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solve 2x + y – z = 3, 4x – y + 4z = 0 and -3y + 2z = 6

0 votes

Systems of Equations with Three Variables

asked Nov 28, 2013 in ALGEBRA 2 by andrew Scholar
reshown Nov 28, 2013 by goushi

1 Answer

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2x+y-z = 3 ----------> (1)

4x-y+4z = 0 ----------> (2)

-3y+2z = 6 -----------> (3)

Multiple to each side to equation (1) by 2.

4x+2y-2z = 6 ------------> (4)

To eliminate x value subtract equation (4) from equation(2).

4x-y+4z = 0

4x+2y-2z = 6

(-)  (-) (+)   (-)

____________

-3y+6z = -6 ---------> (5)

To eliminate y value subtract (5) from (3)

-3y+2z = 6

-3y+6z = -6

(+) (-)     (+)

__________

-4z = 12

Divide to each side by negitive 4.

-4z/-4 = 12/-4

z = -3

Substitute the z value in (3).

-3y+2(-3) = 6

-3y-6 = 6

Add 6 to each side.

-3y-6+6 = 6+6

-3y = 12

Divide to each side by negitive 3.

-3y/-3 = 12/-3

y = -4

Substitute the y,z in (1)

2x-4+3 =3

2x-1 = 3

Add 1 to each side.

2x-1+1 =3+1

2x = 4

Divide to each side by 2.

2x/2 =4/2

x = 2

Solution of given system is x= 2,y = -4, z = -3.

answered Nov 28, 2013 by william Mentor

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