How to solve a system of equations by addition?

Question

Example: 2x+3y=1
5x+3y=16

Answer 1

-2x - 3y = - 1
5x + 3y = 16----ADD
3x = 15
x = 5

25 + 3y = 16
3y = - 9
y = - 3

x = 5 , y = - 3

Answer 2

2x+3y=1 (1)
5x+3y= 16 (2) ok first multiply the first or the second equation by -1 so u can cancel out y varialbes
let say i do the first one

-2x-3y=-1
5x+3y=16
ok next u add the two equations, when u do that the y term will be 0. and u will be left out with 3x=15 then u know x = 5 after this u substitute x in the first equations and solve for y
2(5)+ 3y= 1
10+3y=1 -> 3y=-9 --> y = -3
so x=5 and y = -3

Answer 3

First find out which one are same in bout equation (3Y)now subtract large equality from small one .result will be 3X+0=15
3X=15 and X= 5 . Now 2(5)+3Y=1 so 10+3Y=1 and 3Y=9 result for Y=3 .Generally in these kind of equations you should find a way to eliminate one of X or Y at first time to solve the problem.

Answer 4

2x + 3y = 1

5x + 3y = 16

5x + 3y - 2x - 3y = 16 - 1 (we add the 2nd equation with the negative of the 1st equation)

3x = 15

x = 5

Substituting that in the first equation,

2x + 3y = 1

10 + 3y = 1

3y = - 9

y = -3

Remember that addition and subtraction are not different really.

Answer 5

Well you could do this:
(5x+3y=16)
-(2x+3y=1)
_________
3x+0y=15
therefore x=15/3=5

and then plug x into any equation (I chose the bottom because it is easier)
2(5)+3y=1
10+3y=1
3y=-9
y=-3
x=5 and y=-3

Answer 6

In this case subtract one eq from the other because they both have 3y, resulting

3x = 15 x = 5
sub x = 5 back in either of the eq and y = -3

Answer 7

Subtract them (inverse of addition).

3x = 15
x = 5
So 2(5) + 3y = 1
or 3y = -9
so y = -3

Answer 8

Those are simultaneous equations..Solve it using scientific caculator...x=5 and y=-3 is the solution to above set of equations