Is there any solutions for this set of equation?

Question

Is there any solutions？ Why?How? $$ \begin{array}{cccc} 3z &+ &4x + 3y &= 12\\ z &+ &4x + 3y &= 12\\ 2z &+ &4x + 3y &= 12\\ \end{array} $$

Answer 1

Let $$z+4x+3y=12$$ $$3z+4x+3y=12$$ $$2z+4x+3y=12$$ Multiplying the first equation by $$-3$$ and Adding to the second $$-8x-6y=-24$$ and by the same way we get $$-4x-3y=-12$$ Can you proceed?

Answer 2

Whatever the values of $x$ and $y$, $3z=z=2z$ must hold, and $$z=0.$$ Then the system reduces to

$$4x+3y=12.$$

Answer 3

Yes!! There are many solutions to this set of equations 3z +4x +3y = 12z +4x +3y => 3z = 12z => z= 0 Now, since 12 2z + 4x+3y =12 => 4x +3y =12 ( as z=0) => 4*0 + 3*4 =12 ie x=0,y=4 ,z=0 Or, 4*6/4 +3*2 =12 ie, x=3/2, y=2, z=0 Or, 4*0 + 3*4 = 12 ie, x=0, y= 4, z= 0 This way, x,y,z can hold infinite sets of values

Answer 4

Assume there is a solution $x_0, y_0$, and $z_0 \not = 0$.

Then :

1) $az_0= 12-(4x_0+3y_0)$, where $a =1,2,3.$

Contradiction (Why?).

Hence $z_0=0.$

2) The 3 equations reduce to 1 equation:

$4x+3y=12.$

In 2D (recall $z_0=0$) this is a line.

Combining:

3) Solutions:

$z_0=0$; and $4x+3y=12$ i.e.

$z_0=0$, and all points $x_0,y_0$ that satisfy $4x_0+3x_0= 12$.

Answer 5

The solutions also depend on the "field". For a field of characteristic $p=2$ the system reduces to a single equation $z=y$. Otherwise, for $p\neq 2$, we obtain $2z=0$ and hence $z=0$, and the single equation $4x+3y=12$.