To solve a linear-quadratic system using the elimination method, let's first outline what a typical linear-quadratic system looks like. It consists of one linear equation and one quadratic equation.
Here's a general example of such a system:
- Linear equation: \( y = mx + b \)
- Quadratic equation: \( y = ax^2 + bx + c \)
For this example, let's consider the following specific equations:
- \( y = 2x + 3 \) (linear equation)
- \( y = x^2 - 4 \) (quadratic equation)
Step 1: Set the equations equal to each other
Since both equations are equal to \( y \), we can set them equal to each other:
\[ 2x + 3 = x^2 - 4 \]
Step 2: Rearrange the equation
Rearranging gives us:
\[ 0 = x^2 - 2x - 4 - 3 \]
This simplifies to:
\[ x^2 - 2x - 7 = 0 \]
Step 3: Solve the quadratic equation
We can solve for \( x \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 1, b = -2, c = -7 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \] \[ x = \frac{2 \pm \sqrt{4 + 28}}{2} \] \[ x = \frac{2 \pm \sqrt{32}}{2} \] \[ x = \frac{2 \pm 4\sqrt{2}}{2} \] \[ x = 1 \pm 2\sqrt{2} \]
So we have two values for \( x \):
- \( x = 1 + 2\sqrt{2} \)
- \( x = 1 - 2\sqrt{2} \)
Step 4: Find the corresponding \( y \) values
Now we substitute these \( x \) values back into the linear equation to find the corresponding \( y \) values.
For \( x = 1 + 2\sqrt{2} \):
\[ y = 2(1 + 2\sqrt{2}) + 3 \] \[ y = 2 + 4\sqrt{2} + 3 \] \[ y = 5 + 4\sqrt{2} \]
For \( x = 1 - 2\sqrt{2} \):
\[ y = 2(1 - 2\sqrt{2}) + 3 \] \[ y = 2 - 4\sqrt{2} + 3 \] \[ y = 5 - 4\sqrt{2} \]
Step 5: Write the solution
The solutions to the linear-quadratic system are:
- \( \left(1 + 2\sqrt{2}, 5 + 4\sqrt{2}\right) \)
- \( \left(1 - 2\sqrt{2}, 5 - 4\sqrt{2}\right) \)
These are the two points at which the linear and quadratic graphs intersect.