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Question

Solve the following pair of equations:
5x1+1y2=2
6x13y2=1
(where x1,y2)

A
x=4, y=5
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B
x=5, y=4
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C
x=13,y=13
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D
None of the above
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Solution

The correct option is A x=4, y=5
If we substitute 1x1 as p and 1y2 as q in the given equations. (As x1,y2)
We get the equations as
5p+q=2
6q3q=1
Now we can solve the pair of equation by method of elimination.
15p+3q=6
6p3q=1
On adding both the above equations, we get
p=13
Now by substituting the value in one of the equation we find q=13
As we have assumed p=1x1
Therefore, p=1x1=13
x=4
SImilarly we assumed q=1y2
Hence 1y2=13
y2=3
Thus x=4 and y=5

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