The solutions of the equations {y = 2x, x + y = 3 are

The solutions of the equations {y = 2x, x + y = 3 are

x=1 y=2
y=2x (1)
x+y=3 (2)
Y = 2x substituting (2)
3x=3
x=1
y=2x=2
x=1 y=2

The solution of the system {x + 2y-5 = 0,2x-y-5 = 0} is

x+2y-5=0 ①
2x-y-5=0 ②
① X 2 - 2 was obtained
5y-5=0
y=1
Substituting y = 1 into (1) yields:
x+2-5=0
x=3
So the solution of the equations is x = 3, y = 1

{x-2y = 11,2x-y = 10} to solve the equations

From x-2y = 11, we can get the following results
x=2y+11
By substituting x = 2Y + 11 into 2x-y = 10, we get the following result:
2(2y+11)-y=10
4y+22-y=10
3y=-12
y=-4
x=2y+11=2*(-4)+11=-8+11=3
x=3,y=-4

Solve the equations 2x + y + x = 10 x + 2Y + Z = - 6 x + y + 2Z = 8
2x+y+x=10
x+2y+z=-6
x+y+2z=8
2x+y+z=10
x+2y+z=-6
x+y+2z=8
Forget the first three equations. The next three are correct

2x+y+x=10 (1)
x+2y+z=-6 (2)
x+y+2z=8 (3)
(1)+(2)+(3)
4x+4y+4z=12
x+y+z=3 (4)
(1)-(4),x=7
(2)-(4),y=-9
(3)-(4),z=5