document.write( "Question 71813: please try to solve for all, but if you don't have time try for b for sure. :D
\n" ); document.write( "(0.01a+0.05b+0.10c+0.25d)/(a+b+c+d)=(0.01(a-1)+0.05b+0.10c+0.25d)/((a-1)+b+c+d) \n" ); document.write( "
Algebra.Com's Answer #51347 by venugopalramana(3286)\"\" \"About 
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please try to solve for all, but if you don't have time try for b for sure. :D
\n" ); document.write( "(0.01a+0.05b+0.10c+0.25d)/(a+b+c+d)=(0.01(a-1)+0.05b+0.10c+0.25d)/((a-1)+b+c+d)
\n" ); document.write( "CROSS MULTIPLYING..
\n" ); document.write( "[0.01a+0.05b+0.10c+0.25d)(a+b+c+d-1)]=[0.01(a-1)+0.05b+0.10c+0.25d][a+b+c+d)
\n" ); document.write( "[0.01a+0.05b+0.10c+0.25d][a+b+c+d]-[0.01a+0.05b+0.10c+0.25d]=
\n" ); document.write( "=[0.01a+0.05b+0.10c+0.25d)[a+b+c+d]-[a+b+c+d)
\n" ); document.write( "0.01a+0.5b+0.1c+0.25d=a+b+c+d
\n" ); document.write( "0.99a+0.5b+0.9c+0.75d=0
\n" ); document.write( "hence you can have any value for 3 of the 4 variables a,b,c and d....,the 4th.variable,dependent on the other 3...say
\n" ); document.write( "you can have a=b=c=0 say
\n" ); document.write( "then d=0
\n" ); document.write( "if a=0=b,c=1
\n" ); document.write( "d=-0.9/0.75=-0.3/0.25 =-1.2..and in general
\n" ); document.write( "d=-(0.99a+0.5b+0.9c)/0.75,for any value of a,b, and c will be a solution set. \n" ); document.write( "
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