Math Problem Statement

Solve by elimination and substitution.

Solution

The system of equations given in the image is:

9x^2 - 4y^2 + 54x + 45 = 0 \quad \text{(Equation 1)} \\ (x + 3)^2 = 4y + 4 \quad \text{(Equation 2)} \end{cases}$$ ### Step 1: Simplify Equation 2 First, let's simplify Equation 2: $$(x + 3)^2 = 4y + 4$$ Expanding the left-hand side: $$x^2 + 6x + 9 = 4y + 4$$ Now, subtract 4 from both sides: $$x^2 + 6x + 5 = 4y \quad \text{(Equation 3)}$$ ### Step 2: Substitute Equation 3 into Equation 1 We can express $$4y$$ from Equation 3 as: $$4y = x^2 + 6x + 5$$ Substitute this into Equation 1: $$9x^2 - (x^2 + 6x + 5) + 54x + 45 = 0$$ Simplify this equation: $$9x^2 - x^2 - 6x - 5 + 54x + 45 = 0$$ $$8x^2 + 48x + 40 = 0$$ Now, divide the entire equation by 8 to simplify: $$x^2 + 6x + 5 = 0$$ ### Step 3: Solve the Quadratic Equation for $$x$$ The equation $$x^2 + 6x + 5 = 0$$ is a quadratic equation. We can solve it using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a = 1$$, $$b = 6$$, and $$c = 5$$: $$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(5)}}{2(1)}$$ $$x = \frac{-6 \pm \sqrt{36 - 20}}{2}$$ $$x = \frac{-6 \pm \sqrt{16}}{2}$$ $$x = \frac{-6 \pm 4}{2}$$ Thus, the solutions for $$x$$ are: $$x = \frac{-6 + 4}{2} = -1 \quad \text{and} \quad x = \frac{-6 - 4}{2} = -5$$ ### Step 4: Find Corresponding $$y$$ Values Now, substitute these $$x$$ values back into Equation 3 to find the corresponding $$y$$ values. For $$x = -1$$: $$4y = (-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0$$ $$y = 0$$ For $$x = -5$$: $$4y = (-5)^2 + 6(-5) + 5 = 25 - 30 + 5 = 0$$ $$y = 0$$ ### Final Answer The solutions to the system of equations are: $$(x, y) = (-1, 0) \quad \text{and} \quad (x, y) = (-5, 0)$$ Do you want more details or have any questions? Here are 5 related questions: 1. How do you verify the solution of a system of nonlinear equations? 2. What methods can be used to solve a system of quadratic equations? 3. How does the quadratic formula help in solving quadratic equations? 4. What are the conditions for a system of equations to have no solution, one solution, or infinitely many solutions? 5. How can graphical methods be used to solve a system of equations? **Tip:** Always check your solutions by substituting them back into the original equations to ensure they satisfy both equations.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Substitution Method
Elimination Method

Formulas

Quadratic formula

Theorems

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Suitable Grade Level

Grades 9-12