Lösung 2.3:3d

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Because both terms,
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Because both terms, <math>x(x+3)</math> and <math>x(2x-9)</math>, contain the factor <math>x</math>, we can take out <math>x</math> from the left-hand side and collect together the remaining expression,
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<math>x\left( x+3 \right)</math>
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and
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<math>x\left( 2x-9 \right)</math>
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contain the factor
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<math>x</math>, we can take out
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<math>x</math> from the left-hand side and collect together the remaining expression:
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<math>\begin{align}
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& x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\
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& =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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x(x+3)-x(2x-9)
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&= x\bigl((x+3)-(2x-9)\bigr)\\[5pt]
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&= x(x+3-2x+9)\\[5pt]
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&= x(-x+12)\,\textrm{.}
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\end{align}</math>}}
The equation is thus
The equation is thus
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{{Displayed math||<math>x(-x+12) = 0</math>}}
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<math>x\left( -x+12 \right)=0</math>
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and we obtain directly that the equation is satisfied if either <math>x</math> or <math>-x+12</math> is zero. The solutions to the equation are therefore <math>x=0</math> and <math>x=12</math>.
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and we obtain directly that the equation is satisfied if either
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<math>x</math>
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or
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<math>-x+\text{12}</math>
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is zero. The solutions to the equation are therefore
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<math>x=0\text{ }</math>
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and
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<math>x=\text{12}</math>.
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Here, it can be worth checking that
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Here, it can be worth checking that <math>x=12</math> is a solution (the case
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<math>x=\text{12 }</math>
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<math>x=0</math> is obvious)
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is a solution (the case
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<math>x=0</math>
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is obvious):
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LHS
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{{Displayed math||<math>\text{LHS} = 12\cdot (12+3) - 12\cdot (2\cdot 12-9) = 2\cdot 15 - 12\cdot 15 = 0 = \text{RHS.}</math>}}
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<math>=12\centerdot \left( 12+3 \right)-12\centerdot \left( 2\centerdot 12-9 \right)=2\centerdot 15-12\centerdot 15=0=</math>
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RHS
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Version vom 08:39, 29. Sep. 2008

Because both terms, \displaystyle x(x+3) and \displaystyle x(2x-9), contain the factor \displaystyle x, we can take out \displaystyle x from the left-hand side and collect together the remaining expression,

Vorlage:Displayed math

The equation is thus

Vorlage:Displayed math

and we obtain directly that the equation is satisfied if either \displaystyle x or \displaystyle -x+12 is zero. The solutions to the equation are therefore \displaystyle x=0 and \displaystyle x=12.

Here, it can be worth checking that \displaystyle x=12 is a solution (the case \displaystyle x=0 is obvious)

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:3d“