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2 years ago

15

Solve the system by elimination.(show your work)

2 answers:

PilotLPTM [1.2K]2 years ago

6 0

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation by elimination:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

nata0808 [166]2 years ago

5 0

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation:

Add first equation to third one

-2x + 2y + 3z = 0

+

2x + 3y + 3z = 5

----------------- -------

5y + 6z = 5 (4th equation)

Subtract second equation to first one

-2x + 2y + 3z = 0

-

-2x - y + z = -3

------------------ --------

3y + 2z = 3 (5th equation)

Multiply 5th equation by 3

3*(3y + 2z) = 3*3

9y + 6z = 9 (6th equation)

Subtract 6th equation to fourth one

5y + 6z = 5

-

9y + 6z = 9

-----------

-4y = -4

Solve for y

-4y = -4

y = (-4)/(-4)

y = 1

Replace this value in 4th equation and solve for z

5(1) + 6z = 5

6z = 5-5

6z = 0

z = 0

Replace y and z values obtained in first equation and solve for x

-2x + 2y + 3z = 0

-2x + 2(1) + 3(0) = 0

-2x + 2 = 0

-2x = -2

x = (-2)/(-2)

x = 1

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