Question

. Use the elimination method to solve the linear-quadratic system

Answer 1

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:

1. Linear equation: \( y = mx + b \)
2. Quadratic equation: \( y = ax^2 + bx + c \)

For this example, let's consider the following specific equations:

1. \( y = 2x + 3 \) (linear equation)
2. \( y = x^2 - 4 \) (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 \( x \):

1. \( x = 1 + 2\sqrt{2} \)
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:

1. \( \left(1 + 2\sqrt{2}, 5 + 4\sqrt{2}\right) \)
2. \( \left(1 - 2\sqrt{2}, 5 - 4\sqrt{2}\right) \)

These are the two points at which the linear and quadratic graphs intersect.