Parametric Linear Equations: Problems with Solutions

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

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

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

Solution:
We subtract a from both sides of the equation to get
[tex]x=a+1[/tex]. We substitute a=3:
[tex]x=3+1=4[/tex]

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]

Solution:
We directly substitute:
[tex](3^3+1)x=109+3[/tex]
[tex]28x=112[/tex]
[tex]x=4[/tex]

Solution:
We substitute:
[tex](4^4-250)x=4^3+2[/tex]
[tex](256-250)x=64+2[/tex]
[tex]6x=66[/tex]
[tex]x=11[/tex]

Solution:
We substitute:
[tex]((-2)^3+9)x=237+10\cdot(-2)[/tex]
[tex](-8+9)x=237-20[/tex]
[tex]x=217[/tex]

Solution:
We directly substitute:
[tex]5\cdot17x=85[/tex]
[tex]85x=85[/tex]
[tex]x=1[/tex]