(a) In the simultaneous equations : \(px + qy = 5 ; qx + py = -10\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.
(b) Solve : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
(b) Solve : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
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Correct Answer: Option n
Explanation:
(a) \(px + qy = 5 \)
When x = 1 and y = -2, we have
\(p - 2q = 5 .... (1)\)
\(qx + py = -10 \)
When x = 1 and y = -2, we have
\(q - 2p = -10 .... (2)\)
Multiplying (1) by -2, we have
\(-2p + 4q = -10 ... (1a)\)
\(\equiv 4q - 2p = -10\)
(2) - (1a) : \(q - 4q = -10 + 10\)
\(-3q = 0\)
\(q = 0\)
Putting q = 0 in (1a), we have
\(-2p = -10 \implies p = 5\)
Hence, p = 5 and q = 0.
(b) \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\)
\((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\)
\(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\)
\(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\)
\(11r = 11\)
\(r = 1\).
(a) \(px + qy = 5 \)
When x = 1 and y = -2, we have
\(p - 2q = 5 .... (1)\)
\(qx + py = -10 \)
When x = 1 and y = -2, we have
\(q - 2p = -10 .... (2)\)
Multiplying (1) by -2, we have
\(-2p + 4q = -10 ... (1a)\)
\(\equiv 4q - 2p = -10\)
(2) - (1a) : \(q - 4q = -10 + 10\)
\(-3q = 0\)
\(q = 0\)
Putting q = 0 in (1a), we have
\(-2p = -10 \implies p = 5\)
Hence, p = 5 and q = 0.
(b) \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\)
\((4r - 3)(3r + 4) = (2r - 1)(6r + 1)\)
\(12r^{2} + 16r - 9r - 12 = 12r^{2} + 2r - 6r - 1\)
\(12r^{2} - 12r^{2} + 7r + 4r = - 1 + 12\)
\(11r = 11\)
\(r = 1\).