• LGS mit einem Parameter lösen
  • MNWeG
  • 07.02.2022
  • Mathematik
  • Gleichungen
  • Einzelarbeit
  • Arbeitsblatt
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1
Ermittle mithilfe von Proben, ob eine der drei Lösungsmengen zum LGS gehört.
(311-3r+52212r+6111r+5)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \\\left( \begin{array}{rrr|r} 3 & 1 & 1 & \text{-3}r&+&5 \\ 2 & 2 & 1 & 2r&+&6 \\ 1 & 1 & 1 & r&+&5 \\ \end{array} \right)
L1={1r;6r+1;-12r+4}L2={-2r+12;3r+32;2}L3={-2r;3r+1;4}\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} L_1= &\{1r; 6r+1; \text{-}12r+4\}\\ L_2= &\{\text{-}2r+\frac{1}{2}; 3r+\frac{3}{2};2\}\\ L_3= &\{\text{-}2r; 3r+1; 4\}\\ \end{aligned}
I. 31r +1(6r+1) +1(-12r+4)=3r+6r+112r+4=-3r+5 I ⁣I. 21r +2(6r+1) +1(-12r+4)=2r+12r+212r+4= 2r+6 I ⁣I ⁣I. 11r +1(6r+1) +1(-12r+4)=1r+6r+112r+4  r+5I. 3(-2r+12) +1(3r+32) +12=-6r+32+3r+32+2=-3r+5 I ⁣I. 2(-2r+12) +2(3r+32) +12=-4r+1+6r+3+2= 2r+6 I ⁣I ⁣I. 1(-2r+12) +1(3r+32) +12=-2r+12+3r+32+2  r+5I. 3(-2r) +1(3r+1)+14=-6r+3r+1+4=-3r+5 I ⁣I. 2(-2r) +2(3r+1)+14=-4r+6r+2+4= 2r+6 I ⁣I ⁣I. 1(-2r) +1(3r+1)+14=-2r+3r+1+4=  r+5 \gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \begin{aligned} I.\ &3·1r\ +1·(6r+1)\ +1·(\text{-}12r+4) =3r + 6r + 1 -12r+4&&=\text{-}3r+5\ &\checkmark \\ I\!I.\ &2 ·1r\ +2·(6r+1)\ +1·(\text{-}12r+4)=2r + 12r + 2 -12r+4 &&=\ 2r+6 \ &\checkmark \\ I\!I\!I.\ & 1·1r\ +1·(6r+1)\ +1·(\text{-}12r+4)=1 r + 6r + 1 -12r+4&&≠\ \ r +5\\ \\\\ I.\ &3·(\text{-}2r+\frac{1}{2})\ +1·(3r+\frac{3}{2})\ +1·2 =\text{-}6r +\frac{3}{2}+3r+\frac{3}{2}+2&&=\text{-}3r+5\ &\checkmark \\ I\!I.\ &2 ·(\text{-}2r+\frac{1}{2})\ +2·(3r+\frac{3}{2})\ +1·2=\text{-}4r+1+6r+3+2&&=\ 2r+6 \ &\checkmark \\ I\!I\!I.\ & 1·(\text{-}2r+\frac{1}{2})\ +1·(3r+\frac{3}{2})\ +1·2=\text{-}2r +\frac{1}{2}+3r+\frac{3}{2}+2&&≠\ \ r +5\\ \\\\ I.\ &3·(\text{-}2r)\ +1·(3r+1)+1·4 =\text{-}6r+3r+1+4&&=\text{-}3r+5\ &\checkmark \\ I\!I.\ &2 ·(\text{-}2r)\ +2·(3r+1)+1·4=\text{-}4r+6r+2+4&&=\ 2r+6 \ &\checkmark \\ I\!I\!I.\ & 1·(\text{-}2r)\ +1·(3r+1)+1·4=\text{-}2r + 3r + 1 + 4&&=\ \ r +5\ &\checkmark \end{aligned}

L3 gehört zum LGS.
2
Bestimme die Lösungsmenge. Nutze für die Berechnungen dein Heft.
a)
(2-23-1r22-16-3r14-134r1)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \\\left( \begin{array}{rrr|r} 2 & \text{-} 2 & 3 & \text{-1}r&-&2 \\ 2 & \text{-}1 & 6 & \text{-}3r&-&1 \\ 4 & \text{-}1 & 3 & 4r&-&1 \\ \end{array} \right)
b)
(10-1r+42118r41105r)\gdef\cloze#1{{\raisebox{-.05em}{\colorbox{none}{\color{526060}{\large{$\displaystyle #1$}}}}}} \\\left( \begin{array}{rrr|r} 1 & 0 & \text{-}1 & r&+&4 \\ 2 & 1 & 1 & 8r&-&4 \\ 1 & 1 & 0 & 5r \\ \end{array} \right)
a) L = {2r; r + 1; -r}, b) L = { 2r; 3r; r - 4}