Question

$x+y+z=1,2x+2y+3z=6,x+4y+9z=3$.

Answer 1

Given equation

$x+y+z=1----\left(1\right)$

$2x+2y+3z=5----\left(2\right)$

$x+4y+9z=3----\left(3\right)$

$\left(2\right)-2\times \left(1\right)\Rightarrow \frac{2x+2y+3z=62x+2y+2z=2}{z=4}$

$\left(2\right)-2\times \left(1\right)\Rightarrow \frac{x+4y+9z=34x+4y+6z=12}{-3x+3z=-9-x+z=-3x=z+3=7}$

$\left(1\right)\Rightarrow x+y+z=1$

$7+y+4=1$

$y=-10$

$\therefore x=7,y=-10,z=4$