Question

Solve the following equations:
$3x+y-2z=0$,
$4x-y-3z=0$,
$x^3+y^3+z^3=467$

• A. $x=5;y=-1;z=5$
• B. $x=2;y=-1;z=-7$
• C. $x=3;y=-2;z=7$
• D. $x=5;y=-1;z=7$

Answer 1

The correct option is D $x=5;y=-1;z=7$

Let $3x+y-2z=0$ ........(i)

$4x-y-3z=0$ .........(ii)

and $x^3+y^3+z^3=467$ .......(iii)

Solving (i) and (ii) by cross multiplication

$\frac{x}{-3-2}=\frac{y}{-8+9}=\frac{z}{-3-4}=k$ (let)

$\frac{x}{-5}=\frac{y}{1}=\frac{z}{-7}=k\Rightarrow x=-5k,y=k,z=-7k$ .........(iv)

Substituting $x,y,z$ in (iii), we have

${(-5k)}^3+({k)}^3+{(-7k)}^3=467-125k^3+k^3+-343k^3=467-467k^3=467\Rightarrow k=-1$

Substituting $k$ in (iv)

$\Rightarrow x=5,y=-1,z=7$

which is the required solution