+0  
 
0
124
3
avatar+644 

Find all real x where
\(2\cdot\frac{x-5}{x-3} > \frac{2x-5}{x+2} + 5.\)
Give your answer in interval notation.

waffles  Mar 9, 2018
Sort: 

3+0 Answers

 #1
avatar+19352 
+2

Find all real x where

\(\displaystyle 2\cdot \left(\frac{x-5}{x-3}\right) > \frac{2x-5}{x+2} + 5.\)
2\cdot\frac{x-5}{x-3} > \frac{2x-5}{x+2} + 5.

 

1. rearrange:

\(\begin{array}{|rcll|} \hline 2\cdot \left( \dfrac{x-5}{x-3} \right) &>& \dfrac{2x-5}{x+2} + 5 \\\\ \dfrac{2x-10}{x-3} &>& \dfrac{2x-5+5(x+2)}{x+2} \\\\ \dfrac{2x-10}{x-3} &>& \dfrac{2x-5+5x+10}{x+2} \\\\ \mathbf{ \dfrac{2x-10}{x-3} } & \mathbf{>} & \mathbf{\dfrac{7x+5}{x+2}} \\ \hline \end{array}\)

 

2.  reduce / convert fractions to a common :

\(\begin{array}{|rcll|} \hline \dfrac{2x-10}{x-3} - \dfrac{7x+5}{x+2} & > & 0 \\\\ \dfrac{(2x-10)(x+2)-(7x+5)(x-3)}{(x-3)(x+2)} & > & 0 \\\\ \dfrac{2x^2+4x-10x-20-7x^2+21x-5x+15}{(x-3)(x+2)} & > & 0 \\\\ \dfrac{-5x^2+10x-5}{(x-3)(x+2)} & > & 0 \\\\ \dfrac{-5(x^2-2x+1)}{(x-3)(x+2)} & > & 0 \quad | \quad : (-5) \\ && \text{ attention } ">" \rightarrow "<" \\\\ \dfrac{x^2-2x+1}{(x-3)(x+2)} & < & 0 \quad | \quad x^2-2x+1 = (x-1)^2 \\\\ \dfrac{(x-1)^2 }{(x-3)(x+2)} & < & 0 \quad | \quad \cdot (x-3)^2(x+2)^2 \\\\ \dfrac{(x-1)^2(x-3)^2(x+2)^2 }{(x-3)(x+2)} & < & 0 \\\\ \mathbf{(x-1)^2(x-3)(x+2)} & \mathbf{\color{red}<} & \mathbf{ 0 } \\ \hline \end{array}\)

 

3. solve the equation:

\(\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{sign} & (-\infty,-2 ) & -2 & (-2,1) & 1 & (1,3) & 3 & (3,\infty) \\ \hline x-3 & - & - & - & - & - & 0 & + \\ \hline (x-1)^2 & + & + & + & 0 & + & + & + \\ \hline x+2 & - & 0 & + & + & + & + & + \\ \hline \\ \hline \text{sign of } (x-1)^2(x-3)(x+2) & + & 0 & \color{red}- & 0 & \color{red}- & 0 & + \\ \hline \end{array}\)

 

interval notation: \(\mathbf{(-2,1) \text{ and } (1,3)}\)

 

laugh

heureka  Mar 9, 2018
 #2
avatar+644 
0

im not sure thats correct

waffles  Mar 10, 2018
 #3
avatar+26682 
+2

Heureka's result looks right to me:

 

.

Alan  Mar 10, 2018

23 Online Users

avatar
New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy