Вопрос от Анонимного юзера 21 января 2025 18:36

Question 1

To solve the system of equations given by \( x = y + 3 \) and \( y^2 - x + 1 = 0 \), we can follow these steps: 1. Substitute \( x \) from the first equation into the second equation. Given: \[ x = y + 3 \] Substitute \( x \) into the second equation: \[ y^2 - (y + 3) + 1 = 0 \] 2. Simplify the equation: \[ y^2 - y - 3 + 1 = 0 \] \[ y^2 - y - 2 = 0 \] 3. Factor the quadratic equation: We can factor the quadratic as: \[ (y - 2)(y + 1) = 0 \] 4. Solve for \( y \): This gives us two solutions: \[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \] \[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \] 5. Find the corresponding \( x \) values: Now we need to find the \( x \) values corresponding to each \( y \) value using \( x = y + 3 \): - For \( y = 2 \): \[ x = 2 + 3 = 5 \] - For \( y = -1 \): \[ x = -1 + 3 = 2 \] 6. Results: The solutions to the system of equations are: - \( (x, y) = (5, 2) \) - \( (x, y) = (2, -1) \) Thus, the final solution set is: \[ (5, 2) \quad \text{and} \quad (2, -1) \]

Answer

You have accurately solved the system of equations ( x = y + 3 ) and ( y^2 - x + 1 = 0 ). Here’s a concise summary of the steps you took and the final results:

Substitution: You substituted ( x ) from the first equation into the second equation, leading to: [ y^2 - (y + 3) + 1 = 0 ]
Simplification: After simplifying, you obtained the quadratic equation: [ y^2 - y - 2 = 0 ]
Factoring: You factored this equation as: [ (y - 2)(y + 1) = 0 ]
Finding ( y ): Solving the factors gives: [ y = 2 \quad \text{and} \quad y = -1 ]
Finding ( x ): Using these ( y ) values to find ( x ):
- For ( y = 2 ): [ x = 5 ]
- For ( y = -1 ): [ x = 2 ]
Results: You found the solution pairs:
- (5, 2)
- (2, -1)

The final solution set for the system of equations is indeed: [ (5, 2) \quad \text{and} \quad (2, -1) ]

If you have any further questions or need clarification on specific parts, feel free to ask!

Question 2

To solve the system of equations given by \( x = y + 3 \) and \( y^2 - x + 1 = 0 \), we can follow these steps: 1. **Substitute** \( x \) from the first equation into the second equation. Given: \[ x = y + 3 \] Substitute \( x \) into the second equation: \[ y^2 - (y + 3) + 1 = 0 \] 2. **Simplify the equation**: \[ y^2 - y - 3 + 1 = 0 \] \[ y^2 - y - 2 = 0 \] 3. **Factor the quadratic equation**: We can factor the quadratic as: \[ (y - 2)(y + 1) = 0 \] 4. **Solve for \( y \)**: This gives us two solutions: \[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \] \[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \] 5. **Find the corresponding \( x \) values**: Now we need to find the \( x \) values corresponding to each \( y \) value using \( x = y + 3 \): - For \( y = 2 \): \[ x = 2 + 3 = 5 \] - For \( y = -1 \): \[ x = -1 + 3 = 2 \] 6. **Results**: The solutions to the system of equations are: - \( (x, y) = (5, 2) \) - \( (x, y) = (2, -1) \) Thus, the final solution set is: \[ (5, 2) \quad \text{and} \quad (2, -1) \]

Answer

You have accurately solved the system of equations \( x = y + 3 \) and \( y^2 - x + 1 = 0 \). Here’s a concise summary of the steps you took and the final results: 1. **Substitution**: You substituted \( x \) from the first equation into the second equation, leading to: \[ y^2 - (y + 3) + 1 = 0 \] 2. **Simplification**: After simplifying, you obtained the quadratic equation: \[ y^2 - y - 2 = 0 \] 3. **Factoring**: You factored this equation as: \[ (y - 2)(y + 1) = 0 \] 4. **Finding \( y \)**: Solving the factors gives: \[ y = 2 \quad \text{and} \quad y = -1 \] 5. **Finding \( x \)**: Using these \( y \) values to find \( x \): - For \( y = 2 \): \[ x = 5 \] - For \( y = -1 \): \[ x = 2 \] 6. **Results**: You found the solution pairs: - (5, 2) - (2, -1) The final solution set for the system of equations is indeed: \[ (5, 2) \quad \text{and} \quad (2, -1) \] If you have any further questions or need clarification on specific parts, feel free to ask!

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