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Find the value of p and q for which the system of equations represent coincident lines 2x +3y = 7, (p+q+1)x +(p+2q+2)y = 4(p+q)+1
Ans: a1 = 2, b1 = 3, c1 = 7
a2 = p + q + 1 , b2 = p + 2q + 2 , c2 = (p + q )+ 1
For the following system of equation the condition must be
a1/a2 = b1/b2 =c1/c2
=> 2/p+q+1 = 3/ p + q +2 = 7 /4(p+q)+1
=> 2/p+q+1 = 7 /4(p+q)+1
7p +14q + 14 = 12p + 12q + 3
= 5p - 2q - 11 = 0 ----------------(2)
p + q + - 5 = 0
5p - 2q - 11 = 0
From (1) and (2)
5p + 5q - 25 = 0
Solve it, to get q = 2
Substitute value of q in equation (1)
p + q - 5 = 0
On solving we get, p = 3 and q = 2
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