Решите систему уравнений {x+y=10/{x^2-y^2=40

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вопрос

x+y=10 (1)
x^2-y^2=40 (2)

From equation (1), we can express y in terms of x:
y=10-x

Substitute y in equation (2):
x^2 - (10-x)^2 = 40
x^2 - (100 - 20x + x^2) = 40
x^2 - 100 + 20x - x^2 = 40
20x - 100 = 40
20x = 140
x = 7

Substitute x back into equation (1) to find y:
7 + y = 10
y = 3

Therefore, the solution to the system of equations is:
x = 7
y = 3