Question 220967: solve using the elimination method
5r-6s=-5
6r+5s=55
The answer should be an ordered pair and either an integer or fraction. Found 2 solutions by drj, checkley77:Answer by drj(1380) (Show Source): You can put this solution on YOUR website! solve using the elimination method
5r-6s=-5 Equation A
6r+5s=55 Equation B
Multiply Equation A by 5 and Equation B by 6
25r-30s=-25 Equation A1
36r+30s=330 Equation B1
Add equations A1 and B1 to eliminate s-terms.
61r=305 or r=305/61=5
Then Equation A yields 5*5-6s=5 or 25-6s=-5 or s=5
ANSWER: The solution is r=5 and s=5 or (5,5). Checking via substitution method.
Solve: We'll use substitution. After moving -30*s to the right, we get: , or . Substitute that
into another equation: and simplify: So, we know that s=5. Since , r=5.
Answer: .
I hope the above steps were helpful.
And good luck in your studies!
For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
Respectfully,
Dr J
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website! 5r-6s=-5 MULTIPLY BY 5
6r+5s=55 MULTIPLY BY 6 & ADD.
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25R-30S=-25
36R+30S=330
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61R=305
R=305/61
R=5 ANS.
5*5-6S=-5
25-6S=-5
-6S=-5-25
-6S=-30
S=-30/-6
S=5 ANS.
PROOF:
6*5+5*5=55
30+25=55
55=55