Solving equations in three unknowns (x, y and z) requires three independent equations. Since you have only one equation there is no solution.
The equation can be simplified (slightly) by dividing through by 4 to give:
x + 2y + 3z = 11
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-99 = 12z - 3(6)zSimplifying the expression, we have:-99 = 12z - 18z = -6zDivide by -6:z = 16.5Or, if you meant 3 6 as the number, we have:-99 = 12z - 36zSimplify:-99 = -24zDivide by -24:z = 4.125
x=3y y=4z x=12z x+y+z=z+4z+12z=153 now you have the answer
If you mean: 2x-5y+2z = 16 and 3x+2y-3z = -19 and 4x-3y+4z = 18 then the solutions are found as follows:- 3(2x-5y+2z = 16) => 6x-15y+6z = 48 2(3x+2y-3z = -19) => 6x+4y-6z = -38 Adding the above: 12x-11y = 10 thus eliminating z 4(3x+2y-3z = -19) => 12x+8y-12z = -76 3(4x-3y+4z = 18) => 12x-9y+12z = 54 Adding the above: 24x-y = -22 thus eliminating z 2(12x-11y = 10) => 24x-22y = 20 1(24x-y = -22) => 24x-y = -22 Subtracting the above: -21y = 42 thus eliminating x If: -21y = 42 then y = -2 So by substitution: x = -1, y = -2 and z = 4 Check: (2*-1)-(5*-2)+(2*4) = 16 Check: (3*-1)+(2*-2)-(3*4) = -19 Check: (4*-1)-(3*-2)+(4*4) = 18
Numbers are divisible by 6 if they are even and also divisible by 3. examples are 6,12,18,24,30,36,60,54,...... examples as algebric expressions 18xy+12z+6y-a,where a is divisible by 6 i.e 6,12,18...