Processing Math: Done

From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	Start by rewriting the terms on the left-hand side as one term having a common denominator
-	<center> [[Image:2_2_3c-1(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
-	{{NAVCONTENT_START}}	+	\frac{1}{x-1}-\frac{1}{x+1} &= \frac{1}{x-1}\cdot\frac{x+1}{x+1} - \frac{1}{x+1}\cdot\frac{x-1}{x-1}\\[5pt]
-	<center> [[Image:2_2_3c-2(3).gif]] </center>	+	&= \frac{x+1}{(x-1)(x+1)} - \frac{x-1}{(x-1)(x+1)}\\[5pt]
-	{{NAVCONTENT_STOP}}	+	&= \frac{(x+1)-(x-1)}{(x-1)(x+1)}\\[5pt]
-	{{NAVCONTENT_START}}	+	&= \frac{2}{(x-1)(x+1)}\,\textrm{.}
-	<center> [[Image:2_2_3c-3(3).gif]] </center>	+	\end{align}</math>}}
-	{{NAVCONTENT_STOP}}	+
		+	If we also write <math>3x-3=3(x-1)</math>, the equation can be rewritten as
		+
		+	{{Displayed math||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}}
		+
		+	Because <math>x=1</math> cannot be a solution to the equation, the factor
		+	<math>x-1</math> can be removed from the denominator of both sides (i.e. actually, we multiply both sides by <math>x-1</math> and then eliminate it)
		+
		+	{{Displayed math||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}}
		+
		+	Then, both sides are multiplied by 3 and <math>x+1</math>, so that we get an equation without any denominators
		+
		+	{{Displayed math||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}}
		+
		+	Expanding both sides
		+
		+	{{Displayed math||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}}
		+
		+	The ''x''² terms cancel each other out and we obtain a first-order equation,
		+
		+	{{Displayed math||<math>3=5x-1\,,</math>}}
		+
		+	which has the solution
		+
		+	{{Displayed math||<math>x=\frac{4}{5}\,\textrm{.}</math>}}
		+
		+	We check whether we have calculated correctly by substituting <math>x=4/5</math> into the original equation,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\text{LHS} &= \biggl(\frac{1}{\frac{4}{5}-1}-\frac{1}{\frac{4}{5}+1}\biggr)\bigl( \bigl(\tfrac{4}{5}\bigr)^{2}+\tfrac{1}{2}\bigr)
		+	= \biggl(\frac{1}{-\frac{1}{5}}-\frac{1}{\frac{9}{5}}\biggr)\bigl(
		+	\tfrac{16}{25}+\tfrac{1}{2}\bigr)\\[5pt]
		+	&= \bigl(-5-\tfrac{5}{9}\bigr)\cdot\frac{16\cdot 2+25}{2\cdot 25}
		+	= -\frac{50}{9}\cdot\frac{57}{50}
		+	= -\frac{57}{9}
		+	= -\frac{19}{3}\,,\\[15pt]
		+	\text{RHS} &= \frac{6\cdot\frac{4}{5}-1}{3\cdot\frac{4}{5}-3}
		+	= \frac{\frac{24}{5}-\frac{5}{5}}{\frac{12}{5}-\frac{15}{5}}
		+	= \frac{\frac{1}{5}\cdot (24-5)}{\frac{1}{5}\cdot (12-15)}
		+	= \frac{19}{-3}
		+	= -\frac{19}{3}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

Start by rewriting the terms on the left-hand side as one term having a common denominator

1x−1−1x+1=1x−1x+1x+1−1x+1x−1x−1=x+1(x−1)(x+1)−x−1(x−1)(x+1)=(x−1)(x+1)(x+1)−(x−1)=2(x−1)(x+1).

If we also write 3x−3=3(x−1), the equation can be rewritten as

2(x−1)(x+1)x2+21=6x−13(x−1).

Because x=1 cannot be a solution to the equation, the factor x−1 can be removed from the denominator of both sides (i.e. actually, we multiply both sides by x−1 and then eliminate it)

2x+1x2+21=36x−1.

Then, both sides are multiplied by 3 and x+1, so that we get an equation without any denominators

6x2+21=(6x−1)(x+1).

Expanding both sides

6x2+3=6x2+5x−1.

The x² terms cancel each other out and we obtain a first-order equation,

3=5x−1

which has the solution

x=54.

We check whether we have calculated correctly by substituting x=45 into the original equation,

LHSRHS=154−1−154+1542+21=1−51−1592516+21=−5−95225162+25=−9505057=−957=−319=354−3654−1=524−55512−515=51(24−5)51(12−15)=19−3=−319.