Answer:
Follows are the solution to this question:
Step-by-step explanation:
please find the complete question in the attached file:
In view of their choices, they could solve this problem by linking x and k finding values that meet the value.
a) In the family for equations they have, the plugin first x value.
[tex]\to 3 \times O +k-5=k \times 0 -k+1 \\\\\to k-5 =-k+1 \\\\\to 2k =6 \\\\\to k= \frac{6}{2} \\\\[/tex]
[tex]=3\\[/tex]
It is also obvious that k = 3 makes x = 0 for that equation family.
b) See now the second x value:
It is also obvious that k=3 renders a solution for this equation family of x=1.
c)
We plug in the 3rd of x, finally.
[tex]\to 3 \times 2 + k-5=k \times 2-k+1 \\\to 6+ k-5= 2k-k+1 \\\to k+1= k+1 \\\to k-k= 1-1\\\ \ \ \ 0=0[/tex]
So k could be any real value and x=2 could be a solution for the equations family.
a) 3
b) 3
c) Any real value
