# Gaussian Elimination (Systems of Equations) Word Problem

• Sep 17th 2013, 08:31 PM
kjtruitt
Gaussian Elimination (Systems of Equations) Word Problem
[vendors] sell an herbal tea blend. By weight, Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile?

I set it up like this: I: x, II: y, III: z

.3x + .4y + 3.5z = 2/3
.4x + .2y + .3z = 2/3
.3x + .4y + 3.5z = 2/3

The way I set it up is that E1 is all the peppermint, E2 is all the rose hips, and E3 is all the chamomile. The question stipulates that these must be in equal proportions in the final 2 pound mix: 3 * 2/3 = 6/3 or 2 pounds. To work the problem I multiplied each through by 100 to clear most of the fractions--leaving 200/3 for the constants. Then proceeded to put the equations in "triangular form". You are ok with E1 and E2 but at E3 you end up with 0 = 0 which tells you that Z is a free variable and you need a parametric solution as explained above. My final equations looked like this:

x + 4/3y + 7/6z = 20/9
y + 1/2z = 26/15
0 = 0

You see that E1 were originally E3 are identical, and where I'm from that leads to a "parametric solution". z is a "free variable" so the equation is "dependent". You set z equal to t, then express x and y in terms of t, then give your solution set via the "set builder" method of describing a set: (-4/45-1/2t, 26/15 - 1/2t, t | 0 < t < 4/3). The first three expressions separated by commas represent x, y and z respectively, and the second part just says "such that t is a real number between 0 and 4/3" or something like that. I actually said -infinity < t < infinity. But my answer is wrong. Book says (4/3 - 1/2t, 2/3 - 1/2t, t) for 0 < t < 4/3).

That's all I have done. Where did I go wrong?
• Sep 17th 2013, 09:02 PM
chiro
Re: Gaussian Elimination (Systems of Equations) Word Problem
Hey kjtruitt.

By using MATLAB, we get the following row-reduced echelon matrix:

>> A = [0.3, 0.4, 3.5; 0.4, 0.2, 0.3; 0.3, 0.4, 3.5]

A =

0.3000 0.4000 3.5000
0.4000 0.2000 0.3000
0.3000 0.4000 3.5000

>> rref(A)

ans =

1.0000 0 -5.8000
0 1.0000 13.1000
0 0 0

This confirms your solution of needing an extra parameter. To solve this we solve the augmented system and we get the following solution:

b = 0.6667 0.6667 0.6667
>> B = [A , transpose(b)]
B =
0.3000 0.4000 3.5000 0.6667
0.4000 0.2000 0.3000 0.6667
0.3000 0.4000 3.5000 0.6667

>> rref(B)
ans =
1.0000 0 -5.8000 1.3333
0 1.0000 13.1000 0.6667
0 0 0 0

This translates into x - 5.8z = 4/3 and y + 13.1z = 2/3. If you set z = t as the book has done you get:

x = 4/3 + 5.8t
y = 2/3 - 13.1t
z = t

This is closer to what you said but its not exactly the same. Given what you said though, this should be answer (this was calculated by a computer and not by hand).
• Sep 18th 2013, 01:15 AM
BobP
Re: Gaussian Elimination (Systems of Equations) Word Problem
First and third equations, the third term should be 0.35z rather than 3.5z.
• Sep 18th 2013, 04:00 AM
kjtruitt
Re: Gaussian Elimination (Systems of Equations) Word Problem
thanks I'll fix it
• Sep 18th 2013, 04:04 AM
kjtruitt
Re: Gaussian Elimination (Systems of Equations) Word Problem
Can't edit my original post anymore so here is the corrected problem:

[vendors] sell an herbal tea blend. By weight, Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile, Type II has percents 40%, 20% and 40%, respectively, and Type III has percents 35%, 30% and 35%, respectively. How much of each Type of tea is needed to make 2 pounds of a new blend of tea that is equal parts peppermint, rose hips and chamomile?

I set it up like this: I: x, II: y, III: z

.3x + .4y + .35z = 2/3
.4x + .2y + .3z = 2/3
.3x + .4y + .35z = 2/3

The way I set it up is that E1 is all the peppermint, E2 is all the rose hips, and E3 is all the chamomile. The question stipulates that these must be in equal proportions in the final 2 pound mix: 3 * 2/3 = 6/3 or 2 pounds. To work the problem I multiplied each through by 100 to clear most of the fractions--leaving 200/3 for the constants. Then proceeded to put the equations in "triangular form". You are ok with E1 and E2 but at E3 you end up with 0 = 0 which tells you that Z is a free variable and you need a parametric solution as explained above. My final equations looked like this:

x + 4/3y + 7/6z = 20/9
y + 1/2z = 26/15
0 = 0

You see that E1 were originally E3 are identical, and where I'm from that leads to a "parametric solution". z is a "free variable" so the equation is "dependent". You set z equal to t, then express x and y in terms of t, then give your solution set via the "set builder" method of describing a set: (-4/45-1/2t, 26/15 - 1/2t, t | 0 < t < 4/3). The first three expressions separated by commas represent x, y and z respectively, and the second part just says "such that t is a real number between 0 and 4/3" or something like that. I actually said -infinity < t < infinity. But my answer is wrong. Book says (4/3 - 1/2t, 2/3 - 1/2t, t) for 0 < t < 4/3).

