Question

Solve the following equations:
$xy+x+y=23$,
$xz+x+z=41$,
$yz+y+z=27$.

• A. $x=4,-2;y=2;6;z=6,-5$
• B. $x=2,-4;y=2,4;z=2,-6$
• C. $x=5,-7;y=3,-5;z=6,-8$
• D. $x=3,4;y=2,-5;z=2,-7$

Answer 1

The correct option is C $x=5,-7;y=3,-5;z=6,-8$

Given equations are $xy+x+y=23$

$\Rightarrow x(y+1)=23-y$

$\Rightarrow x=\frac{23-y}{y+1}$ ........(i),

$yz+y+z=27$

$\Rightarrow z(y+1)=27-z$

$\Rightarrow z=\frac{27-y}{y+1}$ ...........(ii)

and $xz+x+z=41$ .....(iii)

Substituting (i) and (ii), we get

$\left(\frac{23-y}{y+1}\right)\left(\frac{27-y}{y+1}\right)+\frac{23-y}{y+1}+\frac{27-z}{y+1}=41\Rightarrow \frac{621+y^2-50y}{{(y+1)}^2}+\frac{50-2y}{y+1}=41\Rightarrow \frac{621+y^2-50y+(50-2y)(y+1)}{{(y+1)}^2}=41\Rightarrow 621+y^2-50y+50y-50-{2y}^2-2y=41y^2+41+82y\Rightarrow 42y^2+84y-630=0\Rightarrow y^2+2y-15=0\Rightarrow y^2-3y+5y-15=0\Rightarrow y(y-3)+5(y-3)=0\Rightarrow (y+5)(y-3)=0\Rightarrow y=-5,3$

Substituting $y$ in (i), we get

$\Rightarrow x=-7,5$

Substituting $y$ in (ii), we get

$\Rightarrow z=-8,6$