document.write( "Question 1112145: Separate 800 into three parts, such that the sum of the first, 1/2 of the second, and 2/5 of the third is 400; and the sum of the second, 3/4 of the first, and 1/4 of the third is 400. \n" ); document.write( "
Algebra.Com's Answer #727206 by greenestamps(13203)\"\" \"About 
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\n" ); document.write( "The given information tells us...
\n" ); document.write( "\"x%2By%2Bz+=+800\" (1)
\n" ); document.write( "\"x%2B%281%2F2%29y%2B%282%2F5%29z+=+400\" (2)
\n" ); document.write( "\"%283%2F4%29x%2By%2B%281%2F4%29z+=+400\" (3)

\n" ); document.write( "Solve the system by first eliminating one variable to reduce the problem to two equations in two variables. With the given coefficients, eliminating y first looks to be the easiest.

\n" ); document.write( "\"x%2By%2Bz+=+800\"; \"2x%2By%2B%284%2F5%29z+=+800\"; --> \"-x%2B%281%2F5%29z+=+0\"

\n" ); document.write( "\"x%2By%2Bz+=+800\"; \"%283%2F4%29x%2By%2B%281%2F4%29z+=+400\"; --> \"%281%2F4%29x%2B%283%2F4%29z+=+400\" --> \"x%2B3z+=+1200\"

\n" ); document.write( "Adding the last two equations,
\n" ); document.write( "\"%2816%2F5%29z+=+1600\" --> \"z+=+500\"

\n" ); document.write( "Plugging back into earlier equations, x=100 and y=200.

\n" ); document.write( "CHECK:
\n" ); document.write( "\"100%2B200%2B500+=+800\"
\n" ); document.write( "\"100%2B%281%2F2%29200%2B%282%2F5%29500+=+100%2B100%2B200+=+400\"
\n" ); document.write( "{{(3/4)(100)+200+(1/4)(500) = 75+200+125 = 400}}}
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