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1. -x-y-z=-8

-4x+4y+5z=7

2x+2z=4

solve by substitution.


2. -2x+2y+3z=0

-2x-y+z=-3

2x+3y+3z=5

solve by elimination.


please help, I've been stuck for a week now!

1 Answer

1 vote

Answer:1) x =19

y = -13

z = 2

2) x= 1

y= 1

z = 0

Explanation:

1)

-x-y-z=-8--------1

-2x-y+z=-3 --------2

2x+3y+3z=5--------3

From eqn 1,

x= 8 - y-z

Put x in eqn 2 and 3

- 2(8-y-z) - y +z = -3

-16+2y+2z-y+z = -3

y+ 3z = 13 -----4

2(8-y-z)+3y+3z=5

16-2y-2z+3y +3z = 5

y+z=-11 -------5

Using elimination method for eqn 4 and 5

2z = 24

z = 2

y = -11-2 = -13

Put x and y into eqn 1

x = 8- -13-2 =19

2)

-2x+2y+3z=0--------1

-2x-y+z=-3-------------2

2x+3y+3z=5-----------3

Subtract eqn 2 from eqn 1

2y - -y +2z = -3

3y +2z = 3----------4

Add eqn 2 and eqn 3

2y +4z = 2---------5

Multiply eqn 4 by 2 and eqn 5 by 3

6y + 4z =6 --------6

6y + 12z = 6-------7

Subtract eqn 7 from eqn 6

-8z = 0

z = 0

2y +4z = 2

2y + 4*0=2

2y = 2

y = 1

Putting y = 1 and z = 0 in eqn 1

-2x+2*1+3*0=0

-2x + 2+ 0 = 0

-2x = -2

x = 1

User Mabahamo
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