# Thread: Possible to find "w" in this equation?

1. ## Possible to find "w" in this equation?

My equation is:

1. w(x+y+z)= 276.40
2. w(x+1.5y+z)= 327.53
3. w(1.5x+y+z)= 355.06
4. w(3x+2.7y+3z)= 798.54

the 4th equation can be swapped to w(2x+1.5y+z)= 484.83 if needed

Is it possible to find "w" from the equation above?

3. ## Re: Possible to find "w" in this equation?

I tried to delete in Pre-University forum subjects but could not be succeeded.

4. ## Re: Possible to find "w" in this equation?

The first thing I would do is divide
(1) the first equation by the second equation.
(2) the third equation by the fourth equation.
(3) The fourth equation by the second equation.

Each division eliminates w so you get three linear equations in x, y, and z. Once you have found x, y, and z, put those values into any one of the original equations to solve for w.

5. ## Re: Possible to find "w" in this equation?

$w(x + y + z) = 276.4 \implies w \ne 0,\ (x + y + z) \ne 0,\ and\ w = \dfrac{276.40}{x + y + z} \implies \dfrac{1}{w} = \dfrac{x + y + z}{276.40}$

$\therefore w(x + 1.5y + z) = 327.53 \implies x + 1.5y + z = 327.53 * \dfrac{1}{w} = 327.53 * \dfrac{x + y + z}{276.40} \implies$

$x + 1.5y + z = \dfrac{327.53x}{276.40} + \dfrac{327.53y}{276.40} + \dfrac{327.53z}{276.40} \implies$

$276.4x + 404.6y + 276.4z = 327.53x + 327.53y + 327.53z.$

Simplify that, and repeat the process for the two remaining equations. You are left with a somewhat ugly system of 3 linear equations in 3 unknowns. If that system has a solution, then you can solve for w.

EDIT: To all intents and purposes, this is identical to Halls' answer.

6. ## Re: Possible to find "w" in this equation?

In my previous attempts I took 3 equations into consideration and reached wx, wy & wz shown below:

1/2wy: 327.53-276.40= 51.13 --> wy: 51.13*2= 102.26
1/2wx: 355.06-276.53= 78.66 --> wx: 78.66*2= 157.32 (3rd equation simplified from w(3x+2y+2z)= 710.12 to w(1.5x+y+z)= 355.06 dividing into 2)
wz: 276.40-(102.57+157.32)= 16.83

These values are OK with "w" multiplier but I can not find "w" itself without assigning an estimated value to x, y or z. If 15 which is the right value for "z" is assigned then 16.83/15= 1.12 is true for "w" to calculate x & y as well.
The required values:
y: 102.26/1.12= 91.30 & x: 157.32/1.12= 140.46 & z: 16.83/1.12= 15.02

So still wondering to find "w" without putting an estimated value even in 4 equations.

7. ## Re: Possible to find "w" in this equation?

18 times the first equation minus 3 times the second equation minus 5 times the 4th equation leads to

$0 = -0.09$

so the system has no solution

Thanks Idea.

9. ## Re: Possible to find "w" in this equation?

Originally Posted by zrs
1. w(x+y+z)= 276.40
2. w(x+1.5y+z)= 327.53
3. w(1.5x+y+z)= 355.06
4. w(3x+2.7y+3z)= 798.54
In case helpful, here's how I would attack that problem:
Example:
Assign easy values: w=5, x=2, y=3, z=4
Change some coefficients to make it easier:
(1.5 to 2, 2.7 to 4)

So equations become:
1. w(x+y+z) = 45
2. w(x+2y+z) = 60
3. w(2x+y+z) = 55
4. w(3x+4y+3z) = 150

Now solving above is easier, plus you can check
if you're "correct so far": eliminates loads of
calculations where one slight error
(like using + instead of -)!
results in erroneous solutions.

Once you got it down correct, then simply apply
the same steps to the actual "messy!" equations.