document.write( "Question 582312: the width of a rectangle is 15m shorter than its length. if its length were increased by 8 m and its width were diminished by 6 m its area will be decreased by 50 sq. m. find its dimensions. \n" ); document.write( "

Algebra.Com's Answer #372263 by KMST(5328) $\"\"$ $\"About$

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$\"Width+=x=length-15\"$ --> $\"Length=x%2B15\"$
\n" ); document.write( " $\"Area=x%28x%2B15%29=x%5E2%2B15x\"$
\n" ); document.write( " $\"New+length=%28x%2B15%29%2B8=x%2B23\"$ , $\"New+width=x-6\"$
\n" ); document.write( " $\"New+area+=%28x%2B23%29%28x-6%29=x%5E2%2B17x-138=Area-50\"$
\n" ); document.write( " $\"x%5E2%2B17x-138=+x%5E2%2B15x-50\"$ --> $\"17x-138=15x-50\"$ --> $\"17x-138-15x=15x-50-15x+\"$ --> $\"2x-138=-50\"$ --> $\"2x-138%2B138++=-50++%2B138+\"$ --> $\"2x=88+\"$ --> $\"x=88%2F2\"$ --> $\"x=44\"$ --> $\"length=+44%2B15=59\"$
\n" ); document.write( "So $\"highlight%28width=44%29\"$ and $\"highlight%28length=59%29\"$
\n" ); document.write( "VERIFICATION
\n" ); document.write( " $\"Area+=44%2A59=2596\"$
\n" ); document.write( "New length = $\"59%2B8=67\"$ New width = $\"44-6=38\"$ New area= $\"67%2A38=2546=2596-50\"$
\n" ); document.write( " \n" ); document.write( "

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