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Question

Solve 2x+23y=16 and 3x+2y=0 and hence find 'a' for which, y=ax4.

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Solution

As, 2x+23y=16 and 3x+2y=0 are not in the general form of linear equations. To make them into linear equaions,
Take 1x=u and 1y=v. The given system of equations become

2u+23v=16

12u+4v=1 (i)
and, 3u+2v=0 .(ii)

Multiplying equation (ii) by 2 and subtracting from equation (i), we get
6u1u=16

Putting u=16 in (i), we get

2+4v=1v=14

Hence, x=1u=6 and y=1v=4

So, the solution of the given system of equations is x=6,y=4

Putting x=6,y=4 in y=ax4, we get

4=6a4a=0.


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