document.write( "Question 701038: (1)/(3)X + (1)/(5)Y =7\r
\n" ); document.write( "\n" ); document.write( "(1)/(6)X - (2)/(5)Y = -4
\n" ); document.write( "using elimination method. \n" ); document.write( "
Algebra.Com's Answer #432171 by checkley79(3341)\"\" \"About 
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(1)/(3)X+(1)/(5)Y=7 MULTIPLY BY 2 & ADD.
\n" ); document.write( "(1)/(6)X-(2)/(5)Y=-4
\n" ); document.write( "(2)/(3)X+(2)/(5)Y=14 ADD.
\n" ); document.write( "----------------------------------
\n" ); document.write( "(1/6+2/3)X=14-4
\n" ); document.write( "(1/6+4/6)X=10
\n" ); document.write( "(5/6)X=10
\n" ); document.write( "5X/6=10 CROSS MULTIPLY
\n" ); document.write( "5X=10*6
\n" ); document.write( "5X=60
\n" ); document.write( "X=60/5
\n" ); document.write( "X=12 ANS.
\n" ); document.write( "(1/3)12+(1/5)Y=7
\n" ); document.write( "4+Y/5=7
\n" ); document.write( "Y/5=7-4
\n" ); document.write( "Y/5=3
\n" ); document.write( "Y=15 ANS.
\n" ); document.write( "PROOF:
\n" ); document.write( "(1)/(3)*12+(1)/(5)*15=7
\n" ); document.write( "4+3=7
\n" ); document.write( "7=7\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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