[ y = -2x + any other number ] is parallel to [ y = -2x + 6 ].
x=0 y=6
x = 2y = 4
2x + 3y = 9x = 3, y = 1 therefore 6 + 3 = 9
The answer is - not are - (-6, 9).
No but y = -1/2x+3 is perpendicular to y = 2x+6
[ y = -2x + any other number ] is parallel to [ y = -2x + 6 ].
If: y = x+6 and y = 2x Then: x+6 = 2x And: 6 = 2x-x => 6 = x So: x = 6 and y = 12
6y
x = -4 and y = 2
x=0 y=6
x = 2y = 4
If: 12 = 2y+x then y = -1/2x+6 So: y = 2x+4 and y = -1/2x+6 which means that they are perpendicular lines
2x + 3y = 9x = 3, y = 1 therefore 6 + 3 = 9
The answer is - not are - (-6, 9).
3x-y = 3y3x = 4yy = 0.75x3x-0.75x = 2x+22.25x = 2x+20.25x = 2x = 8y = 6
eqn 1 Y = 2X + 6 eqn 2 2X - Y = 2 Rearrange eqn2 to isolate Y eqn2 Y = 2X - 2 so 2X + 6 = 2X - 2 subtracting 2X from both sides we get 6 = -2 As this is impossible this set of equations can not be solved.