Solving a System of Equations with 3 Variables

Jimmy84
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Homework Statement



A system of two equations with 3 variables is given.

x + 2y +z = 1
3x - 4y = 2

calculate a thrird equation and add it to the system. the resulting system must have one and only one solution.

Homework Equations







The Attempt at a Solution



my guess is that I must solve the matrix

x + 2y +z = 1
3x - 4y = 2
ax + by+ cz =d

for x, y, z but I have no idea how to show that this system has only one solution.
I really need help with this please. Any advice is welcome, thanks in advance.
 
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Here's a suggestion: write up the augmented matrix. Reduce it to row echelon form. Note that the coefficient of z in the ref must not be zero. Solve the system of 3 equations.

Here's what i got:z=\frac{-8a-b+10d}{-4a-3b+10c}You can take any combinations of a,b,c,d as long as z is not equal to zero.
 
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I got z = (2b +4c)/(b+2d) I am having some difficulties to calculate the augmented matrix with the variables. how did you came to your result?

assuming that I find x, y and z then is that all that the one solution I am ask to find ?
 
Jimmy84 said:

Homework Statement



A system of two equations with 3 variables is given.

x + 2y +z = 1
3x - 4y = 2

calculate a thrird equation and add it to the system. the resulting system must have one and only one solution.

Homework Equations




The Attempt at a Solution



my guess is that I must solve the matrix

x + 2y +z = 1
3x - 4y = 2
ax + by+ cz =d

for x, y, z but I have no idea how to show that this system has only one solution.
I really need help with this please. Any advice is welcome, thanks in advance.

Just do Gaussian elimination. From the first equation you can get x in terms of y and z. Plugging that expression into the second equation, you have an equation containing y and z alone. You can use that equation to solve for y in terms of z (or for z in terms of y if you prefer). Let's say you have y in terms of z; from before, you also have x in terms of y and z, so can use your expression for y(z) to get x in terms of z. So now you have x and y both expressed in terms of z. Substituting your x and y expressions into the third equation gives you an equation involving z alone. You want that equation to have a unique solution for z, so that will give you restrictions on your parameters a, b, c and d.

RGV
 
by solving for z would it mean that the system would have only one solution? after finding z should I also find x and y in terms of a b c n d?

Thanks
 
sharks said:
Here's a suggestion: write up the augmented matrix. Reduce it to row echelon form. Note that the coefficient of z in the ref must not be zero. Solve the system of 3 equations.

Here's what i got:z=\frac{-8a-b+10d}{-4a-3b+10c}You can take any combinations of a,b,c,d as long as z is not equal to zero.

Do I need to solve for x and y in terms of z ?

I'm being ask to calculate a third equation and to add it to the system.

Any advice please?
 
sharks said:
You can take any combinations of a,b,c,d as long as z is not equal to zero.
Try simple substitutions. Let a, b = 0 and let c, d = 1. What do you get for z? Then, what does ax + by+ cz =d become?
 
sharks said:
Try simple substitutions. Let a, b = 0 and let c, d = 1. What do you get for z? Then, what does ax + by+ cz =d become?

why should z not be zero? So 0(x) 0(y) +1(z) =1 where a, b =0 and c, d =1

thanks for the help.
 
Jimmy84 said:
Do I need to solve for x and y in terms of z ?

I'm being ask to calculate a third equation and to add it to the system.

Any advice please?

What part of my previous response did you not understand? Did you actually sit down and DO what I suggested?

RGV
 
  • #10
Ray Vickson said:
What part of my previous response did you not understand? Did you actually sit down and DO what I suggested?

RGV

sorry I was confused justifying how does a,b=0 c,d=1 make a single solution to the system.

Thanks for your time.
 
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