1+12+10

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

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ddx

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

simplify solve for inverse tangent line

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Steps Examples

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x+2z=8, x−2y−2z=4, 2x+5y−6z=6

Solution

x=6, y=0, z=1

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Solve by:

Solve by substitution

Solve by elimination

Solve by Cramer's rule

Solve by Gaussian elimination

One step at a time

[

x+2z=8

x−2y−2z=4

2x+5y−6z=6

]

Isolate x for x+2z=8: x=8−2z

Substitute x=8−2z

[

8−2z−2y−2z=4

2(8−2z)+5y−6z=6

]

Simplify

[

−4z−2y+8=4

−10z+5y+16=6

]

Isolate z for −4z−2y+8=4: z=−y−22

Substitute z=−y−22

[

−10(−y−22 )+5y+16=6

]

Simplify

[

10y+6=6

]

Isolate y for 10y+6=6: y=0

For z=−y−22

Substitute y=0

z=−0−22

−0−22 =1

z=1

For x=8−2z

Substitute y=0, z=1

x=8−2· 1

Simplify

x=6

The solutions to the system of equations are:

x=6, y=0, z=1

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Notebook

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

1+12+10

Solution

Solution steps

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