Questions: Solve the system. 4x^2 + 4y^2 = 36 2x + 3y = 6 (3,0) and (-[?]/[], [?]/[?])

Solve the system. 4x^2 + 4y^2 = 36 2x + 3y = 6 (3,0) and (-[?]/[], [?]/[?])
Transcript text: Solve the system. \[ \left\{\begin{array}{c} 4 x^{2}+4 y^{2}=36 \\ 2 x+3 y=6 \end{array}\right. \] $(3,0)$ and $\left(-\frac{[?]}{[]}, \frac{[]}{[]]}\right)$
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Solution

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Solution Steps

To solve the system of equations, we can follow these steps:

  1. Solve the linear equation for one variable in terms of the other.
  2. Substitute this expression into the quadratic equation.
  3. Solve the resulting quadratic equation for the remaining variable.
  4. Substitute back to find the corresponding value of the other variable.
Step 1: Solve the Linear Equation

We start with the linear equation \( 2x + 3y = 6 \). Solving for \( y \), we get: \[ y = 2 - \frac{2}{3}x \]

Step 2: Substitute into the Quadratic Equation

Next, we substitute \( y \) into the quadratic equation \( 4x^2 + 4y^2 = 36 \): \[ 4x^2 + 4\left(2 - \frac{2}{3}x\right)^2 = 36 \]

Step 3: Solve the Resulting Quadratic Equation

Expanding and simplifying the equation leads to: \[ 4x^2 + 4\left(4 - \frac{8}{3}x + \frac{4}{9}x^2\right) = 36 \] This simplifies to: \[ \frac{40}{9}x^2 - \frac{32}{3}x + 4 = 0 \] Solving this quadratic equation yields the solutions: \[ x = -\frac{15}{13} \quad \text{and} \quad x = 3 \]

Step 4: Find Corresponding \( y \) Values

Substituting \( x = 3 \) back into the equation for \( y \): \[ y = 2 - \frac{2}{3}(3) = 0 \] For \( x = -\frac{15}{13} \): \[ y = 2 - \frac{2}{3}\left(-\frac{15}{13}\right) = 2 + \frac{10}{13} = \frac{36}{13} \]

Final Answer

The solutions to the system of equations are: \[ (3, 0) \quad \text{and} \quad \left(-\frac{15}{13}, \frac{36}{13}\right) \] Thus, the final answer is: \[ \boxed{(3, 0) \text{ and } \left(-\frac{15}{13}, \frac{36}{13}\right)} \]

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