Question 983243
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Instead of doing your problem for you, I will do one EXACTLY like
it so you can do your problem using this as a model to go by.
I will do this problem instead:
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one positive number is 6 more than three times another. if their product is 189,
find the numbers.

>>one positive number is 6 more than three times another. 
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Let x = one positive number
Let y = the other number.

x = 3y + 6
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>>if their product is 189 , find the numbers.
<pre>
xy = 189

The first one is already solved for x, so we substitute
(3y + 6) for x in the second equation.

(3y + 6)y = 189
y(3y + 6) = 189
3y² + 6y = 189
3y² + 6y - 189 = 0 

Divide through by 3

y² + 2y - 63 = 0

(y+9)(y-7) = 0

y+9 = 0;  y-7 = 0
  y = -9    y = 7

Since the number must be positive we ignore -9

So y = 7

The other number = x = 3y + 6 = 3(7) + 6 = 21 + 6 = 27

Now you do yours exactly the same way.

Edwin</pre>