Question on: JAMB Mathematics - 1997
Solve the simultaneous equations \(\frac{2}{x} - {\frac{2}{x}}\) = 2, \(\frac{4}{x} + {\frac{3}{y}}\) = 10
A
x = \(\frac{3}{2}\), y = \(\frac{3}{2}\)
B
x = \(\frac{1}{2}\), y = \(\frac{3}{2}\)
C
x = \(\frac{-1}{2}\), y = \(\frac{-3}{2}\)
D
x = \(\frac{1}{2}\), y = \(\frac{3}{2}\)
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Correct Option: B
\(\frac{2}{x} - {\frac{2}{x}}\) = 2.....(1)
\(\frac{4}{x} + {\frac{3}{y}}\) = 10
\(\frac{6}{x}\) = 12 \(\to\) x = \(\frac{6}{12}\)
x = \(\frac{1}{2}\)
put x = \(\frac{1}{2}\) in equation (i)
= 4 - \(\frac{3}{y}\) = 2
= 4 - 2
= \(\frac{3}{y}\)
therefore y = \(\frac{3}{2}\)
\(\frac{4}{x} + {\frac{3}{y}}\) = 10
\(\frac{6}{x}\) = 12 \(\to\) x = \(\frac{6}{12}\)
x = \(\frac{1}{2}\)
put x = \(\frac{1}{2}\) in equation (i)
= 4 - \(\frac{3}{y}\) = 2
= 4 - 2
= \(\frac{3}{y}\)
therefore y = \(\frac{3}{2}\)
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