There are infinitely many solutions. These are coordinates of all points on the line given by the equation 2 - y = 2x - 1 or 2x + y = 3.
Y-2x1 would merely be Y-2.
2x1=1 so 1-Y=0 so Y=0, was this really a question?
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If [ y = x + 2 ], then x is not -1 when y = 5.If [ y = x + 2 ],then when x = -1, y = 1,and when y = 5, x = 3.
3x - y = -1 x = -2 3(-2) -y = -1 -6 - y = -1 -y = 5 y = -5
Y / (2+x)If x = -1 and y=2, thenY / (2+x) = 2/(2-1) = 2/1 = 2
f'(x)xy=yx(y-1) f'(x)2=2x1=2x
x = 1 and y = -2
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The solution is: x = 1 and y = -1
x = 1 and y = 2 or as (1, 2)
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