Question
5. Use the elimination method to solve the linear-quadratic system y=x^2+x-42 & y=-x+6
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Answered by
GPT-4o mini
To solve the linear-quadratic system of equations using the elimination method, we'll start with the two equations you provided:
1. \( y = x^2 + x - 42 \) (Quadratic equation)
2. \( y = -x + 6 \) (Linear equation)
Since both equations are equal to \( y \), we can set them equal to each other:
\[
x^2 + x - 42 = -x + 6
\]
Now, we will rearrange this equation to form a standard quadratic equation:
1. Add \( x \) and subtract \( 6 \) from both sides:
\[
x^2 + x + x - 42 - 6 = 0
\]
This simplifies to:
\[
x^2 + 2x - 48 = 0
\]
Next, we can factor the quadratic equation \( x^2 + 2x - 48 \):
To factor, we need two numbers that multiply to \(-48\) and add up to \(2\). The numbers \(8\) and \(-6\) work because:
\[
8 \cdot -6 = -48 \quad \text{and} \quad 8 + (-6) = 2
\]
Thus, we can write the factored form:
\[
(x + 8)(x - 6) = 0
\]
Setting each factor to zero gives us the possible values for \( x \):
1. \( x + 8 = 0 \) → \( x = -8 \)
2. \( x - 6 = 0 \) → \( x = 6 \)
Now, we have two values for \( x \). Next, we will substitute each of these values back into the linear equation \( y = -x + 6 \) to find the corresponding \( y \) values.
**For \( x = -8 \):**
\[
y = -(-8) + 6 = 8 + 6 = 14
\]
So, one solution is \( (-8, 14) \).
**For \( x = 6 \):**
\[
y = -(6) + 6 = -6 + 6 = 0
\]
So, another solution is \( (6, 0) \).
### Final Solutions
The solutions to the linear-quadratic system are:
\[
(-8, 14) \quad \text{and} \quad (6, 0)
\]
1. \( y = x^2 + x - 42 \) (Quadratic equation)
2. \( y = -x + 6 \) (Linear equation)
Since both equations are equal to \( y \), we can set them equal to each other:
\[
x^2 + x - 42 = -x + 6
\]
Now, we will rearrange this equation to form a standard quadratic equation:
1. Add \( x \) and subtract \( 6 \) from both sides:
\[
x^2 + x + x - 42 - 6 = 0
\]
This simplifies to:
\[
x^2 + 2x - 48 = 0
\]
Next, we can factor the quadratic equation \( x^2 + 2x - 48 \):
To factor, we need two numbers that multiply to \(-48\) and add up to \(2\). The numbers \(8\) and \(-6\) work because:
\[
8 \cdot -6 = -48 \quad \text{and} \quad 8 + (-6) = 2
\]
Thus, we can write the factored form:
\[
(x + 8)(x - 6) = 0
\]
Setting each factor to zero gives us the possible values for \( x \):
1. \( x + 8 = 0 \) → \( x = -8 \)
2. \( x - 6 = 0 \) → \( x = 6 \)
Now, we have two values for \( x \). Next, we will substitute each of these values back into the linear equation \( y = -x + 6 \) to find the corresponding \( y \) values.
**For \( x = -8 \):**
\[
y = -(-8) + 6 = 8 + 6 = 14
\]
So, one solution is \( (-8, 14) \).
**For \( x = 6 \):**
\[
y = -(6) + 6 = -6 + 6 = 0
\]
So, another solution is \( (6, 0) \).
### Final Solutions
The solutions to the linear-quadratic system are:
\[
(-8, 14) \quad \text{and} \quad (6, 0)
\]
Answered by
GPT-4o mini
Certainly! Here's a shorter version:
We have the system:
1. \( y = x^2 + x - 42 \)
2. \( y = -x + 6 \)
Set them equal:
\[
x^2 + x - 42 = -x + 6
\]
Rearranging gives:
\[
x^2 + 2x - 48 = 0
\]
Factoring:
\[
(x + 8)(x - 6) = 0
\]
Thus, the solutions for \( x \) are:
- \( x = -8 \)
- \( x = 6 \)
Now, substitute back to find \( y \):
1. For \( x = -8 \):
\[ y = -(-8) + 6 = 14 \]
→ Solution: \( (-8, 14) \)
2. For \( x = 6 \):
\[ y = -(6) + 6 = 0 \]
→ Solution: \( (6, 0) \)
### Final Solutions:
\[
(-8, 14) \quad \text{and} \quad (6, 0)
\]
We have the system:
1. \( y = x^2 + x - 42 \)
2. \( y = -x + 6 \)
Set them equal:
\[
x^2 + x - 42 = -x + 6
\]
Rearranging gives:
\[
x^2 + 2x - 48 = 0
\]
Factoring:
\[
(x + 8)(x - 6) = 0
\]
Thus, the solutions for \( x \) are:
- \( x = -8 \)
- \( x = 6 \)
Now, substitute back to find \( y \):
1. For \( x = -8 \):
\[ y = -(-8) + 6 = 14 \]
→ Solution: \( (-8, 14) \)
2. For \( x = 6 \):
\[ y = -(6) + 6 = 0 \]
→ Solution: \( (6, 0) \)
### Final Solutions:
\[
(-8, 14) \quad \text{and} \quad (6, 0)
\]
Answered by
GPT-4o mini
Sure! Here’s an even shorter summary:
1. Set equations equal:
\[
x^2 + x - 42 = -x + 6
\]
becomes:
\[
x^2 + 2x - 48 = 0
\]
2. Factor:
\[
(x + 8)(x - 6) = 0
\]
Solutions: \( x = -8 \) and \( x = 6 \)
3. Find \( y \):
- For \( x = -8 \): \( y = 14 \) → \( (-8, 14) \)
- For \( x = 6 \): \( y = 0 \) → \( (6, 0) \)
### Solutions:
\[
(-8, 14) \quad \text{and} \quad (6, 0)
\]
1. Set equations equal:
\[
x^2 + x - 42 = -x + 6
\]
becomes:
\[
x^2 + 2x - 48 = 0
\]
2. Factor:
\[
(x + 8)(x - 6) = 0
\]
Solutions: \( x = -8 \) and \( x = 6 \)
3. Find \( y \):
- For \( x = -8 \): \( y = 14 \) → \( (-8, 14) \)
- For \( x = 6 \): \( y = 0 \) → \( (6, 0) \)
### Solutions:
\[
(-8, 14) \quad \text{and} \quad (6, 0)
\]
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