How do you solve a system by addition in algebra?

Question

Not trying to get the answer but I need for someone to explain to me how to do this. the book I have is not very easy to follow.

Answer 1

well... imagine 2 functions

2x + 3y = 5
-2x + 2y = 10

add the two togather

5y = 15
y=3

plug y into one of the first equations

2x + 3y = 5
2x + 3(3) = 5
2x + 9 = 5
2x = -4
x = -2 and y = 3

you can check it with the other equation too

-2x + 2y = 10
-2(-2) + 2(3) = 10
4 + 6 = 10
10 = 10

and the system is solved...

x = -2 and y = 3

Answer 2

Folken did a great job of explaining how to solve the system if one of the variables will "add out". If they do not you will have to find a muliplier of one of the 2 systems that will make one of the variables "add out".

3x + 2y = 5
x + 3y =4
Multiply the bottom row by -3 to get the x's to add out.
3x +2y = 5
-3x -9y = -12

-7y = -7
y = 1
Substitute y = 1 into either of the original equations, sometimes one is a little easier to solve once the variable is replaced with the number.
x + 3(1) = 4
x = 1

Answer 3

when the coefficients of one of the two variables are additive inverses like 5 and -5,3 and -3 and so on you add the equation so that that variable vanishes.now it will be a linear equation in one variable and so can be easily solved.after getting the value of one variable by substitution the value of the other can be found.you can read this in the light of the example cited by the previous answerer for clinching it

Answer 4

http://www.google.com/search?hl=en&q=solve+a+system+by+addition+in+algebra&btnG=Google+Search

Answer 5

give me your problem please