Question

Solve the following equations:
$x^2+2xy+3xz=50$,
$2y^2+3yz+yx=10$,
$3z^2+zx+2zy=10$.

• A. $x=\pm 4;y=\pm 2;z=\pm 2$
• B. $x=\pm -4;y=\pm -2;z=\pm 2$
• C. $x=\pm 5;y=\pm 1;z=\pm 1$
• D. None of these

Answer 1

The correct option is C $x=\pm 5;y=\pm 1;z=\pm 1$

Given equations are $x^2+2xy+3xz=50$

$\Rightarrow x(x+2y+3z)=50$ ........(i),

$2y^2+3yz+yx=10$

$\Rightarrow y(2y+3z+x)=10$ ........(ii)

and $3z^2+zx+2zy=10$

$\Rightarrow z(3z+x+2y)=10$ ..........(iii)

Dividing (i) by (ii), we get

$\frac{x}{y}=\frac{50}{10}=5\Rightarrow x=5y$ ......(a)

Dividing (ii) by (iii), we get

$\frac{y}{z}=\frac{10}{10}=1\Rightarrow z=y$ ..........(b)

Substituting (a) and (b) in (ii)

$y(2y+3y+5y)=10\Rightarrow 10y^2=10\Rightarrow y=\pm 1$

From (a), we have

$x=\pm 5$

From (b), we have

$z=\pm 1$