Question 947090
x and y for first and second powders


First Powder:  10% B1, 30% B2
Second Powder:  15% B1, 30% B2
Mixture:  80 mg B1, 200 mg B2


Account for B1:
{{{0.1x+0.15y=80}}}


Account for B2:
{{{0.3x+0.3y=200}}}


Multiply B1 equation by 100 and multiply B2 equation by 10 to form a simplified system
of equations.
{{{system(10x+15y=8000,3x+3y=2000)}}}
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Further simplify the first equation, dividing members by 5.
{{{system(2x+3y=1600,3x+3y=2000)}}}
You can use any method you know or want for solving the system.  Elimination would be
easiest to start with.  Just subtract the first equation from the second equation to
eliminate terms of y. Immediatly find:
{{{highlight(x=400)}}}, mg of the first powder.
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{{{3y=1600-2x}}}
{{{y=(1600-2x)/3}}}
{{{y=(1600-2*400)/3}}}
{{{highlight(y=800)}}}, mg of second powder.