Solution 2.1:6b - Förberedande kurs i matematik 1

-                      Welcome to the course
                       
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                        Examinations
                      
                    
                      1. Numerical calculations
                       
                        1.1 Different types of numbers 
                        1.2 Fractions 
                        1.3 Powers
                      
                    
                      2. Algebra
                       
                        2.1 Algebraic expressions 
                        2.2 Linear expressions 
                        2.3 Quadratic expressions
                      
                    
                      3. Roots and logarithms
                       
                        3.1 Roots 
                        3.2 Equations with roots 
                        3.3 Logarithms 
                        3.4 Logarithmic equations
                      
                    
                      4. Trigonometry
                       
                        4.1 Angles and circles 
                        4.2 Functions 
                        4.3 Relations 
                        4.4 Equations
                      
                    
                      5. Written mathematics
                       
                        5.1 Writing formulas 
                        5.2 Mathematical text
                      
                    
                    
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+			Solution 2.1:6b
			
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- {{NAVCONTENT_START}} + The lowest common denominator for the three terms is <math>(x-2)(x+3)</math> and we expand each term so that all terms have the same denominator
- <center> [[Bild:2_1_6b.gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\begin{align}
  + \frac{x}{x-2}+\frac{x}{x+3}-2
  + &= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] 
  + &= \frac{x(x+3)+x(x-2)-2(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] 
  + &= \frac{x^{2}+3x+x^{2}-2x-2(x^{2}+3x-2x-6)}{(x-2)(x+3)}\\[5pt] 
  + &= \frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{(x-2)(x+3)}\,\textrm{.}
  + \end{align}</math>}}
  +  
  + Now, collect the terms in the numerator
  +  
  + {{Displayed math||<math>\begin{align}
  + \frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt] 
  + &= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.
Current revision
The lowest common denominator for the three terms is \displaystyle (x-2)(x+3) and we expand each term so that all terms have the same denominator





\displaystyle \begin{align}
\frac{x}{x-2}+\frac{x}{x+3}-2
&= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] 
&= \frac{x(x+3)+x(x-2)-2(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] 
&= \frac{x^{2}+3x+x^{2}-2x-2(x^{2}+3x-2x-6)}{(x-2)(x+3)}\\[5pt] 
&= \frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{(x-2)(x+3)}\,\textrm{.}
\end{align}



Now, collect the terms in the numerator





\displaystyle \begin{align}
\frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt] 
&= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.} 
\end{align}



Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_2.1:6b"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+The lowest common denominator for the three terms is <math>(x-2)(x+3)</math> and we expand each term so that all terms have the same denominator
-<center> [[Bild:2_1_6b.gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
+\frac{x}{x-2}+\frac{x}{x+3}-2
+&= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt]
+&= \frac{x(x+3)+x(x-2)-2(x-2)(x+3)}{(x-2)(x+3)}\\[5pt]
+&= \frac{x^{2}+3x+x^{2}-2x-2(x^{2}+3x-2x-6)}{(x-2)(x+3)}\\[5pt]
+&= \frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{(x-2)(x+3)}\,\textrm{.}
+\end{align}</math>}}
+Now, collect the terms in the numerator
+{{Displayed math||<math>\begin{align}
+\frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt]
+&= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.}
+\end{align}</math>}}
+Note: By keeping the denominator factorized during the entire calculation, we can see at the end that the answer cannot be simplified any further.
 \displaystyle \begin{align}
\frac{x}{x-2}+\frac{x}{x+3}-2
&= \frac{x}{x-2}\cdot\frac{x+3}{x+3} + \frac{x}{x+3}\cdot\frac{x-2}{x-2} - 2\cdot\frac{(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] 
&= \frac{x(x+3)+x(x-2)-2(x-2)(x+3)}{(x-2)(x+3)}\\[5pt] 
&= \frac{x^{2}+3x+x^{2}-2x-2(x^{2}+3x-2x-6)}{(x-2)(x+3)}\\[5pt] 
&= \frac{x^{2}+3x+x^{2}-2x-2x^{2}-6x+4x+12}{(x-2)(x+3)}\,\textrm{.}
\end{align}
 \displaystyle \begin{align}
\frac{x}{x-2}+\frac{x}{x+3}-2 &= \frac{(x^{2}+x^{2}-2x^{2})+(3x-2x-6x+4x)+12}{(x-2)(x+3)}\\[5pt] 
&= \frac{-x+12}{(x-2)(x+3)}\,\textrm{.} 
\end{align}