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Yes. Then the domain is every real number except -8.

\(\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ -2&4&2&|1 \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&6&6&|3\\ \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&1&1&|\dfrac{1}2\\ \end{pmatrix}\\\)

No solution.

\(\begin{pmatrix} 5&0&3|1\\ -1&1&1|0\\ 0&3&2|5\\ \end{pmatrix}\\ \sim \begin{pmatrix} 5&0&3&|1\\ 0&1&\dfrac{8}5&|\dfrac{1}{5}\\ 0&3&2&|5\\ \end{pmatrix}\\ \sim \begin{pmatrix} 5&0&3&|1\\ 0&1&\dfrac{8}5&|\dfrac{1}{5}\\ 0&0&-\dfrac{14}{5}&|\dfrac{22}{5}\\ \end{pmatrix}\\ -\dfrac{14}{5}z = \dfrac{22}5\\ \boxed{z = \dfrac{-11}{7}}\\ y + \dfrac{8}{5}z = \dfrac{1}{5}\\ \boxed{y = \dfrac{1}{5} + \dfrac{8}{5}\cdot \dfrac{11}{7} = \dfrac{19}{7}}\\ 5x + 3z = 1\\ \boxed{x = \dfrac{1-3(\dfrac{-11}{7})}{5} = \dfrac{8}{7}}\)

Let's call the first equation (1) and the second equation (2).

Now, consider (1) - 2*(2) and you are done.

You are spamming.

\(\sqrt{t} \ge 0 \quad\forall t \in D(\sqrt x),\text{ where }D(\sqrt x) \text{ denotes the domain of }f(x) = \sqrt x.\)

Whenever you see a square root, you can be sure that it is greater than or equal to 0.

Now write f(x) > 0 using interval notation.

(a) Think about what makes the fraction undefined. What value(s) of x makes the denominator 0?

(b) Notice that the function has a vertical asymptote. Substitute points around the vertical asymptote to find out if the range is \((-\infty,\infty)\), \((-\infty ,0)\),\((0,\infty)\),\([0,\infty)\), or \((-\infty ,0]\).

(c) Notice that the function has a vertical asymptote. Substitute points around the vertical asymptote to find out if the range is \((-\infty,\infty)\), \((-\infty ,0)\),\((0,\infty)\),\([0,\infty)\), or \((-\infty ,0]\)​.

Thank you, CPhill !

Wow... That must have taken a while!

Great explantation with latex coding! (howd you do that... ?)

OHHHH wow... I just checked here and BOOOOMMMM THX GUYS!

418 or 480? 

And did you mean that whether 3 integers (a, b, and c) can be found such that \(a< b< c<180\) and \(a + b + c = 418\text{(or 480)}\)?

Exactly!.

Nice. Is this a linear search on the set of integers?

a=1; b=1;c= 1/a + 1/b; if(c==1/12, goto4, goto5);printa, b; a++;if(a<200, goto2, 0);a=1;b++;if(b<200, goto2, discard=0;

a         b

156    13
84      14
60      15
48      16
36      18
30      20
28      21
24      24
21      28
20      30
18      36
16      48
15      60
14      84
13      156

\(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}\\ \dfrac{x+y}{xy} = \dfrac{1}{12}\\ \text{Let }x + y = k, xy = 12k.\\ y(k-y) = 12k\\ -y^2 +ky = 12k\\ y^2 -ky+12k = 0\\ y = \dfrac{k+\sqrt{k^2 -48k}}{2}\\ \implies k^2 - 48k \text{ is a perfect square.}\\ k^2 - 48k \ge 0\\ k^2 - 48k = t^2\\ k \le 0\text{(rejected)} \vee k\ge48\\ k \ge 48\\ k \text{ cannot be 48. If } k = 48\text{, } x = y = 24.\\ k \ge 49\\ k = 49 \text{ satisfies the condition that } k^2 - 48k \text{ is a perfect square.}\\ \min(k) = 49\\ \boxed{\min(x + y) = \color{red}49}\)