Question

If the system of linear equations $x-4y+7z=g,3y-5z=h,-2x+5y-gz=k$ is consistent, then

• A. $2g+h+k=0$
• B. $g+2h+k=0$
• C. $g+h+k=0$
• D. $g+h+2k=0$

Answer 1

The correct option is A $2g+h+k=0$

$x-4y+7z=g$ ……………$\left(1\right)$

$3y-5z=h$ ……………$\left(2\right)$

$-2x+5y-gz=k$ …………$\left(3\right)$

Given that line $\left(1\right),\left(2\right)$ & $\left(3\right)$ are consistent then

$[\frac{a}{a_1}=\frac{b}{b_1}=\frac{c}{c_1}]$

But we have to make $2$ times from $\left(1\right),\left(2\right),\left(3\right)$

From equation $\left(1\right)$ & $\left(3\right)$ we have

$\left(1\right)\times \left(2\right)+3$

$+2x-8y+14z=+2g$

$-2x+5y-gz=k$

__________________

$-3y+(14-g)z=2g+k$ ………$\left(4\right)$

Now on comparing equation $\left(2\right)$ & $\left(4\right)$

$\frac{3}{-3}=\frac{-5}{14-g}=\frac{h}{2g+k}$

$\frac{h}{2g+k}=1$

$[h+2h+k=0]$.