Question 346583
given
three alloys;  A1,A2,A3
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weight equation 1: A1+A2+A3=10
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make up of each alloy: Gold, silve and lead ratio 4:3:2 in the first alloy, 3:5:1 in the second, and 2:2:5 in the third
want: It is desired to make an alloy containing equal amounts (by weight) of gold, silver, and lead
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Gold:  4/9*A1+3/9*A2+2/9*A3=10*1/3
silver: 3/9*A1+5/9*A2+2/9*A3=10*1/3
lead: 2/9*A1+1/9*A2+5/9*A3=10*1/3
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multiply the gold, silver and lead equations by 9 to eliminate fractions
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Gold equation 2: 4A1+3A2+2A3=30
Silver equation 3: 3A1+5A2+2A3=30
lead equation 4: 2A1+A2+5A3=30
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subtract equation 3 from equ 2:  A1-2A2=0 or A1=2A2
mult eq 3 by 5  15A1+25A2+10A3=150
mult eq 4 by -2  -4A1-2A2-10A3=-60
add these equtions  11A1+23A2=90
substitute A1=2A2 into above equation
22A2+23A2=90 ---> 45A2=90 or  A2=2
since A1=2A2=4
since A1+A2+A3=10 then A3=4
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Validate answers by taking the original equaitons substituting these values
ie lead: 2/9*A1+1/9*A2+5/9*A3=10*1/3
lead: 2/9*4+1/9*2+5/9*4=10*1/3
8/9+2/9+20/9=10/3
30/9=10/3 checks