From Förberedande kurs i matematik 1

Revision as of 15:14, 20 September 2008 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (08:47, 29 September 2008) (edit) (undo) Tek (Talk | contribs) m
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-	In this case, we see that the left-hand side contains the factor	+	In this case, we see that the left-hand side contains the factor <math>x+3</math>, which we can take out to obtain
-	<math>x+\text{3}</math>, which we can take out to obtain	+
-		+
-		+
-	<math>\begin{align}	+
-	& \left( x+\text{3} \right)\left( x-1 \right)-\left( x+\text{3} \right)\left( 2x-9 \right)=\left( x+\text{3} \right)\left( \left( x-1 \right)-\left( 2x-9 \right) \right) \\	+
-	& =\left( x+\text{3} \right)\left( x-1-2x+9 \right)=\left( x+\text{3} \right)\left( -x+8 \right). \\	+
-	\end{align}</math>	+
		+	{{Displayed math||<math>\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}</math>}}
	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>}}
-	<math>\left( x+\text{3} \right)\left( -x+8 \right)=0</math>	+	which has the solutions <math>x=-3</math> and <math>x=8\,</math>.
-		+
-		+
-	which has the solutions	+
-	<math>x=-\text{3}</math>	+
-	and	+
-	<math>x=\text{8}</math>.	+
-	We check the solution	+	We check the solution <math>x=8</math> by substituting it into the equation,
-	<math>x=\text{8 }</math>	+
-	by substituting it into the equation:	+
-	LHS	+	{{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>}}
-	<math>=\left( 8+3 \right)\centerdot \left( 8-1 \right)-\left( 8+3 \right)\centerdot \left( 2\centerdot 8-9 \right)=11\centerdot 7-11\centerdot 7=0=</math>	+
-	RHS	+

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Current revision

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.}