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
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
Answers
GPT-4o mini
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!
---
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!