I got a = 3 and b = 11.
For the first one, substitute 2 in for x:
216 = 16a + 8b + 60 + 18 + 2
216 = 80 + 16a + 8b
27 = 10 + 2a + b (1)
And the next one, -1 for x:
0 = a - b + 8 (2)
So through simultaneous equation,
Eq (1) + Eq (2) gives 27=18 + 3a
a=3 and b=11
Would appreciate it if anyone could help me with these sums; I tried doing them and came up with different answers from the ones in the book. So I'm not sure now if I did them wrong, or there are mistakes in the book's answers:
(y2 - 6)/y < y
|x-3|/(3-x) = x
x2 > a2
|x - 3| +...
Greetings from the equatorial, if not equally sunny, land of Malaysia :wave:
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