1. ## Mixture problem - end up with a negative answer???

One alloy is one part copper to three parts tin. A second alloy is one part copper to four parts tin. How much of a second alloy shold be combined with 24 pounds of the first allow to obtain a new alloy that is two parts copper to seven parts tin?

Code:
            Amount (lbs)  % Copper  Total Copper
Alloy 1   24                .25           .25(24)
Alloy 2   x                  .20           .20x
Mix        24+x             2/7           2/7(24+x)
And then got:

$\displaystyle .25(24) + .20x = .29(24+x)$
$\displaystyle 6 + .20x = 6.96 + .29x$
$\displaystyle -.96 = .09x$
$\displaystyle -10.67 = x$

Where did I go wrong?

2. Originally Posted by cdbmath

Code:
            Amount (lbs)  % Copper  Total Copper
Alloy 1   24                .25           .25(24)
Alloy 2   x                  .20           .20x
Mix        24+x             2/7           2/7(24+x)
And then got:

$\displaystyle .25(24) + .20x = .29(24+x)$ <<<<<< here
$\displaystyle 6 + .20x = 6.96 + .29x$
$\displaystyle -.96 = .09x$
$\displaystyle -10.67 = x$

Where did I go wrong?
The first line must read like this:

$\displaystyle .25(24) + .20x = \dfrac29 \cdot (24+x)$

And $\displaystyle \dfrac29 \neq 0.29$

You should come out with x = 30