Question 223858: solve using elimination method
3r-2s=-15
2r+3s=29
Answer by drj(1380) (Show Source):
You can put this solution on YOUR website! Solve using elimination method
3r-2s=-15 Equation 1
2r+3s=29 Equation 2
Step 1. Multiply Equation 1 by 3 to both sides of the equation to get Equation 1A below and Equation 2 by 2 to get Equation 2A below to eliminate the s terms when we add the following equations
9r-6s=-45 Equation 1A
4r+6s=58 Equation 2A
Step 2. Adding Equations 1A and 2A will eliminate the s terms.
9r+4r=-45+58
13r=`13
Divide by 13 to both sides of the equation yields r=1
Step 3. Substitute r=1 into Equation 1 to find s
3r-2s=-15 or 3-2s=-15 or -2s=-18. Then s=9.
Step 4. With r=1 and s=9, check Equation 2 if true 2r+3s=29 or 2*1+3*9=29 which is a true statement.
Step 5. ANSWER: The solution is r=1 and s=9.
I hope the above steps were helpful.
For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
Good luck in your studies!
Respectfully,
Dr J
|
|
|