solve by quadratic formula 2w(2w-1)=2

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x+y+z=25, 5x+3y+2z=0, y−z=6

Solution

x=−1315 , y=1435 , z=1135

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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=25

5x+3y+2z=0

y−z=6

]

Isolate y for y−z=6: y=6+z

Substitute y=6+z

[

x+6+z+z=25

5x+3(6+z)+2z=0

]

Simplify

[

x+6+2z=25

5x+18+5z=0

]

Isolate x for x+6+2z=25: x=−2z+19

Substitute x=−2z+19

[

5(−2z+19)+18+5z=0

]

Simplify

[

−5z+113=0

]

Isolate z for −5z+113=0: z=1135

For x=−2z+19

Substitute z=1135

x=−2· 1135 +19

−2· 1135 +19=−1315

x=−1315

For y=6+z

Substitute x=−1315 , z=1135

y=6+1135

6+1135 =1435

y=1435

The solutions to the system of equations are:

x=−1315 , y=1435 , z=1135

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

solve by quadratic formula 2w(2w-1)=2

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