AB: Lektion Terme und Gleichungen Einführung (Teil 4)

1.

Wende dein Wissen zum Umstellen von Gleichungen und dem Suchen der Unbekannten x auf die folgenden Bruchterme an. Du solltest dich mit Brüchen auskennen.

a)

\( \frac{x}{2} = 5 \)

\( \frac{x}{2} = 5 \\ \bbox[#e1ffc1,5px]{ \frac{x}{2} = 5 \quad |·2 \\ \frac{x}{2} ·2 = 5 ·2 \\ x = 10 } \)

b)

\( \frac{x-1}{4} = 1 \)

\( \frac{x-1}{4} = 1 \\ \bbox[#e1ffc1,5px]{ \frac{x-1}{4} = 1 \quad |·4 \\ \frac{(x-1)·4}{4} = 1·4 \\ x-1 = 4 \quad |+1 \\ x = 4+1 \\ x = 5 } \)

c)

\( x - \frac{1}{4} = 1 \)

\( x - \frac{1}{4} = 1 \\ \bbox[#e1ffc1,5px]{ x - \frac{1}{4} = 1 \quad | +\frac{1}{4} \\ x = 1 + \frac{1}{4} \\ x = \frac{5}{4} } \)

d)

\( \frac{x}{2} + \frac{1}{2} = \frac{3}{2} \)

\( \frac{x}{2} + \frac{1}{2} = \frac{3}{2} \\ \bbox[#e1ffc1,5px]{ \frac{x}{2} + \frac{1}{2} = \frac{3}{2} \quad |·2 \\ \frac{x·2}{2} + \frac{1·2}{2} = \frac{3·2}{2} \\ x + 1 = 3 \quad |-1 \\ x = 2 } \)

e)

\( x + \frac{2}{5} = \frac{9}{10} \)

\( x + \frac{2}{5} = \frac{9}{10} \\ \bbox[#e1ffc1,5px]{ x + \frac{2}{5} = \frac{9}{10} \quad |-\frac{2}{5} \\ x = \frac{9}{10} - \frac{2}{5} \\ x = \frac{9}{10} - \frac{4}{10} \\ x = \frac{5}{10} \\ x = \frac{1}{2} } \)

f)

\( \frac{x}{3} + \frac{x}{4} + \frac{2x}{12} = 1 \)

\( \frac{x}{3} + \frac{x}{4} + \frac{2x}{12} = 1 \\ \bbox[#e1ffc1,5px]{ \frac{x ·4}{3 ·4} + \frac{x ·3}{4 ·3} + \frac{2·x}{12} = 1 \\ \frac{4·x}{12} + \frac{3·x}{12} + \frac{2·x}{12} = 1 \\ \frac{4·x + 3·x + 2·x}{12} = 1 \\ \frac{9·x}{12} = 1 \quad |·12 \\ 9·x = 1·12 \quad |:9 \\ x = \frac{12}{9} \\ x = \frac{4}{3} } \)

g)

\( \frac{4x+12}{9} = 15 \frac{1}{2} \)

\( \frac{4x+12}{9} = 15 \frac{1}{2} \\ \bbox[#e1ffc1,5px]{ \frac{4x+12}{9} = \frac{31}{2} \quad |·9 \\ \frac{4x+12}{9} ·9 = \frac{31}{2} ·9 \\ 4·x + 12 = \frac{279}{2} \quad |-12 \\ 4·x = \frac{279}{2} - 12 \\ 4·x = \frac{279}{2} - \frac{24}{2} \\ 4·x = \frac{255}{2} \quad |:4 \\ x = \frac{255}{8} \\ x = 31 \frac{7}{8} } \)

h)

\( \frac{x}{8} = \frac{x}{2} + 6 \)

\( \frac{x}{8} = \frac{x}{2} + 6 \\ \bbox[#e1ffc1,5px]{ \frac{x}{8} = \frac{x}{2} + 6 \quad |·8 \\ \frac{x}{8}·8 = \frac{x}{2}·8 + 6·8 \\ x = \frac{x·8}{2} + 48 \\ x = 4·x + 48 \quad |-4·x \\ -3·x = 48 \quad |:(-3) \\ x = 48:(-3) \\ x = -16 } \)

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