Solutions:
Part A:
Let [tex]y_{1}=(4-x) \text { and } y_{2}=(2-x+1)[/tex]
As we know that the values at the point of intersection are the values that satisfy both the equation at that particular point.
So at insertion point, [tex]y_{1}=y_{2}[/tex]
Hence [tex](4-x)=(2-x+1)[/tex]
Part 2:
The problem asked to make tables to find the solution to [tex]4-x=2-x+1[/tex]
If we take the first equation,
[tex]y=4-x[/tex]
then, the table for (x,y) is [tex](-2,6), (-1,5), (0,4), (1,3), (2,2), (3,1), (4,0)[/tex]
if we take the second equation,
[tex]y=2-x+1[/tex]
then the table for (x,y) is[tex](-2,5),(-1,4), (0,3), (1,2), (2,1), (3,0), (4,-1)[/tex]
Part 3:
We can solve the equation [tex]4-x = 2-x+1[/tex] graphically by drawing the line [tex]y = 4-x[/tex] and [tex]y= 2-x+1[/tex] in the graph. The intersecting point of both the lines will be the solution.