World problem Algebra 2. A mom buys 3 different types if t-shirts for her children, long sleeve, polo, and short sleeve. The long sleeve shirts cost $18 each, the polos cost $15 each, and the short sleeve cost $12 each. The mother purchased 3 less polos then she did short sleeve shirts. If the mother purchased 18 shirts at an average of $14 each, find the number of each type of shirt purchased
@satellite73 could you help me me set of the equations?
start by declaring your variable say "let \(x\) = ..."
i set up p= s-3 and 18L + 15p+ 12s = 18 but Ihave a feelings there not right
no it is not right \[18L+15P+12S\] represents the cost, not the total number of shirts bought
if \(L\) is the number of long sleeve shirts, then the cost of all the long sleeve shirts is \(18L\) etc the total number of all the shirts is \[P+L+S\], which you know to be 18 so you can write \[P+L+S=18\] as one equation
now we have to figure out how much was spent all together. she bought 18 shirts at an average of $14 per shirt so that shouldn't be too hard
let me know when you get the answer (by multiplying)
252
ok good, so now we have 3.6*5 \[P+L+S=18\]\[18L+15P+12S=252\]and also \[P=S-3\]
ignore that weird number up top, it is a type
now lets go back and replace all the \(P\) what we see by \(S-3\)
\[S-3+L+S=18\]\[15(S-3)+18L+12S=252\]
we still have a bit more work to do because this problem is annoying first equation become \[2S+L-3=18\] or \[2S+L=21\]
is that our final equation?
oh no we also have the second equation which (after multiplying out and collecting terms) will be \[18L+27S=297\]
whew this is a pain right? we have to solve the two equation \[2S+L=21\]\[18L+27S=297\] so we can do this by rewriting the first equation as \[L=21-2S\]substitute in the second one to get \[18(21-2S)+27S=297\] and FINALLY we have one equation to solve
hopefully after a bit of algebra you will get \(S=9\)
and of course we know S is short, so P is \(9-3=6\) and the rest are \(L\)
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