Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 2.2:2a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (13:44, 23 September 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
If we divide up the denominators that appear in the equation into small integer factors
+
If we divide up the denominators that appear in the equation into small integer factors <math>6=2\cdot 3</math>, <math>9=3\cdot 3</math> and 2, we see that the lowest common denominator is <math>2\cdot 3\cdot 3=18</math>. Thus, we multiply both sides of the equation by <math>2\cdot 3\cdot 3</math> in order to avoid having denominators in the equation
-
<math>6=2\centerdot 3</math>,
+
-
<math>9=3\centerdot 3</math>
+
-
and 2, we see that the lowest common denominator is
+
-
<math>2\centerdot 3\centerdot 3=18</math>. Thus, we multiply both sides of the equation by
+
-
<math>2\centerdot 3\centerdot 3</math>
+
-
in order to avoid having denominators in the equation:
+
 +
{{Displayed math||<math>\begin{align}
 +
& \rlap{/}2\cdot{}\rlap{/}3\cdot 3\cdot\frac{5x}{\rlap{/}6} - 2\cdot{}\rlap{/}3\cdot{}\rlap{/}3\cdot\frac{x+2}{\rlap{/}9} = \rlap{/}2\cdot 3\cdot 3\cdot \frac{1}{\rlap{/}2} \\[5pt]
 +
&\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\
 +
\end{align}</math>}}
-
<math>\begin{align}
+
We can rewrite the left-hand side as <math>3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4</math>, so that we get the equation
-
& 2\centerdot 3\centerdot 3\centerdot \frac{5x}{6}-2\centerdot 3\centerdot 3\centerdot \frac{x+2}{9}=2\centerdot 3\centerdot 3\centerdot \frac{1}{2} \\
+
-
& \Leftrightarrow 3\centerdot 5x-2\centerdot \left( x+2 \right)=3\centerdot 3 \\
+
-
\end{align}</math>
+
 +
{{Displayed math||<math>13x-4=9\,\textrm{.}</math>}}
-
We can rewrite the left-hand side as
+
We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get ''x'' by itself on one side:
-
<math>3\centerdot 5x-2\centerdot \left( x+2 \right)=15x-2x-4=13x-4</math>, so that we get the equation
+
 +
<ol>
 +
<li>Add 4 to both sides, <math>\vphantom{x_2}13x-4+4=9+4\,,</math> which gives <math>\ 13x=13\,\textrm{.}</math></li>
 +
<li>Divide both sides by 13, <math>\frac{13x}{13}=\frac{13}{13}\,,</math> which gives the answer <math>\ x=1\,\textrm{.}</math></li>
 +
</ol>
-
<math>13x-4=9</math>
+
The equation has <math>x=1</math> as the solution.
 +
When we have obtained an answer, it is important to go back to the original equation to check that <math>x=1</math> really is the correct answer (i.e. that we haven't calculated incorrectly)
-
We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get
+
{{Displayed math||<math>\begin{align}
-
<math>x</math>
+
\text{LHS}
-
by itself on one side:
+
&= \frac{5\cdot 1}{6}-\frac{1+2}{9} = \frac{5}{6}-\frac{3}{9} = \frac{5}{6}-\frac{1}{3}\\[5pt]
-
 
+
&= \frac{5}{6}-\frac{1\cdot 2}{3\cdot 2} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}
-
1. Add 4 to both sides:
+
= \text{RHS}\,\textrm{.}
-
+
\end{align}</math>}}
-
<math>13x-+4=9+4</math>
+
-
+
-
which gives
+
-
 
+
-
<math>13x=13</math>
+
-
.
+
-
+
-
2. Divide both sides by 13:
+
-
+
-
<math>\frac{13x}{13}=\frac{13}{13}</math>
+
-
 
+
-
which gives the answer
+
-
<math>x=1</math>.
+
-
 
+
-
The equation has
+
-
<math>1</math>
+
-
as the solution.
+
-
 
+
-
When we have obtained an answer, it is important to go back to the original equation to check that
+
-
<math>x=1</math>
+
-
really is the correct answer( i.e. that we haven't calculated incorrectly):
+
-
 
+
-
LHS =
+
-
<math>\frac{5\centerdot 1}{6}-\frac{1\centerdot 2}{9}=\frac{5}{6}-\frac{3}{9}=\frac{5}{6}-\frac{1}{3}=\frac{5}{6}-\frac{1\centerdot 2}{3\centerdot 2}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}</math>
+
-
= RHS
+

Current revision

If we divide up the denominators that appear in the equation into small integer factors 6=23, 9=33 and 2, we see that the lowest common denominator is 233=18. Thus, we multiply both sides of the equation by 233 in order to avoid having denominators in the equation

23365x−2339x+2=2332135x−2(x+2)=33.

We can rewrite the left-hand side as 35x−2(x+2)=15x−2x−4=13x−4, so that we get the equation

13x−4=9.

We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get x by itself on one side:

  1. Add 4 to both sides, 13x−4+4=9+4 which gives  13x=13.
  2. Divide both sides by 13, 1313x=1313 which gives the answer  x=1.

The equation has x=1 as the solution.

When we have obtained an answer, it is important to go back to the original equation to check that x=1 really is the correct answer (i.e. that we haven't calculated incorrectly)

LHS=651−91+2=65−93=65−31=65−3212=65−2=63=21=RHS.
Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_2.2:2a"