Mathematics - Application of Derivative Question with Solution | TestHub

$x+y+z=0$ ...(1) $2x+3y+z=0$ ...(2) $x+2y=0$ ...(3) From equation (3), we have $x=-2y$ Putting this value of $x$ in equations (1), we get $-2\mathrm{y}+\mathrm{y}+\mathrm{z}=0\Rightarrow \mathrm{y}=\mathrm{z}$ Hence $x=-2z$ Thus, the solution of the given system of equations is $(-2\mathrm{z},\mathrm{z},\mathrm{z})$, where $\mathrm{z}$ is a parameter $(Z\in R)$. Hence the system has infinite number of solutions including zero solution.