looking at the picture, the region AB is where the inequality of f(x) overlaps the inequality g(x), so if we pick any point on that blue region AB, then f(x) will be TRUE and g(x) will also be TRUE, lemme narrow it down some, let's pick the point on AB of say (0,0), namely x = 0 and y = 0
let's try the 1st set
0 ⩽ - 2(0) + 3
0 ⩽ 3 <--- is that true? well yes, zero is less or equals, actually less than 3
0 ⩽ 0 + 3
0 ⩽ 3 <--- is that true? yes it's
let's try the 2nd inequalities set with the same values for "x" and "y"
0 ⩾ -2(0) + 3
0 ⩾ 3 <--- is that true? nope, no dice
let's not bother checking the other, since it already gave FALSE
let's try the 3rd inequalities set, same values
0 ⩽ -3(0) + 2
0 ⩽ 2 <--- is that true? yeap
0 ⩽ -(0) + 2
0 ⩽ 2 <---- is that true? yeap
not to bother you with the 4th set one, the 4th one fails and gives FALSE
so we know the 1st and 3rd one give TRUE on for f(x) and g(x), so which one is it?
take a peek at the picture below, notice the rise and run and therefore the slopes for each, g(x) has a slope of 2/2 = 1, and f(x) has a slope of 3/1 = 3, both going downwards, so both are negative, namely -1 and -3.
that means that g(x) will look like -1x + something, or -x + something.
and means that f(x) will look like -3x + something.
so that means the only set with those slopes will be [tex]\begin{cases} y\leqslant -3x + 2\\ y \leqslant -x + 2 \end{cases}[/tex]
