Question 174330
let x be the # of radio ads and y be the number of newspaper ads
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we know that a radio add reaches 1500 people 
we know that a newspaper ad reaches 500 people  we need a combination of these two that reaches 45000 people. so lets write and equation to fit that
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1500(x)+500(y)=45000...eq 1
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The other stipulation regarding these adds is the cost. we know that the cost of each radio ad is 500 and the cost of each newspaper ad is 200. we have a limit of 16000 dollars to spend altogether so lets write an equation to fit that.
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500(x)+200(y)=16000....eq 2
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we now have 2 equations with 2 unknowns....a linear system. Lets solve by elimination
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1500x+500y=45000...eq 1
 500x+200y=16000...eq 2
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if we multiply eq 2 by -3 we can eliminate the x terms by adding the 2 equations together
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1500x+500y=45000....eq 1
-1500x-600y=-48000....eq 2(multiplied by -3)
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now adding the 2 equations together....the x terms are eliminated because 1500x-1500x=0.  We are left with 
500y-600y=45000-48000---->-100y=-3000
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{{{highlight(y=30)}}}newspaper spots
take y's found value and plug it into either of eq 1 or 2
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500x+200(30)=16000---->500x+6000=16000
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500x=10000
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{{{highlight(x=20)}}}radio spots
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The second part is a bit more difficult and I hope I have done this correctly
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We know that the total of radio spots must equal 20 and the number of newspaper spots have to equal 30

on the weekend we have the number of radio and newspaper ads run as 1 each
during the week we have the number of radio and newspaper ads run at 2 radio and 4 newspaper.
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The number of week and weekend radio ads have to total 20
the number of week and weekend newspaper ads have to total 30
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lets call the number of week and weekend spots, w and wd, respectively
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1wd+2w=20....eq 1
1wd+4w=30....eq 2
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again lets use elimination method
subtract eq1 from eq 2....the wd terms will be eliminated. We are left with
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4w-2w=10
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2w=10
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{{{highlight(w=5)}}}the number of weekly spots
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plug w's value into eq 1 or 2 and solve for wd
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1wd+2(5)=20
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{{{highlight(wd=10)}}}the number of weekend spots