number line 5/4

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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Steps Examples

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x+y−z=0, x+2y−3z=−6, −2x+3y−4z=10

Solution

x=−4, y=14, z=10

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Solve by:

Löse mit Eliminierung

Löse mit Substitution

Löse mit der Cramerschen Regel

Löse durch Gaußsche Eliminierung

One step at a time

[

x+y−z=0

x+2y−3z=−6

−2x+3y−4z=10

]

Subtrahiere die Gleichungen

x+2y−3z=−6

−

x+y−z=0

y−2z=−6

[

x+y−z=0

y−2z=−6

−2x+3y−4z=10

]

Multipliziere x+y−z=0 mit 2

[

2x+2y−2z=0

y−2z=−6

−2x+3y−4z=10

]

Addiere die Gleichungen

−2x+3y−4z=10

2x+2y−2z=0

5y−6z=10

[

2x+2y−2z=0

y−2z=−6

5y−6z=10

]

Multipliziere y−2z=−6 mit 5

[

2x+2y−2z=0

5y−10z=−30

5y−6z=10

]

Subtrahiere die Gleichungen

5y−6z=10

−

5y−10z=−30

4z=40

[

2x+2y−2z=0

5y−10z=−30

4z=40

]

Stelle 4z=40 nach z um: z=10

Setze z=10 in 5y−10z=−30 ein

Stelle 5y−10· 10=−30 nach y um: y=14

Setze y=14, z=10 in 2x+2y−2z=0 ein

Stelle 2x+2· 14−2· 10=0 nach x um: x=−4

Die Lösungen für das Gleichungssystem sind:

x=−4, y=14, z=10

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Hallo, eine Frage stellen zu x+y−z=0,x+2y−3z=−6,−2x+3y−4z=10 oder wählen Sie aus den folgenden Optionen

Lösen durch: Löse mit Substitution

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Notebook

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

number line 5/4

Solution

Solution steps

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