Question 1062359
The system of equations I need to solve is:
7x - 8y = 24
xy^2 = 1

In the second, it's not (xy)^2, but just the y is squared. I get to substitution and replace y in the second equation with y = (7/8)x + 3 and end up with (23/32)x^3 + (21/4)x^2 + 9x - 1 = 0. My professor warned that this problem is pure evil and said at this point I would need to use rational root theorem and that there IS a rational root. Issue is I don't know how to find a rational root from a fraction as a leading coefficient (23/32).... Unless I screwed up before that and that's not what I'm looking at. 

Any help to get me beyond this point will be greatly appreciated. 
<pre>It's not that bad as your professor professed.
{{{xy^2 = 1}}} ------- eq (ii) ==========> You got up to this point
You have: {{{(7/8)x + 3}}}, but it should be: {{{matrix(1,7, (7/8)x - 3, which, is, the, same, as, 7x/8 - 3)}}}
{{{x(7x/8 - 3)^2 = 1}}} ----- Substituting {{{7x/8 - 3}}} for y in eq (ii)
{{{x(49x^2/64 - 2(21x/8) + 9) = 1}}} ----- FOILing {{{(7x/8 - 3)^2}}}	
{{{x(49x^2/64 - 42x/8 + 9) = 1}}}
{{{x((49x^2 - 336x + 576)/64) = 1}}} ---- Multiplying TRINOMIAL in parentheses by LCD, 64
{{{x(49x^2 - 336x + 576) = 64}}} ------ Cross-multiplying
{{{49x^3 - 336x^2 + 576x = 64}}} ------- Distributing left-side
{{{49x^3 - 336x^2 + 576x - 64 = 0}}} ---- Subtracting 64 from each side
Now is the time you use the rational root theorem with factors of 49 or 64. You will find that 4 is a root, so: x = 4, and x - 4 is a factor.

Using the factor x - 4, and SYNTHETIC DIVISION or LONG DIVISION of POLYNOMIALS, you will find that the QUOTIENT when {{{49x^3 - 336x^2 + 576x - 64}}} is divided by x - 4 is: {{{49x^2 - 140x + 16}}}. 
Therefore, the factors of {{{49x^3 - 336x^2 + 576x - 64}}} are {{{(x- 4)(49x^2 - 140x + 16)}}}. 
You just need to use the quadratic equation formula or COMPLETING THE SQUARE to get the final 2 factors of {{{49x^2 - 140x + 16}}}, which happen to be REAL also (2.73787877 and 0.1193).
You have 3 values for x, so substitute each one into the SIMPLER original equation [eq (i) or (ii)] to find each CORRESPONDING y-value. 

That's it!! The trick here is to do things in STEPS, or LITTLE by LITTLE. That way, you won't be confused and make mistakes.