Question

Determine the values of 'a' for which the system has no solutions, exactly one solution, or infinitely many solutions:
x + 2y = 1
2x + (a^2 - 5)y = a-1

Answer 1

solve \(a^2-5=4\) should do it

Answer 2

how did you arrive to the = 4?

Answer 3

by inspection i can tell that there is a solution when a not equal plus/minus three, and there is infinitely many solution when a=+3 and no solution when a=-2
@satellite73

Answer 4

i arrived at the 4 because the first one has \(x\) and the second one has \(2x\)
the first one has \(2y\) so if the second one has \(4y\) then they will be parallel

Answer 5

if \(a=3\) you get
\[x+2y=1\\
2x+4y=2\] and that is the same line, so infinitely many solutions

Answer 6

Thanks. I understand what you did there.

Answer 7

if \(x=-3\) you get
\[x+2y=1\\
2x+4y=-4\] and that has no solution

Answer 8

yw