Math Problem Statement
Solve the system of equations: y = x^2 + 2x + 5 and y = 5x + 15.
Solution
The system of equations provided in the image is:
y = x^2 + 2x + 5 \\ y = 5x + 15 \end{cases}$$ To solve this system, we can use substitution since both equations are equal to $$y$$. ### Step 1: Set the two equations equal to each other: $$x^2 + 2x + 5 = 5x + 15$$ ### Step 2: Rearrange the equation to set it equal to zero: $$x^2 + 2x + 5 - 5x - 15 = 0$$ $$x^2 - 3x - 10 = 0$$ ### Step 3: Solve the quadratic equation: We can factor the quadratic: $$(x - 5)(x + 2) = 0$$ This gives the solutions: $$x = 5 \quad \text{or} \quad x = -2$$ ### Step 4: Substitute these $$x$$-values into one of the original equations to find $$y$$: - When $$x = 5$$: $$y = 5(5) + 15 = 25 + 15 = 40$$ So, one solution is $$(5, 40)$$. - When $$x = -2$$: $$y = 5(-2) + 15 = -10 + 15 = 5$$ So, the other solution is $$(-2, 5)$$. ### Final Answer: The solutions are $$(5, 40)$$ and $$(-2, 5)$$. Let me know if you would like more details on any step! Here are a few related questions to explore: 1. How do you determine whether a system of equations has one, two, or infinite solutions? 2. What is the geometric interpretation of the solutions to this system? 3. How do you factor a quadratic equation if it can't be easily factored like this one? 4. Can you solve the same system using the method of substitution or elimination? 5. What would the solution be if the quadratic had no real roots? **Tip:** Always double-check the discriminant in a quadratic equation to see if real solutions exist before factoring.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Quadratic Equations
Systems of Equations
Formulas
y = x^2 + 2x + 5
y = 5x + 15
Quadratic equation formula: ax^2 + bx + c = 0
Factoring method for solving quadratic equations
Theorems
Zero Product Property
Quadratic Formula
Suitable Grade Level
Grades 8-10