Improve Your Algebra Skills with Practice Problems

$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$
Please help me to find $a,b,c,d$ such that $\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$ ? I take $a,b$ for example $2,3$ to find $c,d$ but I get stuck on this. I know that is diophantine equation , but I am unable to find suitable $a,b,c,d$
$a,b,c,d\in <!--\; \in \; -->\mathbb{Z},a,c\ne <!--\; \ne \; -->0$
Thanks in advance.

Represent improper fraction as a sum of unique unit fractions
Is it possible to represent an improper fraction as a finite sum of unique unit fractions (Egyptian fractions)?

Can we find integers x and y such that f,g,h are strictely positive integers
Let $a>2$ and $b>2$ two strictely positive integers. Let us consider the following quantities:
$$f={\displaystyle \frac{xy+ay+a^2}{by}}$$
$$g=a(y+a){\displaystyle \frac{xy+ay+a^2}{b{y}^{2}x}}$$
$$h={\displaystyle \frac{y+a}{b}}$$
My question is:
Can we find integers x and y (not necessarly positive) such that f,g,h are strictely positive integers. Or at lest how one can proves that they are exist.

How can this expression be calculated? ${\displaystyle \frac{1+{\displaystyle \frac{3\cdots <!--\; \cdots \; -->}{4\cdots <!--\; \cdots \; -->}}}{2+{\displaystyle \frac{5\cdots <!--\; \cdots \; -->}{6\cdots <!--\; \cdots \; -->}}}}$
I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...
${\displaystyle \frac{1+{\displaystyle \frac{3}{4}}}{2+{\displaystyle \frac{5}{6}}}}\approx <!--\; \approx \; -->0.6176$
${\displaystyle \frac{1+{\displaystyle \frac{3+{\displaystyle \frac{7}{8}}}{4+{\displaystyle \frac{9}{10}}}}}{2+{\displaystyle \frac{5+{\displaystyle \frac{11}{12}}}{6+{\displaystyle \frac{13}{14}}}}}}\approx <!--\; \approx \; -->0.6175$
Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge.
Is it possible to find a value for this? Will it have a closed form solution?

Simplifying Fractions Containing Variables (Basic)
I appreciate this is very simple, but I'm experiencing a very basic problem with fractions containing variables and I'd just like to check I'm along the right lines. In the following instance:
$\frac{2(x+7)(3x+1)}{2}=\frac{2}{2}\cdot <!--\; \cdot \; -->\frac{2(x+7)(3x+1)}{1}=(x+7)(3x+1)$
Does the $\frac{2(x+7)(3x+1)}{1}$ have a denominator of 1 because we have factored out the 2 in the prior step?

Suppose that y is inversely proportional to x, and that y=0.6 when x=2.0. Find y when x=4.2

Problem proving inequality $\frac{4^n}{n+1}\le <!--\; \le \; -->\frac{(2n)!}{(n!{)}^2}$
I skip the base case $n=0$ because it's obvious.
I know that this is very equivalent to: Prove by induction: $\frac{2^{2n}}{n+1}<\frac{(2n)!}{(n!{)}^2},n>1$
But I try to learn some tricks so maybe you can help me?
Assumption: $\frac{4^n}{n+1}\le <!--\; \le \; -->\frac{(2n)!}{(n!{)}^2}⟺<!--\; ⟺\; -->4^n\le <!--\; \le \; -->\frac{(n+1)(2n)!}{(n!{)}^2}$ for some n.
Step: $n\to <!--\; \to \; -->n+1$
$4^{n+1}=\frac{(n+2)(2n+2)!}{((n+1)!{)}^2}$
I begin with manipulating the LHS:
$4^{n+1}=4\cdot <!--\; \cdot \; -->4^n\le <!--\; \le \; -->???$
$???\le <!--\; \le \; -->\frac{(2n)!(2n+2)(2n+1)(n+1)}{(n!(n+1))}$
$=\frac{(n+1)(2n+2)!}{((n+1)!{)}^2}$
Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.
Can you help me?

Earth's diameter at the equator is 7,926 miles. Find the distance around Earth at its equator to the nearest tenth.

