x+y+z = 20 3x+4y+z = 48 4x+3y+6z = 96 Values of x, y, z are respectively

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22 Jun 2012 solve x+y+z = 20 3x+4y+z = 48 4x+3y+6z = 96 Values of x, y, z are respectively: OPtion
1) 5, 6, 9
2) 4, 6, 9
3) 6, 5, 9
4) 2, 8, 12
5) 1, 6, 10
6) 2, 7, 6
7) 4, 8, 3
8) 5, 8, 9
9) no solution possible
10) none of these; Correct Option: 9 Best Solution (10)

Niraj 12 years AGO

Adding all 3 given equations,
8x+8y+8z = 164 ----- Equation A
From equation 1,
x+y+z = 20
=> 8x+8y+8z = 160 ----- Equation B
For same LHS, RHS are different.
so no solutions possible.

Like? Yes (3) | No (1)

manudixit 12 years AGO

No solution is possible as we back substitute eq1 in other eqn it same LHS with different RHS.

Like? Yes (1) | No (1)

Shiv Kumar 12 years AGO

from the equation 1, Z = 20-X-Y ------ a
After substitute this in equation 2,
3X+4Y+20-X-Y = 48
2X+3Y=28 -------- B
After substitute equation a in equation 3,
4X+3Y+6Z = 96
4X+3Y+120-6X-6Y = 96
2X+3Y = 24 ----- C

Equation B and C on left hand side they are same. However righ hand side is different. So solution is not possible.

Like? Yes (1) | No (1)

Narender Bhardwaj 12 years AGO

Det(or Area) of above equatons is=|1 1 1|
|3 4 1|
|4 3 6|
=1(4*6-3*1)-1(6*3-4*1)+1(3*3-4*4);
=(24-3)-(18-4)+(9-16)=21-21=0,Since Det(or Area) of above equations is zero,Hence it has no solutions.

Like? Yes (1) | No (1)

Ankit 12 years AGO