Question

Solve the following simultaneous linear equations by method of reduction.
$x+2y+z=8$
$2x+3y-z=11$
$3x-y-2z=5$

Answer 1

$x+2y+z=8.....\left(1\right)$

$2x+3y-z=11.....\left(2\right)$

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

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$(x+2y+z)+(2x+3y-z)=8+11$

$\Rightarrow 3x+5y=19.....\left(4\right)$

Subtracting equation $\left(2\right)$ from $\left(3\right)$, we have

$(3x-y-2z)-(2x+3y-z)=5-11$

$\Rightarrow x-4y-z=-6.....\left(5\right)$

Now adding equation $\left(1\right)\&\left(5\right)$, we have

$(x+2y+z)+(x-4y-z)=8-6$

$\Rightarrow 2x-2y=2$

$\Rightarrow x-y=1.....\left(6\right)$

Multiplying ${eq}^n\left(6\right)$ by $5$, we get

$5x-5y=5.....\left(7\right)$

Adding ${eq}^n\left(4\right)\&\left(7\right)$, we have

$(3x+5y)+(5x-5y)=19+5$

$\Rightarrow 8x=24$

$\Rightarrow x=3$

Substituting the value of $x$ in ${eq}^n\left(6\right)$, we get

$3-y=1$

$\Rightarrow y=3-1=2$

Substituting the value of $x$ aand $y$ in ${eq}^n\left(1\right)$, we get

$3+(2\times 2)+z=8$

$\Rightarrow z=8-7=1$

Hence $x=3,y=2,z=1$.