Parametric Linear Equations: Difficult Problems with Solutions

Solution:
An equation of the type mx=b may have no solutions only if m=0. In our case, this is a-2=0, or a=2. By substituting a=2 into the equation, we get 0x=5, which has no solutions.

Solution:
In order for the equation to not have any solutions, the coefficient standing before x must be zero. This means either a=5 or a=1. By plugging in a=5, we get [tex]0x=5^2-2\cdot5+1=25-10+1[/tex], or 0x=16, which has no solutions. By plugging in a=1, we get 0x=1-2+1, or 0x=0, which has any x for solution. Therefore the only value of a for which the equation has no solutions is a=5.

Solution:
We substitute with a=1: [tex](1-2)x=1^2-4[/tex], or -x=-3, which yields the solution x=3.

Solution:
[tex]a(a^2+a+1)x=186(a^2+a+1)[/tex]
[tex]ax=186[/tex]. Now we substitute
[tex]31x=186[/tex]
[tex]x=6[/tex]