# Why does $r = \frac 23 - \frac 12 T$ rather than $r = \frac 23$?

I have the following question. Using Gaussian elimination, I got an answer different from the book's. Can anyone please tell me what I'm doing wrong?

At The Crispy Critter's Head Shop and Patchouli Emporium, along with their dried up weeds, sunflower seeds and astrological postcards, they sell an herbal tea blend. By weight,

• Type I herbal tea is 30% peppermint, 40% rose hips and 30% chamomile.

• Type II herbal tea is 40% peppermint, 20% rose hips and 40% chamomile.

• Type III herbal tea is 35% peppermint, 30% rose hips and 35% chamomile.

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?

Here are my steps:

Matrix:

$$(E1) \frac{3}{10}p +\frac{4}{10}r + \frac{3}{10}c = \frac{2}{3}$$

$$(E2) \frac{4}{10}p +\frac{2}{10}r + \frac{4}{10}c = \frac{2}{3}$$

$$(E3) \frac{35}{100}p +\frac{3}{10}r + \frac{35}{100}c = \frac{2}{3}$$

I now make the leading coefficient in E1 a 1:

$$(new E1) p +\frac{4}{3}r + c = \frac{20}{9}$$

I eliminate p from E2:

$$(E1) -\frac{4}{10}p -\frac{16}{30}r - \frac{4}{10}c = -\frac{8}{9}$$

$$(E2) \frac{4}{10}p +\frac{2}{10}r + \frac{4}{10}c = \frac{2}{3}$$

and get new E2:

$$(new E2) r = \frac{2}{3}$$

I eliminate p from E3:

$$(E3) -\frac{35}{100}p -\frac{140}{300}r - \frac{35}{100}c = -\frac{7}{9}$$

$$(E3) \frac{35}{100}p +\frac{90}{300}r + \frac{35}{100}c = \frac{6}{9}$$

and get new E3:

$$(new E3) r = \frac{2}{3}$$

So I now have the following matrix:

$$(new E1) p +\frac{4}{3}r + c = \frac{20}{9}$$

$$(new E2) r = \frac{2}{3}$$

$$(new E3) r = \frac{2}{3}$$

Now this leaves c as a free variable so I set it equal to T and I get c = t, $$r = \frac{2}{3}$$, and $$p = \frac{4}{3} - T$$

But the book says the answer is c = t, $$r = \frac{2}{3} - \frac{1}{2}T$$, and $$p = \frac{4}{3} - \frac{1}{2}T$$

Any ideas what I did wrong?

• Where did the 2/3 on the Right-hand-side come from? – NoChance Sep 23 '19 at 11:24
• Related to: math.stackexchange.com/questions/2202515/… – NoChance Sep 23 '19 at 11:27
• 2/3 on the right side is there because we add T1 and T2 and T3 together in equal parts to get 2 pounds of the new blend. So each one is 2/3rds of a pound. – maybedave Sep 23 '19 at 11:46
• Yes, I read that other stackexchange article but it doesn't show the actual answers they got. I'm getting a different answer from the book. For some reason the book is getting r = 2/3 - 1/2T but c is eliminated at the same time p is as I show in the conversion to "new E2" above. – maybedave Sep 23 '19 at 11:48
• But you are not using equal parts to form the blends, you are given specific percentages for each type. You are supposed to use equal parts ONLY for the new blend. Hence the 2/3 on the R.H.S is not clear to me. – NoChance Sep 23 '19 at 12:06

My approach is similar to Eriins. First of all we have to define the variables:

a: amount of tea mixture type 1 in pounds

b: amount of tea mixture type 2 in pounds

c: amount of tea mixture type 3 in pounds

Next we have to notice that the equations are dependent. That means we can omit one equation (here: the third equation) and replace it by $$a+b+c=2$$

$$\begin{eqnarray} \textrm{peppermint constraint} \\ \frac{3}{10}a + \frac{4}{10}b + \frac{35}{100}c = 2/3\\ \textrm{hips constraint} \\ \frac{4}{10}a + \frac{2}{10}b + \frac{3}{10}c = 2/3\\ \textrm{The sum of all three tea mixtures is 2 pounds} \\ a + b + c = 2\\ \end{eqnarray}$$

The solution is $$a=\frac43-\frac{c}2, b=\frac23-\frac{c}2, c=c$$

Since we have the additional condition that $$a,b,c\geq 0$$, we can deduce

$$c\leq \frac43$$

• When setup like that, I'm struggling to understand what the coefficients represent. You have 'a' is an entire tea type comprised of 30% peppermint, 40% rose hip and 30% chamomile. So what does the 3/10s represent of tea 'a' and what does the 4/10s represent of tea 'b'...? I'm just trying to see how you came up with this matrix. – maybedave Sep 23 '19 at 13:40
• What do you mean by 3/10s for instance? The first equation is the peppermint-constaint. With the ombination of the three types of tea mixtures of a,b and c we have to fulfil the condition that the amount of peppermint is 2/3 (pounds). – callculus Sep 23 '19 at 13:51
• ok, there are teas and there are ingredient. You defined variable 'a' as the entire Tea mixture 1 so by that definition a = 3/10p + 4/10r + 3/10c. But I see now you are defining a = to the percentage of peppermint in Tea mixture 1 (in the first equation). Great! But why do you have 35/100c in the first equation when Tea mixture 1 contains 3/10c? – maybedave Sep 23 '19 at 14:08
• a = 3/10p + 4/10r + 3/10c this is not true, no matter how you define p,r and c. Please concentrate on the amount of peppermint at the first equation. We have 3/10 peppermint in mixture 1. And a is the amount of mixture 1. Thus $3/10a$ is the amount of peppermint which we get from mixture a. We have 4/10 peppermint in mixture 2. And b is the amount of mixture 2. Thus $4/10b$ is the amount of peppermint which we get from mixture b. We have 35/100 peppermint in mixture 3. And c is the amount of mixture 1. Thus $35/100c$ is the amount of peppermint which we get from mixture c. – callculus Sep 23 '19 at 14:29
• Then finally the sum is the sum of peppermint we get. I can comprehend that it might by difficult to think in this kind of way. But there is no other way. – callculus Sep 23 '19 at 14:29

Your equations are the wrong way round, it should be $$\begin{eqnarray} \frac{3}{10}a + \frac{4}{10}b + \frac{35}{100}c = 2/3\\ \frac{4}{10}a + \frac{2}{10}b + \frac{3}{10}c = 2/3\\ \frac{3}{10}a + \frac{4}{10}b + \frac{35}{100}c = 2/3 \\ \end{eqnarray}$$ Where $$a$$ is tea type 1, $$b$$ is tea type 2 and $$c$$ is tea type 3.

This way you sum up the herbs in the different teas and get 2/3 pounds of each.