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Question

For what value of k, the system of equations
kx+2y=5, 3x-4y=10
has (i) a unique solution, (ii) no solution?

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Solution

The given system of equations is:
kx + 2y = 5
⇒ kx + 2y − 5= 0 ...(i)
3x − 4y = 10
⇒ 3x − 4y − 10 = 0 ...(ii)
These equations are of the forms:
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
where, a1 = k, b1= 2, c1 = −5 and a2 = 3, b2 = −4, c2 = −10
(i) For a unique solution, we must have:
a1a2b1b2, i.e., k32-4k-32
Thus for all real values of k other than -32, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
a1a2=b1b2c1c2
k3=2-4-5-10
k3=2-4 and k312
k=-32,k32
Hence, the required value of k is -32.

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