Literal Equations: Very Difficult Problems with Solutions

Solution:
In order for an equation to have all [tex]x \in R[/tex] as solutions, it must be of the form [tex]0x=0[/tex]. Which means that
[tex]\begin{array}{|l}a^2-7a+10=0 \\ a^2-8a+15=0\end{array}[/tex]
[tex]\begin{array}{|l}a^2-5a-2a+10=0 \\ a^2-5a-3a+15=0 \end{array}[/tex]
[tex]\begin{array}{|l}(a-5)(a-2)=0 \\ (a-5)(a-3)=0\end{array}[/tex]
The common solution to both equations is [tex]a=5[/tex], which is also the answer to the problem.

Solution:
Before we can talk about cases we must bring the equation to the form [tex]mx=b[/tex]. We multiply both sides by [tex]4[/tex]:
[tex](ax-3)-2(x-a)=4[/tex]
[tex]ax-3-2x+2a=4[/tex]
We keep the terms with [tex]x[/tex] on the left and move the rest to the right:
[tex](a-2)x=7-2a[/tex]
Now the analysis is the usual one. The coefficient is zero for [tex]a=2[/tex], and then the right-hand side is [tex]7-4=3[/tex], which is not zero. The equation becomes [tex]0x=3[/tex] and has no solution.
(For [tex]a\neq 2[/tex] there is exactly one solution, [tex]x=\frac{7-2a}{a-2}[/tex].)
Answer: [tex]a=2[/tex]

Solution:
We open the brackets on both sides:
[tex]ax-a=2ax+6-x[/tex]
We move every term containing [tex]x[/tex] to the left and everything else to the right:
[tex]ax-2ax+x=a+6[/tex]
[tex](1-a)x=a+6[/tex]
Be careful with the sign here — the coefficient is [tex]1-a[/tex], not [tex]a-1[/tex].
It vanishes for [tex]a=1[/tex], and then the right-hand side is [tex]1+6=7[/tex], which is not zero. The equation becomes [tex]0x=7[/tex] and has no solution.
Answer: [tex]a=1[/tex]

Solution:
We factor both sides:
[tex](a-1)(a+1)x=a(a+1)[/tex]
The coefficient vanishes for two values, [tex]a=1[/tex] and [tex]a=-1[/tex], and they behave differently, so we need three cases.
Case 1: [tex]a\neq 1[/tex] and [tex]a\neq -1[/tex]. We cancel [tex]a+1[/tex] and then divide by [tex]a-1[/tex]: [tex]x=\frac{a}{a-1}[/tex].
Case 2: [tex]a=-1[/tex]. Both sides vanish, the equation is [tex]0x=0[/tex] and every real number is a solution.
Case 3: [tex]a=1[/tex]. The left side is zero but the right side is [tex]1+1=2[/tex], so the equation is [tex]0x=2[/tex] and has no solution.
This is the point of the problem: the two roots of the coefficient do not give the same answer, and you cannot tell which is which without substituting.

Solution:
We factor both sides:
[tex]a(a-2)(a+2)x=a(a-2)[/tex]
The coefficient vanishes for three values: [tex]a=0[/tex], [tex]a=2[/tex] and [tex]a=-2[/tex]. Each one has to be checked separately in the right-hand side.
[tex]a=0[/tex]: the right-hand side is [tex]0[/tex], the equation is [tex]0x=0[/tex] — infinitely many solutions.
[tex]a=2[/tex]: the right-hand side is [tex]4-4=0[/tex], the equation is [tex]0x=0[/tex] — infinitely many solutions.
[tex]a=-2[/tex]: the right-hand side is [tex]4+4=8[/tex], the equation is [tex]0x=8[/tex] — no solution, so this value does not belong to [tex]A[/tex].
Hence [tex]A=\{0, 2\}[/tex] and the sum of its elements is [tex]0+2=2[/tex].
Answer: [tex]2[/tex]

Solution:
Let us first inspect the set [tex]A[/tex]. If it has one zero element, then the product of all its elements is zero, since any number, multiplied by zero is zero. We check if [tex]a=0[/tex] belongs in the set. By substituting it into the equation, we get [tex]0x=10[/tex], which has no solution. Therefore, [tex]a=0[/tex] belongs in the set [tex]A[/tex] and the product of its elements is zero.

