Question 58109
Amount of 70% Al: {{{Al[70]}}}
Amount of 54% Al: {{{Al[54]}}}
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The problem tells us to mix the two.  
The total amount of pure aluminum in the mix will be:  (Eq. 1) {{{Al[tot]=Al[70]*(.70)+Al[54]*(.54)}}} 

The total amount of material in the mix {{{Al[70]+Al[54]}}},must equal 640 pounds ==> (Eq. 2) {{{Al[70]+Al[54]=640}}}

We are looking for a final mix of 640 pounds of 65% Al.  Thus, the total aluminum content of the final is:
{{{Al[tot]=640*.65}}}
{{{Al[tot]=416}}}.

Plugging that back into Eq. 1 gives:
{{{416=Al[70]*(.70)+Al[54]*(.54)}}}

Solving Eq. 2 for {{{Al[70]}}}==> {{{Al[70]=640-Al[54]}}}
Plug that into the previous equation:

{{{416=(640-Al[54])*.7+Al[54](.54)}}}
{{{416=448-.7*Al[54]+.54*Al[54]}}}
{{{416=448-.16*Al[54]}}}
{{{416-448=-.16*Al[54]}}}
{{{Al[54]=(416-448)/-.16}}}
{{{highlight(Al[54]=200)}}} pounds.
Plug this into Eq. 2: {{{Al[70]=640-200}}} ==> {{{highlight(Al[70]=440)}}}pounds.