Question 703754
If Andy has just 9 minutes to study for his Spanish exam, Andy is in trouble.
Andy NEEDS HELP!
 
Writing {{{e}}} practice essays would take Andy {{{6e}}} minutes.
Reviewing the vocabulary in {{{s}}} lessons would take Andy {{{9s}}} minutes.
Doing both would require {{{6e+9s}}} minutes.
 
If all the time Andy has to study is 9 minutes,
a) {{{6e+9s<=9}}}
 
b) Not sure what is meant by "these values", but
{{{s>=0}}} and {{{e>=0}}}
If {{{e=0}}} (Andy writes no essay) we get {{{9s<=9}}}
so {{{s=0}}} or {{{s=1}}}
(Andy can review vocabulary for 1 lesson, or review nothing).
I{{{e=1}}} (Andy writes 1 essay) we get {{{6+9s<=9}}}--> {{{9s<=3}}} --> {{{s<+1/3}}} --> {{{s=0}}}
(There's no time to review vocabulary).
I could graph that as
{{{drawing(300,300,-0.5,1.5,-0.5,1.5,
grid(1),
locate(1.3,-0.02,"e="),locate(-0.2,1.45,"s="),
red(circle(0,0,0.04)),red(circle(0,1,0.04)),
red(circle(1,0,0.04))
)}}}
 
If there was a typo, and Andy really had 90 minutes, then we are looking for pairs of integers (e,s) that are solutions to
{{{6e+9s<=90}}}
That would be the integer pairs in the red solution zone:
{{{graph(300,300,-2,18,-2,18,6x+9y<=90)}}} or {{{drawing(300,300,-2,18,-2,18,
grid(1), green(line(-2,34/3,18,-2)),
red(circle(0,0,0.2)),red(circle(1,0,0.2)),red(circle(2,0,0.2)),
red(circle(3,0,0.2)),red(circle(4,0,0.2)),red(circle(5,0,0.2)),
red(circle(6,0,0.2)),red(circle(7,0,0.2)),red(circle(8,0,0.2)),
red(circle(9,0,0.2)),red(circle(10,0,0.2)),red(circle(11,0,0.2)),
red(circle(12,0,0.2)),red(circle(13,0,0.2)),red(circle(14,0,0.2)),
red(circle(0,1,0.2)),red(circle(1,1,0.2)),red(circle(2,1,0.2)),
red(circle(3,1,0.2)),red(circle(4,1,0.2)),red(circle(5,1,0.2)),
red(circle(6,1,0.2)),red(circle(7,1,0.2)),red(circle(8,1,0.2)),
red(circle(9,1,0.2)),red(circle(10,1,0.2)),
red(circle(10,2,0.2)),red(circle(11,2,0.2)),red(circle(11,1,0.2)),
red(circle(12,1,0.2)),red(circle(13,1,0.2)),red(circle(12,2,0.2)),
red(circle(9,4,0.2)),red(circle(10,3,0.2)),red(circle(9,4,0.2)),
red(circle(7,5,0.2)),red(circle(6,6,0.2)),red(circle(4,7,0.2)),
red(circle(3,8,0.2)),red(circle(1,9,0.2)),red(circle(0,10,0.2))
)}}} Too many red circles to draw,
but I guess you get the idea. (By now, I am not willing to give Andy 90 minutes).