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21. Find two numbers where three times the smaller number exceeds
the larger by 5 and the sum of the numbers is 11.
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Answer :

Answers:  4 and 7

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Work Shown

p = smaller number

q = larger number

3p = q+5 since "three times the smaller number exceeds the larger by 5".

That solves to q = 3p-5 which will be useful in the substitution step shown below.

We're told "the sum of the numbers is 11", so,

p+q = 11

p+3p-5 = 11 ...... substitution; replace q with 3p-5

4p-5 = 11

4p = 11+5

4p = 16

p = 16/4

p = 4 is the smaller number.

q = 3p-5 = 3*4-5 = 12-5 = 7 is the larger number.

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Check:

  • The condition 3*smaller = larger + 5 is met since 3*4 = 7+5 is true.
  • smaller + larger = 4+7 = 11 showing the two items sum to 11.

Both conditions have been verified.

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