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So here's the first question I need to be confirmed:

Suppose $u = a + bx + cs$ and $m$ be a fixed number. What does $x$ need to be (as a function of $a, b, c, s$ and $m$) if you need $u > m$ ?

so I've never encountered such a problem but I reckon I might just need to do some inequality:

$u > m$

$a+bx+cs > m$

$\frac {bx}b > \frac{m-a-cs}b$

$ \bf x > \frac {m-a-cs}b$

I'm not really sure if this is the right answer, so I need confirmation, if wrong pls tell me what's wrong or what is the question really asking.

The other question is Suppose $f=\Large{ \frac {t-s}{\frac {s}{n-2}}}$ and $p = \Large\frac {t-s}t$. Express $f$ as a function of $p$ and $n$.

So since $p =\Large \frac{t-s}t$ and $f$ have $(t-s)$ in the numerator, I decided to replace $(t-s)$ in $f$ with $p*t$. so no $f$ is

$ \bf f(p,n) = \Large\frac{(p*t)}{\frac {s} {(n-2)}}$

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I assume you mean $x > ...$ in your last inequality. Anyways, by dividing by $b$ you implicitly assumed that $b \neq 0$. Unless that actually is an unmentioned precondition, you should investigate that case separately. Furthermore take into account that for negative values of $b$ the inequality becomes $bx/b < ...$ after dividing. – mkl Jan 25 '13 at 8:42
Your second approach as you defined the function and the relation looks correct. I assume you are allowed to keep both of $s$ and $t$ in the result. – Babak S. Jan 25 '13 at 8:49
up vote 3 down vote accepted

If $u=a+bx+cs$ wants to be greater than $m$, so we have $$a+bx+cs>m$$ then $$bx+cs>m-a$$ then $$bx>m-a-cs$$ If $b>0$ then $$x>\frac{m-a-cs}{b}$$ and if $b<0$ then $$x<\frac{m-a-cs}{b}$$ Of course $b\neq0$ because we are establishing a definite fraction.

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+1 Nicely done. Step-by-step with every step very easy to follow!, and comprehend! – amWhy Jan 27 '13 at 15:38

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