OK first convert one of the equations into Y=MX+b form
Y-2x = 3
Add 2X
Y = 2x +3
Now substitute this equation in the other one.
So it would be
3X - 2Y = 5
3X-2(2x+3) = 5
Now solve for y
3X - 4X - 6 = 5
-1X - 6 = 5
Add 6
-1X = 11
X = -11
Now substitute this into one of the equations
Y - 2X = 3
Y -2(-11) = 3
Y +22 = 3
y = 3-22
y = -19
Answer:
883037%
Step-by-step explanation:
i thank thats the we way we did it i am not shure
The answer is 1.48226407 hope it helps
Direction vector of line of intersection of two planes is the cross product of the normal vectors of the planes, namely
p1: x+y+z=2
p2: x+7y+7z=2
and the corresponding normal vectors are: (equiv. to coeff. of the plane)
n1:<1,1,1>
n2:<1,7,7>
The cross product n1 x n2
vl=
i j l
1 1 1
1 7 7
=<7-7, 1-7, 7-1>
=<0,-6,6>
Simplify by reducing length by a factor of 6
vl=<0,-1,1>
By observing the equations of the two planes, we see that (2,0,0) is a point on the intersection, because this points satisfies both plane equations.
Thus the parametric equation of the line is
L: (2,0,0)+t(0,-1,1)
or
L: x=2, y=-t, z=t
