Question

Solve the following equations:
$x^2+y^2-z^2=21$,
$3xz+3yz-2xy=18$,
$x+y-z=5$.

• A. $(2,3,5)$
• B. $(3,4,2)$
• C. $(1,2,5)$
• D. $(2,5,3)$

Answer 1

Answer: (B)

$(3,4,2)$

Given equations are $x^2+y^2-z^2=21$ ......(i),

$3xz+3yz-2xy=18$

$\Rightarrow 3z(x+y)-2xy=18$ ........(ii)

and $x+y-z=5$

$\Rightarrow x+y=z+5$ ..........(iii)

$\Rightarrow x^2+y^2+2xy=z^2+25+10z\Rightarrow x^2+y^2-z^2+2xy=25+10z\Rightarrow 21+2xy=25+10z\Rightarrow 2xy=4+10z$ .........(iv)

On susbstituting (iii) and (iv) in (ii), we have

$3z(z+5)-(4+10z)=18\Rightarrow 3z^2+5z-22=0\Rightarrow 3z^2-6z+11z-22=0\Rightarrow 3z(z-2)+11(z-2)=0\Rightarrow (3z+11)(z-2)=0\Rightarrow z=2,-\frac{11}{3}$

Substituting $z$ in (i) and (iii),

(1) $z=2$

$x^2+y^2-4=21\Rightarrow x^2+y^2=25$ ........(a)

$x+y-2=5\Rightarrow x+y=7\Rightarrow y=7-x$ ...........(b)

Solving (a) and (b)

$x^2+49+x^2-14x=252x^2-14x+24=0x^2-7x+12=0x^2-4x-3x+12=0\Rightarrow (x-3)(x-4)=0x=3,4$

WE have $y=7-x$

$\Rightarrow y=4,3$

(2) $z=-\frac{11}{3}$

Substituting in (i) and (iii), we get

$x^2+y^2-\frac{121}{9}=21\Rightarrow 9x^2+9y^2=310$ ..........(c)

$x+y=5-\frac{11}{3}\Rightarrow 3x+3y=4\Rightarrow y=\frac{4-3x}{3}$ ........(d)

Solving (c) and (d), we get

$9x^2+9{\left(\frac{4-3x}{3}\right)}^2=310\Rightarrow 9x^2+16+9x^2-24x-310=0\Rightarrow 18x^2-24x-294=0\Rightarrow 6x^2-8x-98=0\Rightarrow x=\frac{8\pm \sqrt{64-4\left(6\right)(-98)}}{12}=\frac{8\pm \sqrt{2416}}{12}\Rightarrow x=\frac{8\pm 4\sqrt{151}}{12}=\frac{2\pm \sqrt{151}}{3}$

We have $y=\frac{4-3x}{3}$

$\Rightarrow y=\frac{4}{3}-x\Rightarrow y=\frac{4}{3}-\left(\frac{2\pm \sqrt{151}}{3}\right)\Rightarrow y=\frac{2\mp \sqrt{151}}{3}$

So, the values of $x$ are $3,4,\frac{2\pm \sqrt{151}}{3}$ values of $y$ are $4,3,\frac{2\mp \sqrt{151}}{3}$ and value sof $z$ are $2,-\frac{11}{3}$.