(a) Solve the following pair of simultaneous equations: \(2x + 5y = 6\frac{1}{2} ; 5x - 2y = 9\) (b) If \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\), fin...
Question 1 Report
(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.
Answer Details
(a) \(2x+5y=\tfrac{13}{2}\) and \(5x-2y=9\). Multiply the first by \(2\) and the second by \(5\): \[4x+10y=13,\qquad25x-10y=45.\] Adding: \(29x=58\Rightarrow x=2\). Then \(2(2)+5y=6\tfrac12\Rightarrow5y=2\tfrac12\Rightarrow y=\tfrac12\). So \(x=2,\;y=\tfrac12\).
(b) \[\log_{10}(2x+1)-\log_{10}(3x-2)=1\Rightarrow\log_{10}\frac{2x+1}{3x-2}=1\Rightarrow\frac{2x+1}{3x-2}=10.\] \[2x+1=30x-20\Rightarrow28x=21\Rightarrow x=\frac34.\] (Check: \(3x-2=\tfrac14>0\), valid.)
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