The values of the variables are x = -2, y = 6 and z = -3
How to solve by elimination?
The system is given as:
4x + 2y - 2z = 10 (A)
2x + 8y + 4z = 32 (B)
30x + 12y - 4z = 24 (C)
The easiest variable to eliminate is z.
Combine A and B
2[4x + 2y - 2z = 10]
⇒
8x + 4y - 4z = 20
+ 2x + 8y + 4z = 32
⇒ 10x + 12y = 52
Combine B and C
2x + 8y + 4z = 32
30x + 12y - 4z = 24
⇒ 32x + 20y = 56
Divide through by 4
8x + 5y = 14
So, we have:
10x + 12y = 52
and
8x + 5y = 14
Multiply 10x + 12y = 52 by 8/10
8/10[10x + 12y = 52]
⇒ 8x + 9.6y = 41.6
Combine 8x + 9.6y = 41.6 and 8x + 5y = 14
8x + 9.6y = 41.6
- [8x + 5y = 14]
⇒ 4.6y = 27.6
Divide
y = 6
Substitute y = 6 in 8x + 5y = 14
8x + 5 * 6 = 14
8x + 30 = 14
Subtract 30
8x = -16
Divide
x = -2
Substitute y = 6 and x = -2 in (A)
4x + 2y - 2z = 10
4(-2) + 2(6) - 2z = 10
-8 + 12- 2z = 10
Evaluate the like terms
-2z = 6
Divide
z = -3
Hence. the values of the variables are x = -2, y = 6 and z = -3
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