Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №29316
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Аноним
30 октября 2023 16:18 · публичный вопрос
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Аноним

Great job! Your solution is correct. To confirm, substitute the values of x and y back into the original equations and see if they satisfy both equations.

Нейросеть30 октября 2023 16:18

Thank you! I'm glad my solution is correct. Just to confirm, let's substitute the values of x and y back into the original equations and check if they satisfy both equations.

Original equation 1: 2x - y = 5 Substituting x = 3 and y = 1: 2(3) - 1 = 5 6 - 1 = 5 5 = 5

The left side of the equation matches the right side, so the values of x = 3 and y = 1 satisfy equation 1.

Original equation 2: x + 3y = 9 Substituting x = 3 and y = 1: 3 + 3(1) = 9 3 + 3 = 9 6 = 9

The left side of the equation does not match the right side. Therefore, the values of x = 3 and y = 1 do not satisfy equation 2.

Since these values do not satisfy both equations, they are not the solution to the system of equations. I apologize for the mistake in my previous response.

Great job! Your solution is correct. To confirm, substitute the values of x and y back into the original equations and see if they satisfy both equations.
Thank you! I'm glad my solution is correct. Just to confirm, let's substitute the values of x and y back into the original equations and check if they satisfy both equations. Original equation 1: 2x - y = 5 Substituting x = 3 and y = 1: 2(3) - 1 = 5 6 - 1 = 5 5 = 5 The left side of the equation matches the right side, so the values of x = 3 and y = 1 satisfy equation 1. Original equation 2: x + 3y = 9 Substituting x = 3 and y = 1: 3 + 3(1) = 9 3 + 3 = 9 6 = 9 The left side of the equation does not match the right side. Therefore, the values of x = 3 and y = 1 do not satisfy equation 2. Since these values do not satisfy both equations, they are not the solution to the system of equations. I apologize for the mistake in my previous response.