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Question 1 (iv)
Solve the following pairs of equations by reducing them to a pair of linear equations:
5x1+1y2=2
6x13y2=1

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Solution

5x1+1y2=2
6x13y2=1
Substituting 1x1=p and 1y2=q in the given equations, we obtain,
5p + q = 2 ... (i)
6p - 3q = 1 ... (ii)
Now, by multiplying equation (i) by 3 we get,
15p + 3q = 6 ... (iii)
Now, adding equation (ii) and (iii),
21p = 7
p=13
Substituting this value in equation (ii) we get,
6×133q=1
2-3q = 1
-3q = 1-2
-3q = -1
q=13
Now, p=1x1=13
1x1=13
3=x1
x=4
Also, q=1y2=13
3=y2
y=5
Hence, x = 4 and y = 5 is the solution.

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