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Solving systems of equations in three variables

Solving systems of equations in three variables-example-1
User Uglybugger
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Answer:

Option B: (x= -8, y=10, z=-6)

Explanation:

2x + 4y + 3z = 6 ... eq.(1)

5x + 8y + 6z = 4 ... eq.(2)

4x + 5y + 2z = 6 ... eq.(3)

From eq.(1),

2x= 6 - 4y - 3z

x= (6-4y-3z)/2 ... eq.(4)

Putting x as (6-4y-3z)/2 in both eq.(2) and eq.(3)

In eq.(2),

5[(6-4y-3z)/2] +8y +6z =4

Multiplying through by 2

5(6-4y-3z) + 16y + 12z = 8

30-20y-15z+16y+12z = 8

22= 4y + 3z ... eq(5)

In eq.(3),

4[(6-4y-3z)/2] +5y +2z = 6

2(6-4y-3z) +5y + 2z =6

12-8y-6z + 5y +2z = 6

6= 3y + 4z ... eq.(6)

Multiplying eq.(5) and eq.(6) by 3 & 4 respectively

66=12y + 9z ... eq(8)

24=12y + 16z ... eq(9)

Subtracting eq(8) from eq(9)

42= -7z

z= -42/7

z = -6

Putting z as -6 in eq.(5)

22= 4y + 3(-6)

22= 4y -18

4y = 40

y = 40/4

y = 10

Putting y as 10 and z as -6 in eq(4)

x = [6 - 4(10) - 3(-6)]/2

x = (6 - 40 - 18)/2

x = -16/2

x = -8

x= -8, y= 10 , z= -6

User WeakLearner
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