Literal Equations: Problems with Solutions

Solution:
The coefficient in front of [tex]x[/tex] is [tex]1[/tex] and it is never zero, so whatever the value of the parameter is, the equation has exactly one solution.
We subtract [tex]a[/tex] from both sides:
[tex]x=2a+1-a[/tex]
[tex]x=a+1[/tex]
So for every real [tex]a[/tex] the equation has the unique solution [tex]x=a+1[/tex].

Solution:
If [tex]x=2[/tex] is a solution, then substituting it must give a true equality. We put [tex]x=2[/tex] into the equation:
[tex]a\cdot 2=a+3[/tex]
Now we have an ordinary linear equation, but this time the unknown is the parameter [tex]a[/tex]:
[tex]2a-a=3[/tex]
[tex]a=3[/tex]
Check: for [tex]a=3[/tex] the equation is [tex]3x=6[/tex], whose solution is indeed [tex]x=2[/tex].
Answer: [tex]a=3[/tex]

Solution:
If we can divide by [tex]a[/tex], there is always the solution [tex]x=\frac{1}{a}[/tex]. Let's check the case where we cannot divide by [tex]a[/tex], in other words [tex]a=0[/tex]. We get the equation [tex]0x=1[/tex], which has no solutions. Therefore the answer to the problem is [tex]a=0[/tex].

Solution:
An equation of the form [tex]mx=b[/tex] has no solution only when the coefficient [tex]m[/tex] is zero while [tex]b[/tex] is not. So we first set the coefficient equal to zero:
[tex]a^3+1=0[/tex]
[tex]a^3=-1[/tex]
[tex]a=-1[/tex]
Now we check the right-hand side for [tex]a=-1[/tex]: [tex]109+(-1)=108[/tex], which is not zero. The equation becomes [tex]0x=108[/tex] and it really has no solution.
Answer: [tex]a=-1[/tex].

Solution:
The equation can have no solution only when the coefficient in front of [tex]x[/tex] is zero:
[tex]a^2-1=0[/tex], so [tex]a=1[/tex] or [tex]a=-1[/tex].
We must check both values, because a zero coefficient is not enough — the right-hand side has to be different from zero as well.
For [tex]a=1[/tex]: [tex]0x=2[/tex] — no solution.
For [tex]a=-1[/tex]: [tex]0x=0[/tex] — every [tex]x[/tex] is a solution.
So the only value we are looking for is [tex]a=1[/tex].

Solution:
Here the unknown appears on both sides, so first we bring all terms with [tex]x[/tex] to the left and the rest to the right:
[tex]ax-2x=a-4[/tex]
[tex](a-2)x=a-4[/tex]
Only now can we look at the coefficient. It is zero for [tex]a=2[/tex], and then the right-hand side is [tex]2-4=-2[/tex], which is not zero.
The equation becomes [tex]0x=-2[/tex] and has no solution.
Answer: [tex]a=2[/tex]

Solution:
For any [tex]x[/tex], the value of the left side is zero. We get [tex]b-7=0[/tex], or [tex]b=7[/tex]. By substituting, we get [tex]0x=0[/tex], which has infinitely many solutions.

Solution:
For an equation to have any [tex]x[/tex] as solution, it must be of the form [tex]0x=0[/tex]. Which means that [tex](a-2)=(a-2)^2=0[/tex], or [tex]a=2[/tex].

Solution:
Every [tex]x[/tex] is a solution only if the equation takes the form [tex]0x=0[/tex]. This means that both the coefficient and the right-hand side must be zero:
[tex]a-4=0[/tex] and [tex]a^2-16=0[/tex]
From the first equation [tex]a=4[/tex]. We check it in the second one: [tex]4^2-16=0[/tex] — it is satisfied.
For [tex]a=4[/tex] the equation becomes [tex]0x=0[/tex], so every real number is a solution.
(For [tex]a\neq 4[/tex] we may divide by [tex]a-4[/tex] and we get the single solution [tex]x=a+4[/tex].)
Answer: [tex]a=4[/tex].

Solution:
An equation has infinitely many solutions only when it takes the form [tex]0x=0[/tex]. So the coefficient and the right-hand side must both be zero.
From [tex]a+1=0[/tex] we get [tex]a=-1[/tex]. We check the right-hand side: [tex](-1)^2-1=0[/tex] — it vanishes too.
For [tex]a=-1[/tex] the equation is [tex]0x=0[/tex] and every real number is a solution.
(For [tex]a\neq -1[/tex] we factor [tex]a^2-1=(a-1)(a+1)[/tex] and divide, which gives the single solution [tex]x=a-1[/tex].)
Answer: [tex]a=-1[/tex]

Solution:
Everything depends on the coefficient [tex]a[/tex].
First case: [tex]a\neq 0[/tex]. We divide both sides by [tex]a[/tex] and get the single solution [tex]x=4[/tex]. Note that this solution does not depend on the parameter.
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.

Solution:
We factor the right-hand side:
[tex](a-3)x=(a-3)(a+3)[/tex]
Everything depends on the coefficient [tex]a-3[/tex].
First case: [tex]a\neq 3[/tex]. Then [tex]a-3\neq 0[/tex] and we may divide both sides by it, which gives the single solution [tex]x=a+3[/tex].
Second case: [tex]a=3[/tex]. The equation becomes [tex]0x=0[/tex], which is true for every [tex]x[/tex], so every real number is a solution.
Note that there is no value of [tex]a[/tex] for which this equation has no solution.

Solution:
The right-hand side is [tex]5[/tex] and it is never zero, so the equation can never be of the form [tex]0x=0[/tex] — infinitely many solutions are impossible here.
First case: [tex]a\neq 1[/tex]. Then [tex]a-1\neq 0[/tex] and we divide by it: [tex]x=\frac{5}{a-1}[/tex].
Second case: [tex]a=1[/tex]. The equation becomes [tex]0x=5[/tex], which is false for every [tex]x[/tex], so there is no solution.

Solution:
If the equation is to be not defined, then a denominator must be zero. The only denominator is [tex]a+5[/tex], so we get [tex]a=-5[/tex]