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Solving Systems of Equations by Elimination?

0 votes
line 1: -x - 5y - 5z = 2 line 2: 4x - 5y + 4z = 19 line 3: x + 5y - z = -20 or

line 1: -x - 5y + z = 17 line 2: -5x - 5y + 5z = 5 line 3: 2x + 5y - 3z = -10
asked Nov 15, 2013 in ALGEBRA 2 by johnkelly Apprentice

1 Answer

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1) line 1: -x-5y-5z = 2

line2: 4x-5y+4z = 19

line3: x+5y-z = -20

To eliminate x value add the lines 1 and 3.

-x-5y-5z = 2

x+5y-z = -20

___________

-6z = -18

Divide to each side by negitive 6.

z = 3

Substitute z value in line 1 and 2.

-x-5y-15 = 2

-x-5y = 17----> (1)

4x-5y+4*3 = 19

4x-5y+12 = 19

4x-5y = 7 ----> (2)

To eliminate the y value subtract  (1) from (2).

4x-5y = 7

-x-5y = 17

(+) (+)  (-)

_________

5x =-10

x = -10/5

x = -2

Substitute x, z in line1.

-(-2)-5y-5*3 = 2

2-5y-15 = 2

-5y -13 = 2

-5y = 15

y = 15/-5

y = -3

Solution x = -2, y = -3, z = 3.

2) -x-5y+z = 17 -----> (1)

-5x-5y+5z = 5 ----> (2)

 2x+5y-3z = -10 ----> (3)

Multiple to each side of line 1 by  negitive 5.

5x+25y-5z = -85 ---> (4)

To eliminate x value add  (2)&(5) .

-5x-5y+5z = 5

5x+25y-5z = -85

______________

20y = -80

y = -80/20 = -4

Substitute y value in (1),(3)

-x+20+z = 17

-x+z = -3 -----> (5)

2x-20-3z = -10

2x-3z = 10 ----> (6)

Multiple to each  sdie of (5) by 2.

-2x+2z = -6 ----- (7)

To eliminate x , add (7)&(6)

2x-3z = 10

-2x+2z = -6

__________

-z = 4

z = -4

Substitute y,z in (1).

-x-5*-4-4 = 17

-x+20-4 = 17

-x = 17-16

-x = 1

x = -1

Solution x = -1, y = -4, z = -4.

answered Dec 17, 2013 by friend Mentor

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