From Förberedande kurs i matematik 1

Revision as of 08:45, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 2.2:3b moved to Solution 2.2:3b: Robot: moved page) ← Previous diff		Current revision (07:40, 24 September 2008) (edit) (undo) Tek (Talk | contribs) m
(One intermediate revision not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	First, we move all the terms over to the left-hand side,
-	<center> [[Image:2_2_3b-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\frac{4x}{4x-7}-\frac{1}{2x-3}-1=0\,\textrm{.}</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:2_2_3b-2(2).gif]] </center>	+	Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,
-	{{NAVCONTENT_STOP}}	+
		+	{{Displayed math||<math>\frac{4x}{4x-7}\cdot\frac{2x-3}{2x-3} - \frac{1}{2x-3}\cdot\frac{4x-7}{4x-7} - \frac{(2x-3)(4x-7)}{(2x-3)(4x-7)} = 0</math>}}
		+
		+	and so that we can rewrite the left-hand side giving
		+
		+	{{Displayed math||<math>\frac{4x(2x-3) - (4x-7) - (2x-3)(4x-7)}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
		+
		+	We expand the numerator
		+
		+	{{Displayed math||<math>\frac{8x^{2}-12x-(4x-7)-(8x^{2}-14x-12x+21)}{(2x-3)(4x-7)} = 0</math>}}
		+
		+	and simplify
		+
		+	{{Displayed math||<math>\frac{10x-14}{(2x-3)(4x-7)}=0\,\textrm{.}</math>}}
		+
		+	This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when
		+
		+	{{Displayed math||<math>10x-14=0\,,</math>}}
		+
		+	which gives <math>x=7/5\,</math>.
		+
		+	It can easily happen that we calculate incorrectly, so we check that the answer <math>x=7/5</math> satisfies the equation,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS } &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7} - \frac{1}{2\cdot\frac{7}{5}-3}
		+	= \{\,\text{multiply top and bottom by 5}\,\}\\[5pt]
		+	&= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7}\cdot\frac{5}{5} - \frac{1}{2\cdot \frac{7}{5}-3}\cdot\frac{5}{5}
		+	= \frac{4\cdot 7}{4\cdot 7-7\cdot 5}-\frac{5}{2\cdot 7-3\cdot 5}\\[5pt]
		+	&= \frac{4}{4-5}-\frac{5}{14-15}
		+	= -4-(-5) = 1 = \text{RHS.}
		+	\end{align}</math>}}

---

Current revision

First, we move all the terms over to the left-hand side,

\displaystyle \frac{4x}{4x-7}-\frac{1}{2x-3}-1=0\,\textrm{.}

Then, we multiply the top and bottom of all three terms by appropriate factors so that they have the same common denominator, in the following way,

\displaystyle \frac{4x}{4x-7}\cdot\frac{2x-3}{2x-3} - \frac{1}{2x-3}\cdot\frac{4x-7}{4x-7} - \frac{(2x-3)(4x-7)}{(2x-3)(4x-7)} = 0

and so that we can rewrite the left-hand side giving

\displaystyle \frac{4x(2x-3) - (4x-7) - (2x-3)(4x-7)}{(2x-3)(4x-7)}=0\,\textrm{.}

We expand the numerator

\displaystyle \frac{8x^{2}-12x-(4x-7)-(8x^{2}-14x-12x+21)}{(2x-3)(4x-7)} = 0

and simplify

\displaystyle \frac{10x-14}{(2x-3)(4x-7)}=0\,\textrm{.}

This equation is satisfied when the numerator is zero (provided the denominator is not also zero) and this happens when

\displaystyle 10x-14=0\,,

which gives \displaystyle x=7/5\,.

It can easily happen that we calculate incorrectly, so we check that the answer \displaystyle x=7/5 satisfies the equation,

\displaystyle \begin{align} \text{LHS } &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7} - \frac{1}{2\cdot\frac{7}{5}-3} = \{\,\text{multiply top and bottom by 5}\,\}\\[5pt] &= \frac{4\cdot\frac{7}{5}}{4\cdot\frac{7}{5}-7}\cdot\frac{5}{5} - \frac{1}{2\cdot \frac{7}{5}-3}\cdot\frac{5}{5} = \frac{4\cdot 7}{4\cdot 7-7\cdot 5}-\frac{5}{2\cdot 7-3\cdot 5}\\[5pt] &= \frac{4}{4-5}-\frac{5}{14-15} = -4-(-5) = 1 = \text{RHS.} \end{align}