Aus Online Mathematik Brückenkurs 1

Version vom 08:43, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.2:1d moved to Solution 2.2:1d: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 14:28, 16. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	Move
-	<center> [[Image:2_2_1d.gif]] </center>	+	<math>x</math>
-	{{NAVCONTENT_STOP}}	+	to the left-hand side by subtracting
		+	<math>2x</math>
		+	from both sides,
		+
		+
		+	<math>5x+7-2x=2x-6-2x</math>
		+
		+
		+	which gives
		+
		+
		+	<math>3x+7=-6</math>
		+
		+
		+	Subtract
		+	<math>7</math>
		+	from both sides,
		+
		+
		+	<math>3x+7-7=-6-7</math>
		+
		+
		+	so that the term
		+	<math>3x</math>
		+	alone remains on the left-hand side
		+
		+
		+	<math>3x=-13</math>
		+
		+
		+	Then, divide both sides by
		+	<math>3</math>
		+
		+
		+
		+	<math>\frac{3x}{3}=-\frac{13}{3}</math>
		+
		+
		+	to get x:
		+
		+
		+	<math>x=-\frac{13}{3}</math>

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Version vom 14:28, 16. Sep. 2008

Move \displaystyle x to the left-hand side by subtracting \displaystyle 2x from both sides,

\displaystyle 5x+7-2x=2x-6-2x

which gives

\displaystyle 3x+7=-6

Subtract \displaystyle 7 from both sides,

\displaystyle 3x+7-7=-6-7

so that the term \displaystyle 3x alone remains on the left-hand side

\displaystyle 3x=-13

Then, divide both sides by \displaystyle 3

\displaystyle \frac{3x}{3}=-\frac{13}{3}

to get x:

\displaystyle x=-\frac{13}{3}