SOLUTION: use the addition-subtraction method to solve. 2x+5y=15 3x+5y=10 here,s what i came up with: 15-2x=10-3x 2(-5)+5y=15 x=-5 -10+5y=15

Algebra ->  Expressions-with-variables -> SOLUTION: use the addition-subtraction method to solve. 2x+5y=15 3x+5y=10 here,s what i came up with: 15-2x=10-3x 2(-5)+5y=15 x=-5 -10+5y=15       Log On


   

Question 878846: use the addition-subtraction method to solve.
2x+5y=15
3x+5y=10
here,s what i came up with:
15-2x=10-3x 2(-5)+5y=15
x=-5 -10+5y=15
5y=25
y=5
how do i check to see if my answer is correct?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Your given system is ready for using the Elimination Method, but you want "addition-subtraction" method. This name of method is not well known everywhere, so I will show you how to begin the solution process of the Elimination Method.

Both of your equations are in standard form and both have a term, 5y in their left members.

2x%2B5y-%283x%2B5y%29=15-10
2x%2B5y-3x-5y=5
2x-3x%2B5y-5y=5
-x%2B0=5
-x=5
highlight%28x=-5%29
-
I prefer to continue using substitution.
Start with either original equation, arbitrarily choosing----3x+5y=10
5y=10-3x
y=10%2F5-3x%2F5
y=2-3x%2F5
y=2-3%2A%28-5%29%2F5
highlight%28y=5%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

use the addition-subtraction method to solve.
2x+5y=15
3x+5y=10
here,s what i came up with:
15-2x=10-3x 2(-5)+5y=15
x=-5 -10+5y=15
5y=25
y=5
how do i check to see if my answer is correct?


All these steps are TOTALLY UNNECESSARY!!

 2x + 5y = 15 ------- eq (i)

 3x + 5y = 10 ------- eq (ii)

- x      = 5 ------- Subtracting eq (ii) from eq (i)

       highlight_green%28x+=+-+5%29



2(- 5) + 5y = 15 ------- Substituting - 5 for x in eq (i)

  - 10 + 5y = 15

         5y = 15 + 10

         5y = 25

          y+=+25%2F5, or highlight_green%285%29

Now substitute - 5 for x, and 5 for y in either eq (i) or (ii). If the equation proves true after doing

this, then the solutions are CORRECT.



IT IS THAT SIMPLE!! Don't make things SO DIFFICULT for yourself!!



Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.

Further help is available, online or in-person, for a fee, obviously.