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| Question 69283This question is from textbook Algebra 1
 :  I am having difficulty with this problem - please help me if you can.  The chapter is Solving Linear Systems by Linear Combinations (substitution not allowed).
 2q = 7 - 5p
 4p - 16 = q
 
 I am using the 5 step formula - arrange like terms in columns, obtain coefficients that are opposites for one variable, add equations & eliminate one variable & solve, etc...  I did nine other problems this way and solved them.  I don't see what I'm doing wrong with this one.  Any help will be appreciated.
 This question is from textbook Algebra 1
 
 Answer by stanbon(75887)
      (Show Source): 
You can put this solution on YOUR website! 2q = 7 - 5p 4p - 16 = q
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 Rearrange:
 1st: 5p+2q=7
 2nd: 4p-q=16
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 Multiply 2nd by 2 to get:
 3rd: 8p-2q=32
 Add 1st and 3rd to get:
 13p=39
 p=3
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 Substitute that into 2nd to get:
 4*3-q=16
 q=-4
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 Solution: p=3; q=-4
 Check in 1st: 5p+2q=7
 5*3+2*-4 = 7
 7=7
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 Cheers,
 Stan H.
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