system of equation?

multiply equation (1) by 2 and add to equation (2)

we get

6x - 2y + 4x + 2y = 30 + 20
=> 10x = 50
=> x = 5
Putting in (1) we get 10 + y = 10
=> y =0

Hence solution is x = 5, y = 0

4x + 2y = 20
6x - 2y = 30----ADD

10x = 50
x = 5

10 + y = 10
y = 0

x = 5 , y = 0

2x + y = 10 (solve by substitution)
6x - 2y = 30

2x + y = 10
y = 10 - 2x
2y = 2(10 - 2x)

6x - 2y = 30
6x - 2(10 - 2x) = 30
6x - 2*10 + 2*2x = 30
6x - 20 + 4x = 30
6x + 4x = 30 + 20
10x = 50
x = 50/10
x = 5

2x + y = 10
2(5) + y = 10
10 + y = 10
y = 10 - 10
y = 0

∴ x = 5 , y = 0

So easy!
(2)2x+y=10
6x-2y=30

4x+2y=20
6x-2y=30
the y's cancel

solve for x
x=5 (in both equations obviously)

plug in x into one of the equations
2(5)+y=10
10+y=10
y=0

{5,0}

2x + y = 10 ............... x 3
6x - 2y = 30

6x + 3y = 30
6x - 2y = 30
-------------------- --
5y = 0
y = 0

2x + y = 10
2x = 10
x = 5

divide eq2 and ad to eq1,
answer (5, 0)