f. 4 Maths

y = [(k+ 5)(x)/3] +k
y = [(k+ 5)/3]x +k

hx+ 6y + 22 - h =0
6y = -hx + (h - 22)
y = -(h/6)x + [(h - 22)/6]

The two st. lines intersect at infinitely many points, i.e.
the two st. lines are coincide.
(k + 5)/3 = -h/6 ...... [1]
k = (h - 22)/6 ...... [2]

From [1] :
(k+ 5) = -h/2
h = -2k - 10 ...... [3]

Put [3] into [2] :
k = (-2k - 10 - 22)/6
6k = -2k - 32
8k = -32
k = -4

Put k = -4 into [3] :
h = -2(-4) - 10
h = -2

2014-10-12 22:57:05 補充：
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