Question 1202656
<pre>
We mix x of the weaker solution with 70 of the stronger to get y of the medium
strength solution:

                  amt.| %  |total 
                   in |as  |amt of
                  ccs |dec.|saline
weaker          |  x  |0.09|0.09x
stronger        | 70  |0.17|0.17(70)
medium strength |  y  |0.13|0.13y

The equations comes from the first and last columns:

{{{system(x+70=y,0.09x+0.17(70)=0.13y)}}}

{{{system(x+70=y,0.09x+11.9=0.13y)}}}

Multiply the 2nd equation through by 100 to clear decimals:

{{{system(x+70=y,9x+1190=13y)}}}

Substitute x+70 for y in the 2nd equation:

9x+1190 = 13(x+70)
9x+1190 = 13x+910
    280 = 4x
     70 = x

So as it turns out, we use 70 ccs of each and get 140 ccs. of
the medium strength.  We could have told that because 13% just 
happens to be exactly half-way between 9% and 17%.  But that's 
just a coincidence.  In another problem that won't be the case. 

Edwin</pre>