SOLUTION: How do I solve by elimination for 2r - 3s = -5 and 3r + 2s = 38?
Algebra.Com
Question 589273: How do I solve by elimination for 2r - 3s = -5 and 3r + 2s = 38?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Find a pair of constants such that if you multiply the first equation by the first constant and the second equation by the second constant the result will be that the coefficients on one of the variables will be additive inverses, that is to say if one of the coefficients is
, then the coefficient on the same variable in the other equation is
.
Next add the two equations, term by term. The variable with the additive inverse coefficients will be eliminated because the coefficient on that variable in the sum equation will be zero.
Solve for the remaining variable. Substitute that value back into either of the original equations and then solve for the remaining variable.
John

My calculator said it, I believe it, that settles it
RELATED QUESTIONS
3r-2s=5 and... (answered by mananth)
solve using elimination method
3r-2s=-15... (answered by drj)
2r - 3s =-3
3r+2s=28
solve by the elimination... (answered by Alan3354,rwm,blwinbbbles)
solve by the elimination method:
2r-3s= -14... (answered by jim_thompson5910,Alan3354)
Solve by the elimination method
3r-2s=-4
2r+3s=32
what is the solution of the... (answered by jim_thompson5910)
I have tried for 2 days to solve this one! I have the answers, and they do solve the... (answered by venugopalramana,Fermat)
2r+3s=9
3r+2s=12
(answered by Edwin McCravy)
2r+3s=9
3r+2s=12
(answered by Edwin McCravy)
How do I solve: 2s+3r=8 solve for... (answered by nyc_function)