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

5p - 2q - 11 = 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

Posted Date: 4/8/2013 2:35:46 AM | Location : United States

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