SOLUTION: What is the elimination method 2r-5s=-29 and 5r+2s=58?

Algebra.Com
Question 391673: What is the elimination method 2r-5s=-29 and 5r+2s=58?
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
What is the elimination method 2r-5s=-29 and 5r+2s=58?
...
2r-5s=-29........................1
5r+2s=58.........................2
..
multiply (1)by 5 & (2) by -2
10r-25s=-145
-10r-4s=-116
Add them up
-25s-4s=-145-116
-29s==-261
/-29

..
plug value of s in (1)
2r-5*9=-29
2r-45=-29
2r=45-29
2r=16
/2

m.ananth@hotmail.ca

RELATED QUESTIONS

Solve by the elimination method 5r-2s=0... (answered by Flannery)
Solve by elimination method 2r - 5s = -14 5r + 2s = 52 what is the solution of the... (answered by Maths68)
5r-2s=37 2r+5s=38 Solve using the elimination... (answered by stanbon)
5r-2s=13 2r-5s=11 what is the solution of the system? (answered by jim_thompson5910)
Solve by the elimination method 2r-5s= -1 5r+2s=41 What is the solution of the system? (answered by ewatrrr)
I have been working on a problem using the elimination method. I think I have it right... (answered by scott8148,stanbon)
Solve by the elimination method 2r-5s=-33 5r+2s=48 I think the answer is that... (answered by jim_thompson5910)
Solve by the elimination method. 7r-5s=-12 5r-7s=76 What is the solution of the... (answered by jim_thompson5910)
Solve by the elimination method. 7r-5s=-12 5r-7s=76 What is the solution of the... (answered by checkley77)