document.write( "Question 288318: find the smallest number which leaves remainder 8 and 12 when divided by 28 & 32, respectively. \n" ); document.write( "
Algebra.Com's Answer #208981 by richwmiller(17219)\"\" \"About 
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28x+8=z
\n" ); document.write( "32y+12=z\r
\n" ); document.write( "\n" ); document.write( "{4(7x+2)= z,4(8y+3)=z}
\n" ); document.write( "7x+2=8y+3
\n" ); document.write( "7x=8y+1
\n" ); document.write( "7x-1=8y
\n" ); document.write( "(7x-1)/8=y
\n" ); document.write( "x=(8y+1)/7
\n" ); document.write( "x=8n-1, y=7n-1
\n" ); document.write( "let n=1
\n" ); document.write( "x=8-1
\n" ); document.write( "y=7-1
\n" ); document.write( "x=7 y=6
\n" ); document.write( "28*7+8=32*6+12=204
\n" ); document.write( "check
\n" ); document.write( "204-8=196/28=7
\n" ); document.write( "204-12=192/32=6
\n" ); document.write( "ok
\n" ); document.write( "Answer 204
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