Solution 2.2:3a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (07:03, 24 September 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way:
+
We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way,
 +
{{Displayed math||<math>\frac{x+3}{x-3}\cdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\cdot \frac{x-3}{x-3}=0\,\textrm{.}</math>}}
-
<math>\frac{x+3}{x-3}\centerdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\centerdot \frac{x-3}{x-3}=0</math>
+
Now, the numerators can be subtracted,
 +
{{Displayed math||<math>\frac{(x+3)(x-2)-(x+5)(x-3 )}{(x-2)(x-3)}=0\,\textrm{.}</math>}}
-
Now, the numerators can be subtracted:
+
Expand the brackets in the numerator,
-
 
+
-
 
+
-
<math>\frac{\left( x+3 \right)\left( x-2 \right)-\left( x+5 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x-3 \right)}=0</math>
+
-
 
+
-
 
+
-
Expand the brackets in the numerator
+
-
 
+
-
 
+
-
<math>\frac{x^{2}-2x+3x-6-\left( x^{2}-3x+5x-15 \right)}{\left( x-2 \right)\left( x-3 \right)}=0</math>
+
 +
{{Displayed math||<math>\frac{x^{2}-2x+3x-6-(x^{2}-3x+5x-15)}{(x-2)(x-3)}=0</math>}}
and simplify
and simplify
-
 
+
{{Displayed math||<math>\frac{-x+9}{(x-2)(x-3)}=0\,\textrm{.}</math>}}
-
<math>\frac{-x+9}{\left( x-2 \right)\left( x-3 \right)}=0</math>
+
-
 
+
The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to
The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to
 +
{{Displayed math||<math>-x+9=0\,</math>,}}
-
<math>-x+9=0</math>
+
i.e. <math>x=9</math>.
-
 
+
-
i.e.
+
-
<math>x=9</math>.
+
-
 
+
-
Substituting
+
-
<math>x=9</math>
+
-
into the original equation shows that we have calculated correctly:
+
 +
Substituting <math>x=9</math> into the original equation shows that we have calculated correctly,
-
<math>\text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS}</math>
+
{{Displayed math||<math>\text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS.}</math>}}

Current revision

We multiply the top and bottom of the terms on the left-hand side by appropriate factors so that they have the same common denominator, in the following way,

\displaystyle \frac{x+3}{x-3}\cdot \frac{x-2}{x-2}-\frac{x+5}{x-2}\cdot \frac{x-3}{x-3}=0\,\textrm{.}

Now, the numerators can be subtracted,

\displaystyle \frac{(x+3)(x-2)-(x+5)(x-3 )}{(x-2)(x-3)}=0\,\textrm{.}

Expand the brackets in the numerator,

\displaystyle \frac{x^{2}-2x+3x-6-(x^{2}-3x+5x-15)}{(x-2)(x-3)}=0

and simplify

\displaystyle \frac{-x+9}{(x-2)(x-3)}=0\,\textrm{.}

The left-hand side will be zero only when its numerator is zero (provided the denominator is not also zero), which gives us that the equation's solutions are given by the solutions to

\displaystyle -x+9=0\,,

i.e. \displaystyle x=9.

Substituting \displaystyle x=9 into the original equation shows that we have calculated correctly,

\displaystyle \text{LHS}=\frac{9+3}{9-3}-\frac{9+5}{9-2}=\frac{12}{6}-\frac{14}{7}=2-2=0=\text{RHS.}
Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.2:3a"