Question 1119464
Let x = the larger number

Then  (x)(x-3) = 88
   {{{ x^2 - 3x - 88 = 0 }}}
    {{{ (x-11)(x+8) = 0 }}}

x=11 and/or x=-8  <br>

x=11 —> x-3=8,  11*8 = 88…  {{{ highlight(matrix(1,4,"11 ", " and ", "8"," ")) }}} is a solution<br>

x=-8 —> x-3 = -11, -8*-11=88 …  {{{highlight(matrix(1,4,"-8 ", " and ", "-11"," ")) }}} is a solution<br> 

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Tutor @Alan3354 must have misspoken, as a quadratic equation most certainly may have integer solutions.  
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No.

I said if the 2 solutions are NOT integers, then solving a quadratic is necessary.
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If they are integers, forming a quadratic and then factoring it just adds unneeded lines IF the solutions are integers.
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PS  I intended to edit my answer, but for some reason I'm editing someone else's ???

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To Alan5534 - Yeah, weird, it allowed you to edit my post.   Anyway, I get what you are saying now.  However, I do think forming the quadratic and then solving is a straightforward method that works every time (at least when there is a real answer),  while the guessing at factors may or may not work, depending on the solution.   Thanks.