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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
 #1
avatar+20108 
+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+27128 
+2

Heureka's result looks right to me:

 

.

Alan  Mar 10, 2018

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