Solution:
We'll first determine the set [tex]A[/tex].
[tex]3<2a<15[/tex] is equivalent to
[tex]\begin{array}{|l}2a>3 \\ 2a<15\end{array}[/tex]
[tex]\begin{array}{|l}a>\frac{3}{2}=1\frac{1}{2} \\ a < \frac{15}{2}=7\frac{1}{2} \end{array}[/tex]
So, the numbers [tex]a_1,a_2,\dots,a_n[/tex] are all the integers (all of them natural numbers) between [tex]1\frac{1}{2}[/tex] and [tex]7 \frac{1}{2}[/tex]. These numbers are [tex]2,3,4,5,6,7[/tex] and respectively [tex]a_1=2,a_2=3,a_3=4,a_4=5,a_5=6,a_6=7[/tex].
We'll also find an explicit formula for the solution of the equation.
[tex](a-1)(a+1)x=(a+1)^2[/tex]. Obviously [tex]a \ne -1[/tex], we divide by [tex]a+1[/tex] and get
[tex](a-1)x=a+1[/tex]. Since each [tex]a \ge 2[/tex], [tex]a-1 \ne 0[/tex] and the solutions are
[tex]x=\frac{a+1}{a-1}[/tex]. We substitute with each [tex]a[/tex]: [tex]x_1=\frac{a_1+1}{a_1-1}=\frac{2+1}{2-1}=3[/tex]
[tex]x_2=\frac{a_2+1}{a_2-1}=\frac{3+1}{3-1}=\frac{4}{2}=2[/tex]
[tex]x_3=\frac{a_3+1}{a_3-1}=\frac{4+1}{4-1}=\frac{5}{3}[/tex]
[tex]x_4=\frac{a_4+1}{a_4-1}=\frac{5+1}{5-1}=\frac{6}{4}=\frac{3}{2}[/tex]
[tex]x_5=\frac{a_5+1}{a_5-1}=\frac{6+1}{6-1}=\frac{7}{5}[/tex]
[tex]x_6=\frac{a_6+1}{a_6-1}=\frac{7+1}{7-1}=\frac{8}{6}=\frac{4}{3}[/tex]. We multiply them:
[tex]x_1x_2x_3x_4x_5x_6=3\cdot2\cdot\frac{5}{3}\cdot\frac{3}{2}\cdot\frac{7}{5}\cdot\frac{4}{3}=\frac{3\cdot2\cdot5\cdot3\cdot7\cdot4}{3\cdot2\cdot5\cdot3}=7\cdot4=28[/tex]

Solution:
The coefficient in front of [tex]x[/tex] is [tex]3[/tex] and it is never zero, so the equation always has exactly one solution:
[tex]x=\frac{a+6}{3}[/tex]
Instead of hunting for the values of [tex]a[/tex] directly, it is easier to turn the problem around: choose the solution first. The condition says [tex]x[/tex] may be any of [tex]1, 2, 3, \ldots, 10[/tex].
From [tex]3x=a+6[/tex] we express the parameter: [tex]a=3x-6[/tex]. Every one of the ten choices of [tex]x[/tex] gives exactly one value of [tex]a[/tex], and different values of [tex]x[/tex] give different values of [tex]a[/tex]:
[tex]x=1\Rightarrow a=-3,\quad x=2\Rightarrow a=0,\quad \ldots,\quad x=10\Rightarrow a=24[/tex]
So there are exactly ten such values of the parameter.
Answer: [tex]10[/tex]