Literal Equations: Difficult Problems with Solutions

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

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

Solution:
The coefficient must be zero: [tex]a^2-9=0[/tex], so [tex]a=3[/tex] or [tex]a=-3[/tex]. Both must be checked in the right-hand side.
For [tex]a=3[/tex]: [tex]9-6-3=0[/tex], so the equation is [tex]0x=0[/tex] — every [tex]x[/tex] is a solution.
For [tex]a=-3[/tex]: [tex]9+6-3=12[/tex], so the equation is [tex]0x=12[/tex] — no solution at all.
Only one of the two roots gives infinitely many solutions.
Answer: [tex]a=3[/tex]

Solution:
As long as the coefficient is different from zero we may divide by it and get exactly one solution. So the equation fails to have exactly one solution precisely when the coefficient vanishes:
[tex]a^2-5a+6=0[/tex]
[tex](a-2)(a-3)=0[/tex]
so [tex]a=2[/tex] or [tex]a=3[/tex].
Notice that the question does not ask which case occurs — both are acceptable. Still, it is instructive to look: for [tex]a=2[/tex] we get [tex]0x=0[/tex] (infinitely many solutions), and for [tex]a=3[/tex] we get [tex]0x=1[/tex] (no solution). Neither is "exactly one".
The required sum is [tex]2+3=5[/tex].

Solution:
The parameter stands in a denominator, so first of all [tex]a\neq 0[/tex]. We multiply both sides by [tex]2a[/tex]:
[tex]a(x-a)+2(x-2)=4a[/tex]
[tex]ax-a^2+2x-4=4a[/tex]
We collect the terms with [tex]x[/tex] on the left and everything else on the right:
[tex](a+2)x=a^2+4a+4[/tex]
[tex](a+2)x=(a+2)^2[/tex]
The equation has infinitely many solutions when it takes the form [tex]0x=0[/tex], that is when [tex]a+2=0[/tex], or [tex]a=-2[/tex]. This value is admissible, because [tex]-2\neq 0[/tex].
(For every other [tex]a\neq 0[/tex] the equation has the single solution [tex]x=a+2[/tex].)
Answer: [tex]a=-2[/tex].

Solution:
First we gather the terms with [tex]x[/tex] on the left and the rest on the right:
[tex]ax-3x=a-2[/tex]
[tex](a-3)x=a-2[/tex]
First case: [tex]a\neq 3[/tex]. We divide by [tex]a-3[/tex] and get the single solution [tex]x=\frac{a-2}{a-3}[/tex].
Second case: [tex]a=3[/tex]. The equation becomes [tex]0x=1[/tex], which is false for every [tex]x[/tex], so there is no solution.

Solution:
We factor the left-hand side:
[tex]a(a^2+a+1)x=186(a^2+a+1)[/tex]
The expression [tex]a^2+a+1[/tex] is never zero for a real [tex]a[/tex] — its discriminant is [tex]1-4=-3[/tex], which is negative — so we may divide both sides by it:
[tex]ax=186[/tex]
Now everything depends on [tex]a[/tex].
If [tex]a\neq 0[/tex], we divide by [tex]a[/tex] and get the single solution [tex]x=\frac{186}{a}[/tex].
If [tex]a=0[/tex], the equation becomes [tex]0x=186[/tex], which has no solution.

Solution:
It is tempting to cancel one [tex]a[/tex] from both sides straight away, but that is only allowed when [tex]a\neq 0[/tex] — and the case [tex]a=0[/tex] is exactly the interesting one.
First case: [tex]a\neq 0[/tex]. Then [tex]a^2\neq 0[/tex] and we divide: [tex]x=\frac{a}{a^2}=\frac{1}{a}[/tex].
Second case: [tex]a=0[/tex]. The equation becomes [tex]0x=0[/tex], which is true for every [tex]x[/tex], so every real number is a solution.
Note the difference from [tex]ax=1[/tex]: here the right-hand side vanishes together with the coefficient, so we get infinitely many solutions instead of none.

Solution:
For [tex]a=0[/tex] the equation is [tex]0x=12[/tex] and has no solution at all, so [tex]a=0[/tex] is not counted.
For [tex]a\neq 0[/tex] the solution is [tex]x=\frac{12}{a}[/tex], and it is an integer exactly when [tex]a[/tex] divides [tex]12[/tex].
The positive divisors of [tex]12[/tex] are [tex]1, 2, 3, 4, 6, 12[/tex] — six of them. Do not forget the negative ones: [tex]-1, -2, -3, -4, -6, -12[/tex] work just as well, because dividing by a negative number still gives an integer.
In total [tex]6+6=12[/tex] values.
Answer: [tex]12[/tex]