(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...
Question 1 Report
(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 : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
Answer Details
(a) Substitute \(x=1,\ y=-2\):
\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]
\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]
From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).
Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).
(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):
\[(4r-3)(3r+4)=(2r-1)(6r+1)\]
\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]
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