Solution 2.1:6b - Förberedande kurs i matematik 1

-                      Welcome to the course
                       
                        About the course 
                        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
                                            
                    
                    
		       Search
+			Solution 2.1:6b
			
			  From Förberedande kurs i matematik 1
			  (Difference between revisions)
			  			                            Jump to: navigation, search
                          
		          
			
			
			
			
			
			
				Revision as of 11:16, 31 March 2008 (edit)
Lina  (Talk | contribs)
 (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_6b.gif </center> {{NAVCONTENT_STOP}})
← Previous diff
				Current revision (11:35, 23 September 2008) (edit) (undo)
Tek  (Talk | contribs) 
m 
 
			
		(3 intermediate revisions not shown.)
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.
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/2008/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}