If I have the equations
$27a+30b+33c+36c \geq x$
$a+b+c+d=4$
and want to solve them using only natural numbers (including 0) for both
$x=131 $ and $x=142 $
preferably but not necessarily with
$a\geq b\geq c\geq d$
How do I approach this. Can I use linear algebra? Do I just use algebra? Theres two knowns and four unknowns, but with the extra restrictions there should be a limited number of solutions. Using just algebra I get stuck after reducing like in this question:
$27(a+b+c+d)+3b+6c+9d \geq 131$
$108+3b+6c+9d \geq 131$
$3b+6c+9d \geq 23$
here it gets messy with wanting only natural numbers, I would like to divide by 3 but then probably my results wont be in natural numbers. Can I continue like this having to use real numbers and then later return to natural numbers? Then I get
$b+2c+3d \geq 7+2/3$
Three unknowns, dont know how to continue, there must be many solutions but Im not sure how to find them. Maybe $a+b+c+d=4$ and only natural numbers is so restricting I should just calculate all possible combinations and see which solve the inequality? Then how to calculate all possible combinations?