SOLUTION: 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

SOLUTION: 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

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
--------------
Rearrange:
1st: 5p+2q=7
2nd: 4p-q=16
---------------
Multiply 2nd by 2 to get:
3rd: 8p-2q=32
Add 1st and 3rd to get:
13p=39
p=3
------------
Substitute that into 2nd to get:
4*3-q=16
q=-4
------------
Solution: p=3; q=-4
Check in 1st: 5p+2q=7
5*3+2*-4 = 7
7=7
===================
Cheers,
Stan H.
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