Find the value of x and y in the simultaneous equation: 3x + y = 21; xy = 30
- x = 3 or 7, y = 12 or 8
- x = 6 or 1, y = 11 or 5
-
x = 2 or 5, y = 15 or 6
- x = 1 or 5, y = 10 or 7
The correct answer is: C
Explanation
3x + y = 21 ... (i);
xy = 30 ... (ii)
From (ii), \(y = \frac{30}{x}\). Putting the value of y in (i), we have
3x + \(\frac{30}{x}\) = 21
\(\implies\) 3x\(^2\) + 30 = 21x
3x\(^2\) - 21x + 30 = 0
3x\(^2\) - 15x - 6x + 30 = 0
3x(x - 5) - 6(x - 5) = 0
(3x - 6)(x - 5) = 0
3x - 6 = 0 \(\implies\) x = 2.
x - 5 = 0 \(\implies\) x = 5.
If x = 2, y = \(\frac{30}{2}\) = 15;
If x = 5, y = \(\frac{30}{5}\) = 6.
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