Question

Question on linear inequalities

Answer 1

Find the range of values of a, given that a+b+c=6, 2a-b+c=3, and b≥c≥0

Answer 2

what the hack

Answer 3

Can I help?

Answer 4

YES PLEASE :D

Answer 5

this seems tricky
but from
a+b+c=6
a= 6-(b+c)
they tell us b>=c>=0 (i.e. range from 0 to +infinity)
if b and/or c get large then a becomes very negative, i.e. -infinity
so -infinity is the lower bound on a

Answer 6

oh ok i understand this part but is there a way to find the upper bound?

Answer 7

if we add the two equations
a+b+c=6
2a-b+c=3
---------
3a +2c= 9

we know c>=0 so the smallest c can be is 0. in that case
3a=9
a= 3
any bigger value of c will make a smaller (because a= ⅓*(9-2c) and we subtract off 2c)
so it looks like 3 is the biggest a can be

the range for a is
\[ ( -\infty, 3]\]

Answer 8

OHHHH I UNDERSTAND!! :D This question has been bugging me for quite a while. Thank you!!