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1992 WAEC Mathematics Theory (a) Solve the following pair of simultaneous equations: \(2x + 5y = 6\frac{1}{2} ; 5x...

Mathematics
WAEC 1992

(a) Solve the following pair of simultaneous equations: \(2x + 5y = 6\frac{1}{2} ; 5x - 2y = 9\)

(b) If \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\), find x.

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Explanation

(a) \(2x + 5y = \frac{13}{2} \implies 4x + 10y = 13 ... (1)\)

\(5x - 2y = 9 ... (2)\)

\((2) \times 5 : 25x - 10y = 45 ... (2a)\)

\((1) + (2a) : 29x = 58 \implies x = 2\)

\(4(2) + 10y = 13 \implies 10y = 13 - 8 = 5\)

\(y = \frac{5}{10} = 0.5\)

(b) \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\)

\(\log_{10} (\frac{2x + 1}{3x - 2}) = 1\)

\(\frac{2x + 1}{3x - 2} = 10^{1} = 10\)

\(2x + 1 = 10(3x - 2) \implies 2x + 1 = 30x - 20\)

\(21 = 28x \implies x = \frac{3}{4}\)


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