Solving 3 unknowns!?

Question

Hi,

I'm having lot's of trouble trying to solve a problem.

I have to figure out the 3 unknowns; x, y and z.

Can someone please provide me with a simple way of how to do it?

The problem is listed below:

40x + 50y + 20z = 7000

24x + 10y + 10z = 3150

10x + 15y + 10z = 2000

P.S. thank you in advance! x

Answer 1

I would use elimination so lets label these as equation 1, 2 ...

1) 40x + 50y + 20z = 7000

2) 24x + 10y + 10z = 3150

3) 10x + 15y + 10z = 2000

First I would use elimination process for equation 1 and 2 to eliminate z (the resulting equation will be called 4), then I would use equations 1 and 3 to eliminate z (the resulting equation will be called 5),

Now use equations 4 and 5 to eliminate x (if you want then do y...its really the same thing). So then you are left with a Y value...now go back to Equations 4 and 5 and sub in the y value you got, after that you can get the x value.

Now go to equation 1 and sub in the x and y values and then you can get the z value.

2 unknows = 2 equations,

3 unknowns = 3 equations.

Msg me if you are still confused.

Answer 2

What you need to do is find definitions for one variable in terms of the other two, then substitute these back into two of the equations, to then have two equations and two unknowns.

It sounds worse than it is.....

Step 1)

Subtract equation 3 from equation 2...

24x + 10y + 10z = 3150

10x + 15y + 10z = 2000

-----------------------------------

14x - 5y + 0z = 1150

Now solve for x

14x = 1150 + 5y

x = 575/7 + 5/14y

Plug this value into the first two equations

40(575/7 + 5/14y) + 50y + 20z = 7000

24(575/7 + 5/14y) + 10y + 10z = 3150

23000/7 + 100/7y + 50y + 20z = 7000

23000 + 100y + 350y + 140z = 49000

450y + 140z = 26000

45y + 14z = 2600

24(575/7 + 5/14y) + 10y + 10z = 3150

13800/7 + 60/7y + 10y + 10z = 3150

13800 + 60y + 70y + 70z = 22050

130y + 70z = 8250

13y + 7z = 825

So now you have two equations and two unknowns...

45y + 14z = 2600

13y + 7z = 825

I would double the second equation and subtract

45y + 14z = 2600

26y + 14z = 1650

-------------------------

19y + 0z = 950

19y = 950

y = 50

remember that you have the following equation...

x = 575/7 + 5/14y

x = 575/7 + 5/14*50

x = 575/7 + 125y

x = 700/7

x = 100

So x = 100, y = 50

Plug these back into the third equation...

10(100) + 15(50) + 10z = 2000

1000 + 750 + 10z = 2000

1750 + 10z = 2000

10z = 250

z = 25

So x = 100, y = 50 and z = 25

Check these values with each equation...

40(100) + 50(50) + 20(25) = 7000

4000 + 2500 + 500 = 7000

7000 = 7000 CHECK

24(100) + 10(50) + 10(25) = 3150

2400 + 500 + 250 = 3150

3150 = 3150 CHECK

10(100) + 15(50) + 10(25) = 2000

1000 + 750 + 250 = 2000

2000 = 2000 CHECK

So your answer is confirmed...

x = 100, y = 50 and z = 25