Question

Solve the following equations:
$x+2y-z=11$,
$x^2-4y^2+z^2=37$,
$xz=24$.

• A. $x=2,-5;y=2;z=2,-4$
• B. $x=8,-3;y=3;z=-3,-8$
• C. $x=-3,5;y=4;z=2,5$
• D. $x=2,4;y=3;z=3,-5$

Answer 1

Answer: (B)

$x=8,-3;y=3;z=-3,-8$

$x^2-4y^2+z^2=37$ ......(i)

$xz=24$ .......(ii)

$x+2y-z=11$ .....(iii)

$\Rightarrow x-z=11-2y$

On squaring both sides, we have

$x^2+z^2-2xz=121+4y^2-44y\Rightarrow x^2+z^2-4y^2-2xz=121-44y\Rightarrow 37-2\left(24\right)=121-44y\Rightarrow -44y=-132\Rightarrow y=3$

Substituting $y$ in (iii),

$x+6-z=11\Rightarrow x-z=5$

From (ii), $z=\frac{24}{x}$

Thus $x-\frac{24}{x}=5$

$\Rightarrow x^2-24=5x\Rightarrow x^2-5x-24=0\Rightarrow x^2-3x+8x-24=0\Rightarrow x(x-3)+8(x-3)=0\Rightarrow (x+8)(x-3)=0\Rightarrow x=-8,3$

Putting in $z=\frac{24}{x}$

Thus $z=-3,8$

So, the values of $x$ are $-8,3$, values of $z$ are $-3,8$ and value of $y$ is $3$.