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.

Question

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.
make this like a 6th grader wrote it

Answer 1

Sure! Here’s a simpler way to explain how to solve a linear-quadratic system of equations, like a 6th grader might do it.

What is a Linear-Quadratic System?

A linear-quadratic system has one straight line equation and one curved (quadratic) equation. For example, let’s look at these two equations:

1. Linear equation: \( y = 2x + 3 \) (this is a straight line)
2. Quadratic equation: \( y = x^2 - 4 \) (this makes a curve, like a U-shape)

Step 1: Set the Equations Equal

Both equations equal \( y \), so we can set them equal to each other:

\[ 2x + 3 = x^2 - 4 \]

Step 2: Rearrange the Equation

Now, let’s rearrange the equation to get everything on one side. We want \( 0 \) on the other side:

\[ 0 = x^2 - 2x - 4 - 3 \]

This then simplifies to:

\[ x^2 - 2x - 7 = 0 \]

Step 3: Solve the Quadratic Equation

Next, we can use the quadratic formula to find out what \( x \) values work. The quadratic formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case:

• \( a = 1 \) (from \( x^2 \))
• \( b = -2 \) (from \(-2x\))
• \( c = -7 \)

Let's plug in these numbers:

\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \] \[ x = \frac{2 \pm \sqrt{4 + 28}}{2} \] \[ x = \frac{2 \pm \sqrt{32}}{2} \] \[ x = \frac{2 \pm 4\sqrt{2}}{2} \] \[ x = 1 \pm 2\sqrt{2} \]

So we get two \( x \) values:

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

Step 4: Find the \( y \) Values

Now, let’s plug these \( x \) values back into the linear equation \( y = 2x + 3 \) to find \( y \):

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 Final Answers

So, we found two pairs of points where the line and the curve meet:

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

These points are where the straight line and the U-shaped curve cross each other. And that's how you solve a linear-quadratic system!