Question 107664
You want to know how much 20% copper alloy, so let that be x.  We also don't know the total amount of 30% alloy, so let that be y.


Now, let's figure out what we do know.


The amount of 20% alloy plus the 200 oz. of 50% alloy must equal the amount of 30% alloy, so:


Eq 1) {{{x+200=y}}}


We also know that the amount of copper in the 20% alloy is 0.2x, the amount of copper in the 50% alloy is 50% of 200 oz, or 100 oz, and the amount of copper in the 30% alloy must be 0.3y.  We also know that the copper in the 20% alloy plus the copper in the 50% alloy must equal the copper in the 30% alloy.  Therefore:


Eq 2) {{{0.2x+100=0.3y}}}


Now it is a matter of solving two simultaneous linear equations in two variables.  There are a number of methods, but this is the one I like.  First re-arrange the equations to standard form, {{{ax+by+c=0}}}.  Or, in this case:


{{{x-y+200=0}}} and {{{0.2x-0.3y+100=0}}}


Then I multiply the 2nd equation by -5:  {{{-x+1.5y-500=0}}}


Then add the result to the first equation, term by term:  {{{0x+0.5y-300=0}}}


{{{0.5y=300}}}
{{{y=600}}}


Then since {{{x+200=600}}}


{{{x=400}}}ounces.



Check:

{{{400+200=600}}} and

{{{0.2*400+0.5*200=0.3*600}}}
{{{80 + 100 = 180}}} Check!