Lösung 2.3:3d

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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,
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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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
x(x+3)-x(2x-9)
x(x+3)-x(2x-9)
&= x\bigl((x+3)-(2x-9)\bigr)\\[5pt]
&= x\bigl((x+3)-(2x-9)\bigr)\\[5pt]
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The equation is thus
The equation is thus
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{{Displayed math||<math>x(-x+12) = 0</math>}}
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{{Abgesetzte Formel||<math>x(-x+12) = 0</math>}}
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>.
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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<math>x=0</math> is obvious)
<math>x=0</math> is obvious)
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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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{{Abgesetzte Formel||<math>\text{LHS} = 12\cdot (12+3) - 12\cdot (2\cdot 12-9) = 2\cdot 15 - 12\cdot 15 = 0 = \text{RHS.}</math>}}

Version vom 08:32, 22. Okt. 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,

\displaystyle \begin{align}

x(x+3)-x(2x-9) &= x\bigl((x+3)-(2x-9)\bigr)\\[5pt] &= x(x+3-2x+9)\\[5pt] &= x(-x+12)\,\textrm{.} \end{align}

The equation is thus

\displaystyle x(-x+12) = 0

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)

\displaystyle \text{LHS} = 12\cdot (12+3) - 12\cdot (2\cdot 12-9) = 2\cdot 15 - 12\cdot 15 = 0 = \text{RHS.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:3d“