That's all I have done. Where did I go wrong?
• Sep 18th 2013, 04:39 AM
kjtruitt
Re: Gaussian Elimination (Systems of Equations) Word Problem
Unfortunately we haven't studied matrices (sp?) yet and I don't know what the acronym MATLAB stands for. But given that my solution is similar to the correct solution, I think I must have set the problem up incorrectly. The issue seems to be with the constants. Anyway here is my full working of the problem:

.3x + .4y + .35z = 2/3
.4x + .2y + .3z = 2/3
.3x + .4y + .35z = 2/3 ... (multiply through by 100 to eliminate decimals--my preference)

30x + 40y + 35z = 200/3
40x + 20y + 30z = 200/3
30x + 40y + 35z = 200/3 ... (attempt to put equations in triangular form--divide E1 through by 30 to give the first term a 1 coefficient as required by triangular form)

x + 4/3y + 7/6z = 20/9 ... (multiply E1 through by -40 in order to add it to E2 and eliminate the leading x term in E2)

-40x - 160/3y - 140/3z = -80/9 ... (perform the addition of modified E1 and E2)
+ 40x + 60/3y + 90/3z = 600/9
______________________________
-100/3y - 50/3z = 520/9 ... (make leading coefficient 1 in E2)

y + 1/2z = 26/15

... (Now the system looks like this:)

x + 4/3y + 7/6z = 20/9

y + 1/2z = 26/15

30x + 40y + 35z = 200/3

... (multiply E1 by -30 and add it to E3 to eliminate the x term in E3)

-30x - 40y - 35z = -200/3

+ 30x + 40y + 35z= 200/3
________________________
0 = 0

Z is a "free variable" and the answer must be parameterized. z = t, express x and y in terms of t

x + 4/3(26/15 - 1/2t) + 7/6t = 20/9:
x + 104/45 - 4/6t + 7/6t = 20/9:
x + 104/45 + 1/2t = 20/9:

x = -4/45 - 1/2t
y = 26/15 - 1/2t
z = t

I have found that I had a fractions adding error in the solution I originally posted, but my answer is still definitely incorrect.
• Sep 18th 2013, 04:46 AM
BobP
Re: Gaussian Elimination (Systems of Equations) Word Problem
Your 'final equations', the first one is correct, the second one wrong. (40 times 20 does not equal 80).
Also, you don't need to bother with the third equation, it's the same as the first. You are dealing with two equations in three unknowns.
The book answer is correct, (or at least I got the same answer).
• Sep 18th 2013, 05:18 AM
kjtruitt
Re: Gaussian Elimination (Systems of Equations) Word Problem
Yeah 40 * 20 was it. For some reason when dealing with new concepts my basic math skills diminish.

You mention I don't need the third equation which is identical to the first. Hmm. But if that's true, I could have replaced it with x + y + z = 2, isn't that correct?

• Sep 18th 2013, 05:49 AM
BobP
Re: Gaussian Elimination (Systems of Equations) Word Problem
No, you only ignore the third equation because it's the same as the first, it doesn't give you any extra information.
x + y + z = 2 is a totally different equation and could not be ignored, if it were correct that is.
• Sep 18th 2013, 06:42 AM
kjtruitt
Re: Gaussian Elimination (Systems of Equations) Word Problem
Ok I'm not comfortable with a 'underdetermined' system at this point--and the third equation leading to 0 = 0 tells me that I need to parameterize z in a way that I understand. Anyhow, x, y and z refer to pounds of tea mix of types I, II, and III respectively. If the total resulting mix is to be 2 pounds, the x + y + z = 2 should be a valid equation in this system.
• Sep 18th 2013, 08:12 AM
BobP
Re: Gaussian Elimination (Systems of Equations) Word Problem
x + y + z = 2 is a valid equation in this system, (multiply the first equation by 2 and add the second).
What I was getting at, (and I apologise for the poor phrasing of my response), is that the first and third equations are clearly identical in which case the third (or first) equation can be ignored.
That x + y + z = 2 is a redundant equation may not be immediately apparent, (in which case it couldn't be ignored), and the fact that it is, (giving rise to the 0 = 0), may only become obvious after the reduction to 'triangular form'.