Question 993523
<pre>
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:
</pre>
Find two integers whose product is 329 such that one of the 
integers is nine less than eight times the other integer.
<pre>
Let the integers be x and y
</pre>
>>...whose product is 329...<<
<pre>
So xy = 329
</pre>
>>...one of the integers is nine less than eight times 
the other integer...<<
<pre>
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:

{{{system(xy=329,y=8x-9)}}}

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

{{{x(8x-9)=329}}}

{{{8x^2-9x=329}}}

{{{8x^2-9x-329=0}}}

{{{x = (-(-9) +- sqrt( (-9)^2-4(8)(-329) ))/(2*(8)) }}}

{{{x = (9 +- sqrt(81+10528 ))/16 }}}

{{{x = (9 +- sqrt(10609 ))/16 }}}

{{{x = (9 +- 103)/16 }}}

Using the + sign:

{{{x = (9 + 103)/16 }}}
{{{x = 112/16 }}}
{{{x = 7}}}

Using the - sign:

{{{x = (9 - 103)/16 }}}
{{{x = -94/16 }}}
{{{x = -47/8}}}.  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: 
</pre>
>>...whose product is 329...<<
<pre>
7*47 = 329
</pre>
>>...one of the integers is nine less than eight times 
the other integer...<<
<pre>
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</pre>