Question 1166618
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This one is not quite as obvious as the other similar problem (rectangle area 130 with length 3 more than the width; answer length 13, width 10).<br>
However, note in the solution from the other tutor that in solving the problem using formal algebra you have to factor a quadratic equation by finding two numbers whose sum is 24 and whose product is 135.<br>
But that is exactly what the original problem asks you to do -- so the formal algebra doesn't actually get you anywhere towards solving the problem.<br>
You know length plus width is 24; find two numbers with a sum of 24 that have a product of 135.<br>
Obviously one of the numbers has to end in "5"; a very little bit of trial and error finds the dimensions to be 15 and 9.<br>