If the system of equations x+5y+6z=4, 2x+3y+4z=7, x+6y+az=b has infinitely many solutions, then the point (a, b) lies on the line
- y-x=3
- x-y=3
- x+y=11
- x+y=12
Answer
(2) x-y=3
Solution
For infinitely many solutions, the third equation must be a linear combination of the first two:
x+6y+az-b=p (x+5y+6z-4)+q (2x+3y+4z-7)
Coefficient of x: p+2q=1
Coefficient of y: 5p+3q=6
5(1-2q)+3q=6 q=- 17, p= 97
Coefficient of z: a=6p+4q= 54-47= 507
Constant: b=4p+7q= 36-77= 297
(Check: = vmatrix1&5&6\\2&3&4\\1&6&a vmatrix=50-7a=0 holds)
a-b= 50-297=3
(a, b) lies on x-y=3
Answer: (2)