System of 3 equations with 3 variables

**Booh** · Jan 9th 2011, 09:04 AM

I have this system of 3 equations that contain 3 variables:

$900x^4 + 90y^3 - z^3 - 5000 = 0$

$600x^2 - 20y^3 - z^2 + 13000 = 0$

$500x - 20y^2 + z - 600 = 0$

Approximate solutions are:

$x \approx 5.1293908$

$y \approx 10.1340453$

$z \approx 89.2820764$

OR

$x \approx 37.2632316$

$y \approx -31.0102185$

$z \approx 1201.0572216$

How do I solve this system of equations?
One set of solutions is good enough.
Please show all the steps you make.

Thanks.

**snowtea** · Jan 9th 2011, 09:41 AM

It will be tedious, but you can do it this way.
Solve for x in the 3rd equation in terms of y and z.
Substitue for x in the 1st and 2nd.
Solve for z in the 2nd equation (using quadratic formula) in terms of y.
Substitue for z in the 1st equation.
Solve for y.

**Booh** · Jan 9th 2011, 01:27 PM

From the 3rd equation:

$500x = 20y^2 - z + 600$

$\displaystyle x = \frac{20y^2 - z + 600}{500}$

Substituting x in the first equation:

$\displaystyle 900 * (\frac{20y^2 - z + 600}{500})^4 + 90y^3 - z^3 - 5000 = 0$

Substituting x in the 2nd equation:

$\displaystyle 600 * (\frac{20y^2 - z + 600}{500})^2 - 20y^3 - z^2 + 13000 = 0$

I don't know how to continue from here. How do I use the quadratic formula with this equation?

**snowtea** · Jan 9th 2011, 01:52 PM

Originally Posted by Booh

$\displaystyle 600 * (\frac{20y^2 - z + 600}{500})^2 - 20y^3 - z^2 + 13000 = 0$

I don't know how to continue from here. How do I use the quadratic formula with this equation?

$\displaystyle 600 * (\frac{20y^2 - z + 600}{500})^2 - 20y^3 - z^2 + 13000$
Expand the square:
$\displaystyle = \frac{600}{500^2} (400y^4 - 40y^2z + 24000y^2 + z^2 - 1200z + 360000) - 20y^3 - z^2 + 13000$

Now combine like terms, and solve for z (in terms of y).

**Booh** · Jan 10th 2011, 04:51 AM

After multiplying the brackets with $\frac {600}{500^2}$ , I got:

$\displaystyle \frac {24}{25}y^4 - \frac {12}{125}y^2z + \frac {288}{5}y^2 + \frac {3}{1250}z^2 - \frac {72}{25}z + 864 - 20y^3 - z^2 + 13000 = 0$

After combining what possible and ordering appropriately, I got:

$\displaystyle \frac {24}{25}y^4 - 20y^3 - \frac {12}{125}y^2z + \frac {288}{5}y^2 - \frac {1247}{1250}z^2 - \frac {72}{25}z + 13864 = 0$

After multiplying with -1, I got:

$\displaystyle - \frac {24}{25}y^4 + 20y^3 + \frac {12}{125}y^2z - \frac {288}{5}y^2 + \frac {1247}{1250}z^2 + \frac {72}{25}z - 13864 = 0$

Is this correct so far? But this is not a normal quadratic function.
How do I use the quadratic formula with it? Like this?

$\displaystyle \frac { - \frac {72}{25} \pm \sqrt {(\frac {72}{25})^2 - 4 * \frac {1247}{1250} * (-13864)}}{2 * \frac {1247}{1250}}$

But z is here as well: $\displaystyle \frac {12}{125}y^2z$

**snowtea** · Jan 10th 2011, 05:25 AM

$\displaystyle \underbrace{\frac {1247}{1250}}_Az^2 + \underbrace{(\frac {12}{125}y^2 + \frac {72}{25})}_Bz + \underbrace{(- \frac {24}{25}y^4 + 20y^3 - \frac {288}{5}y^2 - 13864)}_C = 0$

Solve the quadratic for z in terms of y.
It's not pretty.

**Booh** · Jan 10th 2011, 12:16 PM

$\displaystyle z_{1,2} = \frac {- (\frac {12}{125}y^2 + \frac {72}{25}) \pm \sqrt {(\frac {12}{125}y^2 + \frac {72}{25})^2 - 4 * \frac {1247}{1250} * (- \frac {24}{25}y^4 + 20y^3 - \frac {288}{5}y^2 - 13864)}} {2 * \frac {1247}{1250}}$

After substituting z in the 1st equation, it did give me the correct value for y.

$y \approx 10.13404534$

But it only worked when using + in the quadratic formula. When I used minus, it didn't give me anything.
This went way too complex to be solved by hand, so I used Maple to solve it.
Then I gave Maple the command to solve it for y, but it gave me a long line of some rubbish.

This is what it gave me:

y = RootOf(-52936978645430292900*_Z^6-293861552997168000*_Z^8+238000711051843200000*_Z^5 +4505714305341120000*_Z^7-2837670767919*_Z^12+2592194400*_Z^14+1438749685008 00*_Z^11+54414726480*_Z^13-3300393830932800*_Z^10+37486248556335980*_Z^9+1296 0000*_Z^16+2592000*_Z^15-6503192508801875193750000-82136005048011600000000*_Z^2-1728614902681710000000*_Z^4+2814998094954067965000 0*_Z^3, index = 9)

This is only a small part of it though. It continues with multiple y = RootOf(...).
Are there any other programs to solve it for y, or is this the only way and I just don't know what this means?

-- EDIT --

Well, after running all that through the "evalf" command, it gave me all the approximate solutions.

Now I just have to figure out how to calculate it manually.

Thanks for all the help snowtea.

**snowtea** · Jan 10th 2011, 01:29 PM

Originally Posted by Booh

Well, after running all that through the "evalf" command, it gave me all the approximate solutions.

Now I just have to figure out how to calculate it manually.

Don't calculate these by hand if you don't have to. Many times you can't even find closed form solutions in terms of roots, fractions, addition and rational numbers.
Most software have algorithms that can get arbitrarily close with approximations.
If you don't trust the answers (good for you), you can always plug the answers back in to check

**snowtea** · Jan 10th 2011, 02:37 PM

Originally Posted by Booh

I have this system of 3 equations that contain 3 variables:

$900x^4 + 90y^3 - z^3 - 5000 = 0$

$600x^2 - 20y^3 - z^2 + 13000 = 0$

$500x - 20y^2 + z - 600 = 0$

As another note, 3 unknowns are not necessarily solved with 3 equations. Sometimes you need more, and sometimes less. As you already noticed in your example you have multiple solutions.
Solving n equations for n unknowns uniquely generally apply to linear equations.

For example, if you were dealing with real numbers, I bet you can solve the following for all unknowns with just one equation:
$x^2 + y^2 + z^2 = 0$

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