document.write( "Question 951917: The length of a rectangular room is 5 yd more than the width. If the area is 300 yd2, find the length and the width of the room. \n" ); document.write( "
Algebra.Com's Answer #581413 by addingup(3677)\"\" \"About 
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Let's call width \"w\", and the length \"l\" is w+5. Formula for the area is w*l:\r
\n" ); document.write( "\n" ); document.write( "w*(w+5) = 300 Multiply on the left:
\n" ); document.write( "w^2 + 5w = 300 Subtract 300 from both sides, and I'm going to rearrange the terms:
\n" ); document.write( "-300 + 5w + w^2 = 0 Factor and you get:
\n" ); document.write( "(-20 + -1w)(15 + -1w) = 0 Now we have two equations, let's solve for each:\r
\n" ); document.write( "\n" ); document.write( "-20 + -1w = 0 and 15 + -1w = 0 Solve the first equation:
\n" ); document.write( "-20 + -1w = 0 Add 20 to both sides:
\n" ); document.write( "-1w = 20 Divide both sides by -1:
\n" ); document.write( "w = -20 \r
\n" ); document.write( "\n" ); document.write( "Now second equation:
\n" ); document.write( "15 + -1w = 0 subtract 15, both sides
\n" ); document.write( "-1w = -15 Divide both sides by -1:
\n" ); document.write( "w = 15
\n" ); document.write( "Our answer is w = {-20, 15} Since -20 is negative we cannot use it. Our answer has to be 15 for the width, And the length is 5 more:
\n" ); document.write( "Proof: 15 x (15 + 5) = 15 x 20 = 300 Our answer is correct. \n" ); document.write( "
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