Simultaneous equations (Systems of linear equations): Very Difficult Problems with Solutions

Solution:
Subtract the second equation from the first: [tex]|y-4|-2y=1[/tex].
If [tex]y\geq 4[/tex]: [tex]y-4-2y=1 \Rightarrow y=-5[/tex], which does not satisfy [tex]y\geq 4[/tex].
If [tex]y\lt 4[/tex]: [tex]4-y-2y=1 \Rightarrow y=1[/tex], which satisfies the condition.
[tex]|x+3|=8-2y=6 \Rightarrow x=3[/tex] or [tex]x=-9[/tex]
The solutions are [tex](3, 1)[/tex] and [tex](-9, 1)[/tex].

Solution:
Add the first two equations: [tex](x-y+2z)+(2x+3y-z)=3x+2y+z=2+4=6[/tex].
This is exactly the third equation, so it gives no new information. In fact we have two equations with three unknowns.
Let [tex]z=t[/tex]:
[tex]\begin{cases} x-y=2-2t \\ 2x+3y=4+t \end{cases} \Rightarrow y=t,\ x=2-t[/tex]
The system has infinitely many solutions: [tex](2-t, t, t)[/tex], where [tex]t[/tex] is any number.

Solution:
Add the equations and subtract them:
[tex](3x-2y)+(2x-3y)=5x-5y=(a+c)^2-(a-c)^2+6ac=4ac+6ac=10ac \Rightarrow x-y=2ac[/tex]
[tex](3x-2y)-(2x-3y)=x+y=(a+c)^2+(a-c)^2=2a^2+2c^2[/tex]
[tex]x=\frac{(x+y)+(x-y)}{2}=a^2+ac+c^2[/tex], [tex]y=\frac{(x+y)-(x-y)}{2}=a^2-ac+c^2[/tex]

Solution:
Write the second equation in the same form: [tex]\begin{cases} \left(a^2+1\right)x-2y=-5 \\ -4ax+4y=3 \end{cases}[/tex]
The system has exactly one solution when the coefficients of [tex]x[/tex] and [tex]y[/tex] are NOT proportional, i.e.
[tex]\left(a^2+1\right)\times 4-(-2)(-4a)\neq 0 \Leftrightarrow 4a^2-8a+4\neq 0 \Leftrightarrow 4(a-1)^2\neq 0 \Leftrightarrow a\neq 1[/tex]
For [tex]a=1[/tex] the system becomes [tex]2x-2y=-5[/tex], [tex]-4x+4y=3[/tex], i.e. [tex]x-y=-\frac{5}{2}[/tex] and [tex]x-y=-\frac{3}{4}[/tex] — no solution. Note that the condition [tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}[/tex] describes exactly the opposite case (no solution or infinitely many solutions).

Solution:
A system of two linear equations has more than one solution only when it has infinitely many solutions. For this the coefficients of [tex]x[/tex] and [tex]y[/tex] must be proportional:
[tex]\frac{2}{a}=\frac{a}{8} \Rightarrow a^2=16 \Rightarrow a=4[/tex] or [tex]a=-4[/tex]
Check each value:
[tex]a=4:\ \begin{cases} 2x+4y=8 \\ 4x+8y=16 \end{cases}[/tex] — the second equation is the first one multiplied by [tex]2[/tex]; infinitely many solutions.
[tex]a=-4:\ \begin{cases} 2x-4y=8 \\ -4x+8y=16 \end{cases} \Rightarrow \begin{cases} x-2y=4 \\ x-2y=-4 \end{cases}[/tex] — a contradiction; no solution.
Answer: [tex]a=4[/tex].