JAMB - Mathematics (1988 - No. 32)
Given that 3x - 5y - 3 = 0, 2y - 6x + 5 = 0 the value of (x, y) is
(\(\frac{-1}{8}, \frac{19}{24}\))
8, \(\frac{24}{19}\)
-8, \(\frac{24}{19}\)
(\(\frac{19}{24}, \frac{-1}{8}\))
Explanation
3x - 5y = 3, 2y - 6x = -5
-5y + 3x = 3........{i} x 2
2y - 6x = -5.........{ii} x 5
Substituting for x in equation (i)
-5y + 3(\(\frac{19}{24}\)) = 3
-5y + 3 x \(\frac{19}{24}\) = 3
-5y = \(\frac{3 - 19}{8}\)
-5 = \(\frac{24 - 19}{8}\)
= \(\frac{5}{8}\)
y = \(\frac{5}{8 \times 5}\)
y = \(\frac{-1}{8}\)
(x, y) = (\(\frac{19}{24}, \frac{-1}{8}\)
-5y + 3x = 3........{i} x 2
2y - 6x = -5.........{ii} x 5
Substituting for x in equation (i)
-5y + 3(\(\frac{19}{24}\)) = 3
-5y + 3 x \(\frac{19}{24}\) = 3
-5y = \(\frac{3 - 19}{8}\)
-5 = \(\frac{24 - 19}{8}\)
= \(\frac{5}{8}\)
y = \(\frac{5}{8 \times 5}\)
y = \(\frac{-1}{8}\)
(x, y) = (\(\frac{19}{24}, \frac{-1}{8}\)
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