document.write( "Question 328351: The perimeter of a rectangle is 68 cm. If the width is increased by 2cm. and twice th length is decreased by 10cm, the new perimeter is 100 cm. What are the dimensions of the original rectangle? \n" ); document.write( "
Algebra.Com's Answer #235167 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
let width be x
\n" ); document.write( "length = y
\n" ); document.write( "2(x+y)=68
\n" ); document.write( "x+y=34.................1
\n" ); document.write( "..
\n" ); document.write( "x+2 new width
\n" ); document.write( "new length = 2y-10
\n" ); document.write( "2(x+2)+2(2y-10)=100
\n" ); document.write( "2x+4+4y-20=100
\n" ); document.write( "2x+4y-16=100
\n" ); document.write( "2x+4y=100+16
\n" ); document.write( "2x+4y=116
\n" ); document.write( "x+2y=58................2
\n" ); document.write( "subtract the equations
\n" ); document.write( "x+y-x-2y=34-58
\n" ); document.write( "-y=-24
\n" ); document.write( "y=24cm the width.
\n" ); document.write( "plug the value of y in equation 1
\n" ); document.write( "x+24=34
\n" ); document.write( "x=10cm the length\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );