From Förberedande kurs i matematik 1

Revision as of 08:43, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:2a moved to Solution 2.2:2a: Robot: moved page) ← Previous diff		Current revision (13:44, 23 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	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
-	<center> [[Image:2_2_2a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
-	{{NAVCONTENT_START}}	+	& \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]
-	<center> [[Image:2_2_2a-2(2).gif]] </center>	+	&\qquad\Leftrightarrow\quad 3\cdot 5x-2\cdot (x+2) = 3\cdot 3\,\textrm{.}\\
-	{{NAVCONTENT_STOP}}	+	\end{align}</math>}}
		+
		+	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
		+
		+	{{Displayed math||<math>13x-4=9\,\textrm{.}</math>}}
		+
		+	We can now solve this first-degree equation by carrying out simple arithmetical calculations so as to get ''x'' by itself on one side:
		+
		+	<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>
		+
		+	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)
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS}
		+	&= \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}
		+	= \text{RHS}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

If we divide up the denominators that appear in the equation into small integer factors \displaystyle 6=2\cdot 3, \displaystyle 9=3\cdot 3 and 2, we see that the lowest common denominator is \displaystyle 2\cdot 3\cdot 3=18. Thus, we multiply both sides of the equation by \displaystyle 2\cdot 3\cdot 3 in order to avoid having denominators in the equation

\displaystyle \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}

We can rewrite the left-hand side as \displaystyle 3\cdot 5x-2\cdot (x+2) = 15x-2x-4 = 13x-4, so that we get the equation

\displaystyle 13x-4=9\,\textrm{.}

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, \displaystyle \vphantom{x_2}13x-4+4=9+4\,, which gives \displaystyle \ 13x=13\,\textrm{.}
2. Divide both sides by 13, \displaystyle \frac{13x}{13}=\frac{13}{13}\,, which gives the answer \displaystyle \ x=1\,\textrm{.}

The equation has \displaystyle x=1 as the solution.

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

\displaystyle \begin{align} \text{LHS} &= \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} = \text{RHS}\,\textrm{.} \end{align}