Question:
Solve the following systems of equations:
$x-y+z=4$
$x+y+z=2$
$2 x+y-3 z=0$
Solution:
The given equations are:
$x-y+z=4 \ldots(i)$
$x+y+z=2 \quad \ldots($ ii $)$
$2 x+y-3 z=0 \quad \ldots($ iii $)$
First of all we find the value of $x$
$x=4+y-z$
Put the value of $x$ in equation $(i)$, we get
$4+y-z+y+z=2$
$\Rightarrow 2 y=-2$
$\Rightarrow y=-1$
Put the value of $x$ and $y$ in equation in $(i i i)$ we get
$2(4+y-z)+y-3 z=0$
$\Rightarrow 8-2-2 z-1-3 z=0$
$\Rightarrow-5 z=-5$
$\Rightarrow z=1$
Put the value of $y$ and $z$ in equation $(i)$, we get
$x-(-1)+1=4$
$\Rightarrow x=2$
Hence the value of $x=2, y=-1$ and $z=1$.
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