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Q
x is one more than 3 times ‘a’, y is 7 less than twice ‘a’, z is 3 less than ‘a’....
Solution
x = 3a + 1y = 2a − 7z = a − 3
Given 6z2 + 7 = xy⇒ 6(a − 3)2 + 7 = (3a + 1)(2a −...
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