Question

The system of linear equations $x+y+z=6,x+2y+3z=10$ and $x+2y+az=b$ has no solutions when _____

• A. $a=2,b\ne 3$
• B. $a=3,b\ne 10$
• C. $b=2,a=3$
• D. $b=3,a\ne 10$

Answer 1

Answer: (B)

$a=3,b\ne 10$

A System of linear equations has no solution when two of these equations are parallel to each other but not coincident. Here, since the first two lines are already given, we know that they are neither parallel nor coincident. However, the third line can be parallel to the second line by choosing an appropriate value of $a$.

Thus, in order for $x+2y+3z=10$and $x+2y+az=b$ to be parallel and not coincident, $a$ has to be equal to $3$ and $b$ not equal to $10$.