substitution x+y-z=3, 2x+y+z=5, 3x-2y+z=4

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

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√☐

^□√☐

□□

log_□

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ddx

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x+y−z=3, 2x+y+z=5, 3x−2y+z=4

Solution

x=2, y=1, z=0

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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=3

2x+y+z=5

3x−2y+z=4

]

Isolate x for x+y−z=3: x=3−y+z

Substitute x=3−y+z

[

2(3−y+z)+y+z=5

3(3−y+z)−2y+z=4

]

Simplify

[

−y+3z+6=5

−5y+4z+9=4

]

Isolate y for −y+3z+6=5: y=3z+1

Substitute y=3z+1

[

−5(3z+1)+4z+9=4

]

Simplify

[

−11z+4=4

]

Isolate z for −11z+4=4: z=0

For y=3z+1

Substitute z=0

y=3· 0+1

Simplify

y=1

For x=3−y+z

Substitute y=1, z=0

x=3−1+0

Simplify

x=2

The solutions to the system of equations are:

x=2, y=1, z=0

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

substitution x+y-z=3, 2x+y+z=5, 3x-2y+z=4

Solution

Solution steps

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