# Simultaneous Equation with 4 Variables

• Dec 18th 2013, 04:21 PM
Fratricide
Simultaneous Equation with 4 Variables
The system of equations x + y + z + w = 4, x + 3y + 3z = 2, x + y + 2z - w = 6 has infinitely many solutions. Describe this family of solutions (in terms of the parameter t) and give the unique solutions when w = 6.

I got as far as w = (t - 2)/2, but can't seem to go any further.

• Dec 18th 2013, 04:42 PM
romsek
Re: Simultaneous Equation with 4 Variables
Quote:

Originally Posted by Fratricide
The system of equations x + y + z + w = 4, x + 3y + 3z = 2, x + y + 2z - w = 6 has infinitely many solutions. Describe this family of solutions (in terms of the parameter t) and give the unique solutions when w = 6.

I got as far as w = (t - 2)/2, but can't seem to go any further.

could use a bit more of your work here. There's nothing magic you just have to hack away at it getting say z in terms of y and x from the 2nd equation. Then you can use that to reduce the first or third, and then similar with the last.

That will give you everything in terms of x.

Now just replace x with t.

For the last bit you'll have to solve for t when w=6 and then substitute that t in for the expressions for y and z. x of course is just t.

I'm not seeing w=(t-2)/2 from my hack at this so maybe check your work.

Post back some of your cut at this if you like and we can check for any mistakes.
• Dec 19th 2013, 02:49 PM
Fratricide
Re: Simultaneous Equation with 4 Variables
I had another crack at it this morning, but still didn't get any further. I would try substituting w = (t-2)/2 and z = t into the first equation and solving for x and y, but when I substituted those x and y values into the other equations I would always end up with "x - x" or "y - y":

x + y + t + (t-2)/2 = 4
2x + 2y + 2t + t - 2 = 4
2x + 2y + 3t = 6
2x = 6 - 3t - 2y
x = 3 - (3t/2) - y

Substituting back into equation 1 I get:

3 - (3t/2) - y + y + z + w = 4

From there I could only solve for z or w, both of which I already have the values for. I also tried substituting that value for x into the second equation, but turned up nothing.

For our reference, the answers are:

z = t
y = (-3(t + 2))/4
x = (26 - 3t)/4
w = (t - 2)/2

And for the second part of the question:

x = -4, y = -12, z = 14
• Dec 19th 2013, 04:05 PM
romsek
Re: Simultaneous Equation with 4 Variables
Quote:

Originally Posted by Fratricide
I had another crack at it this morning, but still didn't get any further. I would try substituting w = (t-2)/2 and z = t into the first equation and solving for x and y, but when I substituted those x and y values into the other equations I would always end up with "x - x" or "y - y":

x + y + t + (t-2)/2 = 4
2x + 2y + 2t + t - 2 = 4
2x + 2y + 3t = 6
2x = 6 - 3t - 2y
x = 3 - (3t/2) - y

Substituting back into equation 1 I get:

3 - (3t/2) - y + y + z + w = 4

From there I could only solve for z or w, both of which I already have the values for. I also tried substituting that value for x into the second equation, but turned up nothing.

For our reference, the answers are:

z = t
y = (-3(t + 2))/4
x = (26 - 3t)/4
w = (t - 2)/2

And for the second part of the question:

x = -4, y = -12, z = 14

Well let's see. We want to just bang away at it keeping things in terms of z since that's going to become our independent variable.

x + y + z + w = 4
x +3y+3z = 2
x + y +2z - w = 6

taking the 2nd equation

x = 2 - 3y - 3z

plug this into equation 1

(2 - 3y - 3z) + y + z + w = 4

y(-3+1) + z(-3+1) + 2 + w = 4

-2y - 2z + w = 2

w = 2 +2y + 2z

plug both into equation 3

(2-3y-3z) + y + 2z - (2+2y+2z) = 6

(2-2) + y(-3 + 1 + -2) + z(-3 + 2 -2) = 6

-4y -3z = 6

y = -(6+3z)/4

and collecting everything and letting z=t we get

y = -(6+3t)/4

x = 2 - 3y - 3z = 2 -3(6+3t)/4 - 3t = (26-3t)/4

w = 2 + 2y + 2z = 2 -2(6+3t)/4 +2t = (t-2)/2

{x, y, z, t} = {(26-3t)/4, -(6+3t)/4, t, (t-2)/2}

I think you just need to be careful collecting terms.
• Dec 21st 2013, 07:32 PM
Wilmer
Re: Simultaneous Equation with 4 Variables
Quote:

Originally Posted by Fratricide
x + y + z + w = 4
x + 3y + 3z = 2
x + y + 2z - w = 6
give the unique solutions when w = 6.

After substituting 6 for w:
x + y + z = -2 [1]
x + 3y + 3z = 2 [2]
x + y + 2z = 12 [3]

Subtract [1] from [3]:
x - x + y - y + 2z - z = 12 - (-2)
z = 14

Now you're almost done...