how do i plug in x,y values for (8-4xy-y^2-y)i - (4x+y)j + (4x^2 +5xy+x+y+y^2-6)k =0?

Question

values that work?

Answer 1

Each component should be zero, so I would start with the simplest one.
4x + y = 0 so x = -(1/4)y
Plug this into the first one
8 - 4xy - y^2 - y becomes 8 - 4(-1/4)y*y - y^2 - y
= 8 + y^2 - y^2 - y
= 8 - y
Set this equal to zero
8 - y = 0
y = 8
x = -(1/4)y so x = -(1/4)(8) = -2
Plug x = -2 and y = 8 into the third part to see if it equals zero. If it does, then you have your answer.
Answer: x = -2 and y = 8

Answer 2

I'm assuming you're trying to get the zero vector.

4x+y=0, so y=-4x.

Substitute y=-4x into the other components:

8-4x(-4x)-(-4x)^2-(-4x)=0

so

8+16x^2-16x^2+4x=0

which simplifies to 8+4x=0, so x=-2, therefore y=8.

Substitute x=-2 and y=8 into the third component:

4(-2)^2 + 5(-2)(8) + (-2) + 8 + 8^2 - 6 =
16 - 80 - 2 + 8 + 64 - 6 = 0.

Answer 3

sq. x+y+z you get x^2 + y^2 + z^2 + 2xy + 2yz + 2xz = one hundred Subtract off x^2 + y^2 + z^2 You get 2xy + 2yz + 2xz = 2(xy+yz+xy) = one hundred-40 = 60 2(xy+yz+xy) = 60 xy+yz+xy = 30 extremely not too undesirable as quickly as you sq. the two sides of x+y+z=10

Answer 4

http://www.accd.edu/spc/math/faculty/psmith/1314notes7s1.htm

Use the matrices method. I learned this in college. Hope this helps. It is actually easy when you get the hang of it. you can do mathematical equations in minutes when you learn this method