Question

Solve the following equations:
$x+2y+2z=1$
$3x-2y-3z=0$
$x-2y+2z=1$

• A. $x=\frac{1}{2},y=0andz=\frac{1}{3}$
• B. $x=\frac{1}{3},y=0andz=\frac{2}{3}$
• C. $x=\frac{1}{3},y=1andz=\frac{1}{3}$
• D. $x=\frac{1}{3},y=0andz=\frac{1}{3}$

Answer 1

The correct option is D $x=\frac{1}{3},y=0andz=\frac{1}{3}$

We have
$x+2y+2z=1$...........(1)
$3x-2y-3z=0$...........(2)
$x-2y+2z=1$............(3)

Adding equations (1) and (2), we get

$4x-z=1$............(4)

Adding equations (1) and (3), we get

$2x+4z=2$............(5)

Multiplying equation (4) by 4, we have
$16x-4z=4$...............(6)

Adding equations (5) and (6), we get

$18x=6$

$\Rightarrow x=\frac{1}{3}$

Putting $x=\frac{1}{3}$ in equation 4, we get

$\frac{4}{3}-z=1$

$\Rightarrow z=\frac{1}{3}$

Putting $x=\frac{1}{3}andz=\frac{1}{3}$ in equation (1), we get

$\frac{1}{3}+2y+\frac{2}{3}=1$

$\Rightarrow 2y=1-\frac{1}{3}-\frac{2}{3}$

$\Rightarrow y=0$

Thus, we have

$x=\frac{1}{3},y=0andz=\frac{1}{3}$