Aus Online Mathematik Brückenkurs 1

Version vom 08:47, 29. Sep. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:32, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	In this case, we see that the left-hand side contains the factor <math>x+3</math>, which we can take out to obtain		In this case, we see that the left-hand side contains the factor <math>x+3</math>, which we can take out to obtain
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	(x+3)(x-1) - (x+3)(2x-9)		(x+3)(x-1) - (x+3)(2x-9)
	&= (x+3)\bigl((x-1)-(2x-9)\bigr)\\[5pt]		&= (x+3)\bigl((x-1)-(2x-9)\bigr)\\[5pt]
Zeile 10:		Zeile 10:
	This rewriting of the equation results in the new equation		This rewriting of the equation results in the new equation
-	{{Displayed math||<math>(x+3)(-x+8)=0</math>}}	+	{{Abgesetzte Formel||<math>(x+3)(-x+8)=0</math>}}
	which has the solutions <math>x=-3</math> and <math>x=8\,</math>.		which has the solutions <math>x=-3</math> and <math>x=8\,</math>.
Zeile 16:		Zeile 16:
	We check the solution <math>x=8</math> by substituting it into the equation,		We check the solution <math>x=8</math> by substituting it into the equation,
-	{{Displayed math||<math>\text{LHS} = (8+3)\cdot (8-1) - (8+3)\cdot (2\cdot 8 - 9) = 11\cdot 7 - 11\cdot 7 = 0 = \textrm{RHS.}</math>}}	+	{{Abgesetzte Formel||<math>\text{LHS} = (8+3)\cdot (8-1) - (8+3)\cdot (2\cdot 8 - 9) = 11\cdot 7 - 11\cdot 7 = 0 = \textrm{RHS.}</math>}}

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Version vom 08:32, 22. Okt. 2008

In this case, we see that the left-hand side contains the factor \displaystyle x+3, which we can take out to obtain

\displaystyle \begin{align} (x+3)(x-1) - (x+3)(2x-9) &= (x+3)\bigl((x-1)-(2x-9)\bigr)\\[5pt] &= (x+3)(x-1-2x+9)\\[5pt] &= (x+3)(-x+8)\,\textrm{.} \end{align}

This rewriting of the equation results in the new equation

\displaystyle (x+3)(-x+8)=0

which has the solutions \displaystyle x=-3 and \displaystyle x=8\,.

We check the solution \displaystyle x=8 by substituting it into the equation,

\displaystyle \text{LHS} = (8+3)\cdot (8-1) - (8+3)\cdot (2\cdot 8 - 9) = 11\cdot 7 - 11\cdot 7 = 0 = \textrm{RHS.}