Question 1104297
<br>
Both of the other tutors showed the setup for an algebraic solution to the problem.<br>
However, notice that, in the end, you had to do some trial and error to finish solving the problem algebraically.<br>
So if your objective is to find the answer as quickly as possible by any method (as, for example, on a competitive timed test), then trial and error from the beginning might be the best path to the answer.<br>
Simply look for two numbers for the length and width whose product is 540 which satisfy the condition that the length is 15 less than 5 times the width:<br>
10*54 = 540; 5(10)-15 = 35
The value we get for the length using 10 for the width is too small; we need a larger value for the width.<br>
15*36 = 540; 5(15)-15 = 60
Now the value we get for the length using 15 for the width is too big; the width must be between our first guess of 10 and our second guess of 15.<br>
12*45 = 540; 5(12)-15 = 45
AHA! The length of the rectangle is 45.