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?

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
$c\leq \frac43$
