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taylor tan(5x)

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x+y+z=24, x−2y+0=0, 2x+3y+1=12

Solution

x=227 , y=117 , z=1357

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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+y+z=24

x−2y+0=0

2x+3y+1=12

]

Isolate x for x−2y+0=0: x=2y

Substitute x=2y

[

2· 2y+3y+1=12

2y+y+z=24

]

Simplify

[

7y+1=12

3y+z=24

]

Isolate y for 7y+1=12: y=117

Substitute y=117

[

3· 117 +z=24

]

Simplify

[

337 +z=24

]

Isolate z for 337 +z=24: z=1357

For x=2y

Substitute y=117 , z=1357

x=2· 117

2· 117 =227

x=227

The solutions to the system of equations are:

x=227 , y=117 , z=1357

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

taylor tan(5x)

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