inequalities

First draw the line y + 1 = 2x
Conside the original (0,0) does not satisfy y + 1 <= 2x so the inequality is satisfied on the right side of the line.
Draw the line y + 4x – 5 = 0
Conside the original (0,0) does not satisfy y + 4x – 5 >=0 so the inequality is satisfied on the right side of the line.
Draw the circle (x – 1)^2 + (y + 1)^2 = 4, centre (1,1), radius = 2
Consider the origin (0,0) does satisfy the (x – 1)^2 + (y + 1)^2 <= 4, so the inequality is satisfied inside the circle.
The yellow portion satisfies the combined conditions.
y + 1 = 2x => y = 2x – 1
Sub (y = 2x – 1) into y + 4x – 5 = 0 yields 2x – 1 + 4x – 5 = 0 =>x = 1
x = 1 => y = 1 : vertex 1 is (1,1)
Sub (y = 2x – 1) into (x – 1)^2 + (y + 1)^2 = 4 yields
(x – 1)^2 + (2x)^2 = 4
x^2 – 2x + 1 + 4x^2 – 4 = 0
5x^2 – 2x – 3 = 0
(x – 1)(5x + 3) = 0
x = 1 or x = -0.6 (rejected)
x = 1 => y = 1
Vertex is (1, 1) same as vertex 1
y + 4x – 5 = 0 => y = 5 – 4x
Sub into (x – 1)^2 + (y + 1)^2 = 4 yields
(x – 1)^2 + (6 – 4x)^2 = 4
x^2 – 2x + 1 + 36 – 48x + 16x^2 – 4 = 0
17x^2 – 50x + 33 = 0
(x – 1)(17x – 33) = 0
x = 33/17 or x = 1
x = 1 => y = 1 same as vertex 1
x = 33/17 => y = 5 – 132/17 = -47/17
vertex 2 is (33/17, - 47/17) or (1.94, -2.76)

http://img66.imageshack.us/img66/3507/14oct2.png
圖片參考：http://img66.imageshack.us/img66/3507/14oct2.png

這裏可以幫到你
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cross point of y+1=2x and y+4x-5=0 is (1,1)

cross point of (x-1)^2+(y+1)^2=4 and y+1=2x (-3/5,-11/5) or (1,1)

cross point of (x-1)^2+(y+1)^2=4 and y+4x-5=0 are (1,1) and (33/17,-47/17)

so (1,1) is one and only one can satisfy the three condition .