SOLUTION: Find two integers whose product is 396 such that one of the integers is ten less than six times the other integer. Please help I do not understand

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Question 993523: Find two integers whose product is 396 such that one of the integers is ten less than six times the other integer. Please help I do not understand
Found 2 solutions by josgarithmetic, Edwin McCravy:
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
Most of the work is literal translation from wording into numbers.

Two numbers, x and y are expected to be integers.


Next descriptive part is
and initially, not clear which of x and y is the greater number.

Continue from there.

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Instead of doing your problem for you, I'll do one exactly
like yours that takes exactly the same steps.  The problem 
I will do is:

Find two integers whose product is 329 such that one of the
integers is nine less than eight times the other integer.
Let the integers be x and y

>>...whose product is 329...<<
So xy = 329

>>...one of the integers is nine less than eight times
the other integer...<<
So we set y equal to 8 times x with 9 subtracted 
from it to make it nine less:

y = 8x - 9 

So we have this system of equations:



Substitute (8x-9) for y in the first 















Using the + sign:





Using the - sign:



.  That's not an integer so we 
disregard it.

Substitute x=7 in 

y = 8x - 9
y = 8(7) - 9
y = 56 - 9
y = 47

So the integers are 7 and 47.

Checking: 

>>...whose product is 329...<<
7*47 = 329

>>...one of the integers is nine less than eight times
the other integer...<<
8 times 7 is 56, and indeed 47 is 9 less than 56.

So it checks.

Now do yours EXACTLY STEP BY STEP like this one.

Edwin

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