Write each fraction in terms of the LCD.
$$\frac{3}{a{b}^2};\frac{7}{ab}$$

Proving an inequality via the Cauchy-Schwarz Inequality
I apolgize for contributing yet another question asking about an application of CS. Here it is:
Suppose $p_1,\dots <!--\; \dots \; -->,p_n$ and $a_1,...,a_n$ are real numbers such that $p_i\ge <!--\; \ge \; -->0$, $a_i\ge <!--\; \ge \; -->0$ for all $i$, and $p_1+\cdots <!--\; \cdots \; -->+p_n=1$. Then
$(p_1{a}_1+\cdots <!--\; \cdots \; -->+p_n{a}_n)(\frac{p_1}{a_1}+\cdots <!--\; \cdots \; -->+\frac{p_n}{a_n})\ge <!--\; \ge \; -->1$
The author of my textbook gives the following proof: Apply Cauchy's inequality to the sequences $\sqrt{p_1{a}_1}\dots <!--\; \dots \; -->\sqrt{p_n}{a}_n$ and $\sqrt{\frac{p_1}{a_1}}\dots <!--\; \dots \; -->\sqrt{\frac{p_n}{a_n}}$. (Thats it)
In trying to fill in the blanks I obtained the following
$\sqrt{p_1{a}_1}+\cdots <!--\; \cdots \; -->+\sqrt{p_n{a}_n}\le <!--\; \le \; -->\sqrt{p_1+\cdots <!--\; \cdots \; -->+p_n}\sqrt{a_1+...+a_n}=\sqrt{a_1+...+a_n}$
and
$\sqrt{\frac{p_1}{a_1}}+\cdots <!--\; \cdots \; -->+\sqrt{\frac{p_n}{a_n}}\le <!--\; \le \; -->\sqrt{p_1+\cdots <!--\; \cdots \; -->+p_n}\sqrt{\frac{1}{a_1}+...+\frac{1}{a_n}}=\sqrt{\frac{1}{a_1}+...+\frac{1}{a_n}}$
I'm not entirely sure where to go from here. Perhaps I have misunderstood what he meant by "apply cauchy's inequality to the sequences...". Another idea I had was to note that
$(p_1{a}_1+\cdots <!--\; \cdots \; -->+p_n{a}_n)\le <!--\; \le \; -->M_a(p_1+\cdots <!--\; \cdots \; -->+p_n)$
where $M_a$ is the largest $a_i$. And, that
$(\frac{p_1}{a_1}+\cdots <!--\; \cdots \; -->+\frac{p_n}{a_n})\le <!--\; \le \; -->\frac{1}{m_a}(p_1+\cdots <!--\; \cdots \; -->+p_n)$
where $m_a$ is the smallest $a_i$. Therefore, since $\frac{M_a}{m_a}\ge <!--\; \ge \; -->1$ the inequality follows. I am not very confident in the correctness of this method though and would like to understand how to prove the inequality via CS as my book suggests.

I'm reviewing my arithmetic and right now I'm at fractions, I'm just having a little bit of problem "visualizing" why cancellation works in multiplying fractions, I know how it works, it's just the why. Take for example.
$$\frac{2}{3}\cdot <!--\; \cdot \; -->\frac{3}{4}=\frac{1}{3}\cdot <!--\; \cdot \; -->\frac{3}{2}=\frac{1\cdot <!--\; \cdot \; -->1}{1\cdot <!--\; \cdot \; -->2}=\frac{1}{2}.$$
Through cancellation we know this ends in 1/2 because the common factor of the numerator "2" and denomaninator "4" is 2, so we divide both by "2" and end with new numbers in place, we also know that the common factor of numerator "3" and denominator "3" is "3" which equals "1" in both places, so it makes the work of "reducing to lowest terms" non-existent since this method is already a shortcut to that.
But I just can't seem to visualize how it works, can you guys give me a visual model to help me make sense of this? thank you.

Exponent Simplification
I have this expression where u and k are arbitrary constants:
$u^{{\left(\frac{1}{k}\right)}^{{\log }_k<!--\; \; -->({\log }_k<!--\; \; -->(u))}}$
I'm trying to clean up or simplify this expression... how can i go just making this cleaner? I'm forgetting log and exponent rules. Thanks!

Find a nonzero p such that px^2-12x+4=0 has only one solution.

2. Convert $1.0mg\cdot <!--\; \cdot \; -->m{m}^{-<!--\; -\; -->2}$ into$kg\cdot <!--\; \cdot \; -->m{m}^2$. Solution provided by Mr. Awesome:
$$\begin{gathered} 1.0mg\cdot <!--\; \cdot \; -->m{m}^2=\frac{1.0mg}{m{m}^2} \\ =\frac{1.0mg}{m{m}^2}\times <!--\; \times \; -->\frac{{10}^{-<!--\; -\; -->6}kg}{1mg}\times <!--\; \times \; -->\frac{1000mm}{m} \\ =1.0\times <!--\; \times \; -->{10}^{-<!--\; -\; -->3}\frac{kg}{m} \\ =1.0\times <!--\; \times \; -->{10}^{-<!--\; -\; -->3}kg\cdot <!--\; \cdot \; -->m^2 \end{gathered}$$

Find all solutions of the equation $$5x+12y+4=0$$ in $$\mathbb{Z}\times <!--\; \times \; -->\mathbb{Z}$$

What is the APY for money invested at each rate?
13% compounded semiannually

Do there exist positive integers $a,b,c$ such that $1/a+1/b+1/c=47/48$?
So far the only conclusion i have drawn is that $a+b+c$ is greater than 48. However, I cannot make anymore connections that would help me solve the problem. Am I just supposed to "brute force" this by bashing the algebra, or is there some logic I am missing?

Rewrite the expression in terms of the given function.
$$\frac{1}{1+\cos <!--\; \; -->x}+\frac{\cos <!--\; \; -->x}{1-<!--\; -\; -->\cos <!--\; \; -->x};\cot <!--\; \; -->x$$

Find the indicated expression of 1/2m inches in feet

Divide and simplify.
$$\frac{x^2+4}{8(x+3)}÷<!--\; ÷\; -->\frac{x+2}{8(x^2-<!--\; -\; -->9)}$$