From Förberedande kurs i matematik 1

Revision as of 08:59, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.3:3e moved to Solution 2.3:3e: Robot: moved page) ← Previous diff		Current revision (08:47, 29 September 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	In this case, we see that the left-hand side contains the factor <math>x+3</math>, which we can take out to obtain
-	<center> [[Image:2_3_3e.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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
		+
		+	{{Displayed math||<math>(x+3)(-x+8)=0</math>}}
		+
		+	which has the solutions <math>x=-3</math> and <math>x=8\,</math>.
		+
		+	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>}}

---

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.}