(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.
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) \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\)
\(\log_{10} (\frac{2x + 1}{3x - 2}) = 1\)
\(\frac{2x + 1}{3x - 2} = 10^{1} = 10\)
\(2x + 1 = 10(3x - 2) \implies 2x + 1 = 30x - 20\)
\(21 = 28x \implies x = \frac{3}{4}\)