Hi, and welcome to the forum!
Please familiarize yourself with the http://mathhelpboards.com/rules/, especially rules #8 and 11. As you see, there is some conflict between the requirement that you show some effort and your shortage of time before the test. This may not be the perfect resource for the last-minute help, but let us try to find a middle ground.
I am sure you have a textbook that describes how to solve these types of problems, and you have been presented with example solutions. I think the best you can do is to go back and read that material. Why do you think that what someone writes here would be better than explanations in your textbook? Now, if those explanations are unclear, then please describe your difficulty and we would be happy to help. Describing your problem and showing what you do and don't understand definitely counts as an effort.
I believe many formulas you wrote are not what you in fact intended because of the order of operations. For example, in problem 7 you must mean f$(x)= x/(4x+3)$, that is, $f(x)=\dfrac{x}{4x+3}$. As it is written, $f(x)= x/4x+3$ means $f(x)=\dfrac{x}{4}x+3$. Similarly, $3x/10x$ in problem 4 strictly speaking means $\dfrac{3x}{10}x=\dfrac{3x^2}{10}$ because multiplication and division are done left to right. You don't need to know how to write fractions vertically because all formulas can be written correctly as plain text using parentheses. For example, in problem 2 you probably mean √(3x+1) and not (√3)x+1. Please rewrite your problems putting no more than two in one thread.
One solution.
Jinxxerz said:
3. Find the LCM: 35x, 5x^2, 7x^3
First, recall how to divide two monomials. If $a,b$ are real numbers and $m,n$ are integers, then
\[
\frac{ax^m}{bx^n}=\frac{a}{b}\cdot\frac{x^m}{x^n}=\frac{a}{b}x^{m-n}\qquad(*)
\]
If $m\ge n$, then the resulting expression is a monomial, but when $m<n$, we say that $bx^n$ does not divide $ax^m$. For example, $3x^2$ divides $6x^3$ and $(6x^3)/(3x^2)=2x$, but $3x^4$ does not divides $6x^3$. If $a$ and $b$ in (*) are integers and $b$ does not divide $a$ but $m\ge n$, then we can still divide the monomials, but the coefficient would be a fraction: e.g., $(6x^3)/(4x^2)=(3/2)x$.
Thus, to find the least common multiple of several monomials, we need to find the maximum degree of $x$ among those monomials. With respect to the coefficient, it is a bit tricky because I don't know the exact definition you were given. The LCM is divisible by every monomial. But in considering (*) above, do we require that $b$ divides $a$ or not? E.g., do we say that the LCM of $6x^3$ and $4x^2$ is $6x^3$ because it is divisible by $4x^2$, even though the resulting coefficient is $6/4=3/2$? Or do we say that the LCM of $6x^3$ and $4x^2$ is $12x^3$ because $(12x^3)/(6x^3)=2$ and $(12x^3)/(4x^2)=3x$, and in both cases the coefficients are integers?
If we require that coefficients are divided evenly (which is more likely), then the LCM of $35x$, $5x^2$ and $7x^3$ is $35x^3$. Indeed, $(35x^3)/(35x)=x^2$, $(35x^3)/(5x^2)=7x$ and $(35x^3)/(7x^3)=5$. The power of $x$ cannot be less than 3 because the LCM must be divisible by $7x^3$. If we don't require that the coefficients divide evenly, then the LCM is $x^3$ or $cx^3$ for any non-zero real number $c$ because $x^3/(35x)=(1/35)x^2$, $x^3/(5x^2)=(1/5)x$ and $x^3/(7x^3)=1/7$. These are still monomials, though with rational, instead of integer, coefficients.
You see that it would be more efficient to consult a textbook. (Smile)
