Question

To solve the given linear-quadratic system using the substitution method, we have the equations:
1. \( y = x^2 - 4x - 41 \) (quadratic equation)
2. \( y = 4 \) (linear equation)

We will substitute the value of \( y \) from the second equation into the first equation.

### Step 1: Substitute \( y = 4 \) into the quadratic equation

Substituting \( y = 4 \) into \( y = x^2 - 4x - 41 \):

\[
4 = x^2 - 4x - 41
\]

### Step 2: Rearrange the equation

Rearranging the equation gives:

\[
x^2 - 4x - 41 - 4 = 0
\]
\[
x^2 - 4x - 45 = 0
\]

### Step 3: Factor or use the quadratic formula

We need to solve the quadratic equation \( x^2 - 4x - 45 = 0 \). We can factor it:

\[
(x - 9)(x + 5) = 0
\]

### Step 4: Set each factor to zero

Setting each factor equal to zero gives us:

\[
x - 9 = 0 \quad \Rightarrow \quad x = 9
\]

\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]

### Step 5: Find corresponding \( y \) values

Now we substitute \( x = 9 \) and \( x = -5 \) back into the linear equation \( y = 4 \):

- For \( x = 9 \):

\[
y = 4
\]

- For \( x = -5 \):

\[
y = 4
\]

### Step 6: Write the solutions

Thus, the solutions to the linear-quadratic system are:

1. \( (9, 4) \)
2. \( (-5, 4) \)

These are the points where the quadratic graph intersects the line \( y = 4 \).

make this like a 6th grader wrote it

Answer 1

Sure! Let's simplify the explanation so a 6th grader can understand it.

---

We're trying to solve two equations:

1. \( y = x^2 - 4x - 41 \) (this one is a bit curvy called a quadratic)
2. \( y = 4 \) (this one is a straight line)

### Step 1: Plug in \( y = 4 \)

We know \( y \) is 4 from the second equation. So, we put 4 in for \( y \) in the first equation:

\[
4 = x^2 - 4x - 41
\]

### Step 2: Move things around

Now, we want to make it a bit easier to solve. We’ll move 4 over to the other side:

\[
x^2 - 4x - 41 - 4 = 0
\]

So this becomes:

\[
x^2 - 4x - 45 = 0
\]

### Step 3: Solve the equation

Next, we need to solve the equation \( x^2 - 4x - 45 = 0 \). We can make this easier by breaking it down into two parts. We want two numbers that multiply to -45 and add up to -4. Those numbers are:

- 9 and -5

So, we can write it like this:

\[
(x - 9)(x + 5) = 0
\]

### Step 4: Find the answers for \( x \)

Now we set each part to zero:

1. \( x - 9 = 0 \) means \( x = 9 \)
2. \( x + 5 = 0 \) means \( x = -5 \)

### Step 5: Get \( y \)

Both times we found \( x \), we can find \( y \) using the second equation \( y = 4 \):

- When \( x = 9 \), \( y = 4 \)
- When \( x = -5 \), \( y = 4 \)

### Step 6: Write the final answer

So the solutions to our equations, which are the points where the curvy line meets the straight line, are:

1. \( (9, 4) \)
2. \( (-5, 4) \)

And that's it! Now we know where the two equations meet!