document.write( "Question 487611: In 2 years, a boy will be ¾ as old as his sister. 2 years ago, he was 2/3 as old as she. How old is the boy now? \n" ); document.write( "
Algebra.Com's Answer #333098 by ptaylor(2198)\"\" \"About 
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Let x=boy's age now
\n" ); document.write( "And let y=sister's age now
\n" ); document.write( "We are told that:
\n" ); document.write( "x+2=(3/4)(y+2) ---------------------eq1
\n" ); document.write( "We are also told that:
\n" ); document.write( "x-2=(2/3)(y-2)----------------------eq2
\n" ); document.write( "simplifying eq1, we get:
\n" ); document.write( "x+2=(3/4)y+6/4 multiply each term by 4:
\n" ); document.write( "4x+8=3y+6
\n" ); document.write( "or 4x-3y=-2------------------------------------eq1a
\n" ); document.write( "Now we simplify eq2:
\n" ); document.write( "x-2=(2/3)y-4/3 multiply each term by 3:
\n" ); document.write( "3x-6=2y-4
\n" ); document.write( "3x-2y=2--------------------------------------eq2a\r
\n" ); document.write( "\n" ); document.write( "Next, we multiply each term in eq1a by 2 and each term in eq2a by 3:
\n" ); document.write( "8x-6y=-4----eq1b
\n" ); document.write( "9x-6y=6------eq2b
\n" ); document.write( "subtract eq1b from eq2b:
\n" ); document.write( "x=10--------------------------------boy's age now
\n" ); document.write( "substitute x=10 into eq 1a:
\n" ); document.write( "40-3y=-2
\n" ); document.write( "-3y=-42
\n" ); document.write( "y=14----sister's age now\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "10+2=(3/4)(14+2)
\n" ); document.write( "12=12
\n" ); document.write( "and
\n" ); document.write( "10-2=(2/3)(14-2)
\n" ); document.write( "8=8\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r
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