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