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-3x+y+z=2, 5x+2y-4z=21, x-3y-7z=-10

0 votes

solve algebraically for x, y, and z

asked Nov 28, 2013 in ALGEBRA 2 by andrew Scholar

1 Answer

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

5x+2y-4z = 21 ---------> (2)

x-3y-7z = -10 ---------->(3)

Multiple the equation (1) to each side by 3.

-9x+3y+3z = 6 ---------> (4)

To eliminate the y value add the equtions (3) and (4).

-9x+3y+3z = 6

x-3y-7z = -10

____________

-8x-4z = -4

Divide to each side the above equation by 4.

-2x-z = -1-------->(5)

Multiple the equation (1) by 2.

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

To eliminate the y value subtract equation (6) from eqution (2).

5x+2y-4z = 21

-6x+2y+2z = 4

(+)  (-)  (-)    (-)

____________

11x-6z = 17 ---------> (7)

Multiple the equation (5) to each side by 6.

-12x-6z = -6 ----------> (8)

To eliminate the z value subtract the equation (8) from (7).

11x-6z = 17

-12x-6z = -6

(+)  (+)    (+)

___________

23x = 23

x = 23/23 = 1

Substitute the x value in (7).

11(1)-6z = 17

11-6z = 17

Subtract 11 to each side.

11-11-6z = 17-11

-6z = 6

Divide to each side by negitive 6.

-6z/-6 = 6/-6

z = -1

Substitute the x,z valuesin (3).

1-3y+7 = -10

-3y +8 = -10

Subtract 8 from each side

-3y +8-8 = -10-8

-3y =-18

Divide to each side by negitive 3.

-3y/-3 = -18/-3

y = 6

Solution of given system is x =1,y = 6, z = -1.

answered Nov 29, 2013 by william Mentor

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