification by Graphing To confirm, you would graph both equations: 1. The line $$x + y = 4$$. 2. The hyperbola $$x^2 - 2xy - 3 = 0$$. The points of intersection on the graph will correspond to the solutions we found: $$(3, 1)$$ and $$\left(-\frac{1}{3}, \frac{13}{3}\right)$$. Would you like a graphical sketch for verification or further explanation on any of the steps? --- ### Related Questions 1. How would you solve a system of equations if both were quadratic? 2. What is the significance of finding points of intersection in a system of equations? 3. Can a linear equation intersect a hyperbola at more than two points? 4. How can we check the solutions of a nonlinear system using a calculator? 5. What other methods can be used to solve nonlinear systems algebraically? ### Tip When dealing with systems involving a linear and nonlinear equation, substituting from the linear equation often simplifies the solution process significantly.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Systems of Equations
Quadratic Equations
Formulas
Substitution method
Quadratic formula
Theorems
Quadratic formula theorem
Suitable Grade Level
Grades 10-12