MHB Is it possible to solve such a system of equations?

[zrs]

Hi,

Is it possible to solve such an equation used as a must at my job?

1- x + y + z + ((x + y + z) * x1)= 276.26
2- x + 1,5y + z + ((x + 1,5y + z) * x1)= 327.35
3- 2x + 2y + 2z + ((2x + 2y + 2z) * x1)= 553
4- 3x + 2y + 2z + ((3x + 2y + 2z) * x1)= 709.74

The figures depending the equation above are:

1- 140 + 91 + 15 + ((140 + 91 + 15) * 0.123)= 276.26
2- 140 + 136.50 + 15 + ((140 + 136.50 + 15) * 0.123= 327.35
3- 280 + 182 + 30 + ((280 + 182 + 30) * 0.123)= 552.52
4- 420 + 182 + 30 + ((420 + 182 + 30) * 0.123)= 709.74

Thanks for your kindest help, very appreciated

 

[topsquark]

Re: Possible to solve such an equation?

There is no solution.

For starters, note that each equation factors. For example: (x + y + z)(1 + x1) = 276.26

Also note that the third equation becomes: 2(x + y + z)(1 + x1) = 553, so we know that (x + y + z)(1 + x1) = 553/2.

We are looking at the two equations:
(x + y + z)(1 + x1) = 276.26
(x + y + z)(1 + x1) = 553/2

Looking at this note that the LHS of each equation is equal. But on the RHS we need 553/2 = 276 in both equations, which is not true. So the system cannot be solved.

-Dan

 

[Opalg]

Re: Possible to solve such an equation?

Hi,

Is it possible to solve such an equation used as a must at my job?

1- x + y + z + ((x + y + z) * x1)= 276.26
2- x + 1,5y + z + ((x + 1,5y + z) * x1)= 327.35
3- 2x + 2y + 2z + ((2x + 2y + 2z) * x1)= 553
4- 3x + 2y + 2z + ((3x + 2y + 2z) * x1)= 709.74

The figures depending the equation above are:

1- 140 + 91 + 15 + ((140 + 91 + 15) * 0.123)= 276.26
2- 140 + 136.50 + 15 + ((140 + 136.50 + 15) * 0.123= 327.35
3- 280 + 182 + 30 + ((280 + 182 + 30) * 0.123)= 552.52
4- 420 + 182 + 30 + ((420 + 182 + 30) * 0.123)= 709.74

Thanks for your kindest help, very appreciated

As topsquark points out, equations 1. and 3. are incompatible as they stand. But suppose we change the constant on the right side of equation 3. from 553 to 552.52. Then equation 3. is exactly twice equation 1. So 3. is then redundant (it only tells us what we already knew from 1.). We can therefore jettison 3., but then we are left with only three equations for the four unknowns $x,y,z$ and $x_1$.

You can't expect to get a unique solution for a system where there are more unknowns than equations. But suppose that we already knew the value of one of the unknowns. You can then use that information to solve for the other three unknowns. For example, suppose we knew that $x_1 = 0.123$. I prefer to write $w = 1+x_1 = 1.123$, because then we can write equations 1, 2 and 4 as

$1.\quad w(x+y+z) = 276.26,$
$2.\quad w(x+1.5y+z) = 327.35,$
$4.\quad w(1.5x+y+z) = 354.87.$

If you now subtract 1. from 2. you get $\frac12wy = 51.09$, so that $wy = 102.18.$

Similarly, subtract 1. from 4., getting $\frac12wx = 78.61$, so that $wx = 157.22.$

Then substitute those values into 1., and you find that $wz = 16.86.$

Now, if you use the value $w = 1.123$ to divide those three results by $w$, then you come up with the solutions $x = 140$, $y = 90.98$ and $z = 15.01$, which are very close to those that you quote as the desired solution.

 

[zrs]

Re: Possible to solve such an equation?

Thanks for both replies really very helpful but sorry for my mistake in #3 had to be 552.52 instead of 553.

You are right that #3 is double of #1 and redundant as you stated.

The only estimation can be made for z = 15 but rest are always unpredictable. If I rewrite the equation with considered value of 15 as in below within "opalg" advises, how can be the calculation of equation? I am not good at mathematics and apologize for my weird questions.

w(x + y + 15)= 276.26
w(x + 1.5y + 15)= 327.35
w(1.5x + y + 15)= 354.87

Regards

 

[Opalg]

Re: Possible to solve such an equation?

The only estimation can be made for z = 15 but rest are always unpredictable. If I rewrite the equation with considered value of 15 as in below within "opalg" advises, how can be the calculation of equation? I am not good at mathematics and apologize for my weird questions.

w(x + y + 15)= 276.26
w(x + 1.5y + 15)= 327.35
w(1.5x + y + 15)= 354.87

Solving those equations as in my previous comment, you get $wx = 157.22$, $wy = 102.18$ and $wz = 16.86$. (I had a different value for $wz$, but that was a typo that I have since corrected.)

If you know that $z = 15$ then the equation for $wz$ becomes $15w = 16.86$. This tells you that $w = \frac{16.86}{15} = 1.124$. Put that into the other two equations and you get $x = 139.875$, $y = 90.907$ (almost exactly what you had before).

 

[zrs]

Re: Possible to solve such an equation?

Thanks indeed really resqued me from a very big headache.

This is good enough but wonder your thought to solve the revised equation below with 4 unknowns. I checked some 4 unknowns equations on internet solved with matris process and curious to learn the applicability of the same process or another to this one.

1- w(x + y + z)= 276.26
2- w(x + 1.5y + z)= 327.35
3- w(1.5x + y + z)= 354.87
4- w(2x + 1.5y + z)= 484.57

Sincerely