Standard form of the linear equation in two variables: ax + by = c.
The slope: m = -a/b
L1: x + 3y = 3 so m1 = -1/3
L2: 6x + 2y = 12 so m2 = -6/2 = -3
Since m1 ≠ m2, the two lines are non-perpendicular and intersect.
If 3y = 12, 2y = 8
If you mean 3x+2y = -5 and -2x+3y = -5 then they are straight line equations
5y = 3y + 12 5y - 3y = (3y - 3y) + 12 2y = 12 y = 6
Plot the straight line representing 2y = 12 - x. Plot the straight line representing 3y = x - 2 The coordinates of the point of intersection of these two lines is the solution to the simultaneous equations.
5x + (3y-x) + (10-2y) = 5x + 3y - x + 10 -2y = 5x - x + 3y - 2y + 10 = 4x + y +10
Perpendicular
perpendicular
If 3y = 12, 2y = 8
Perpendiculat straight lines.
If you mean 3x+2y = -5 and -2x+3y = -5 then they are straight line equations
5y = 3y + 12 5y - 3y = (3y - 3y) + 12 2y = 12 y = 6
2y=8..... y=4
if you wish to multiply 3y-2y by 3y-y the answer is 2y^2
2y - 3y = -1y
x = 3y - 3 so 6 (3y - 3) + 2y = -12 ie 18y - 18 + 2y = -12 ie 20y = 6 so y = 0.3 and x = -2.1
3y squared - 2y = 1
It is y+3y-2y = 2y simplified