Question

Solve the following equations:
\[\begin{align}
  & xy+x+y=23, \\
 & xz+x+z=41, \\
 & yz+y+z=27. \\
\end{align}\]
(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

Hint: Observe the following relation $ab+a+b+1=\left( a+1 \right)\left( b+1 \right)$
It means add 1 to both sides in each of the equations and convert each equation of the above form. Now, multiply each equation to get another equation. Take square root to both sides of this equation to get the value of $\left( x+1 \right)\left( y+1 \right)\left( z+1 \right)$ . Now, use the individual equations calculated after adding 1 to each equation. Get $x,y,z$ by using these equations.

Complete step-by-step solution -
Given equations in the problem are
$\begin{align}
  & xy+x+y=23................\left( i \right) \\
 & xz+x+z=41................\left( ii \right) \\
 & yz+y+z=27..................\left( iii \right) \\
\end{align}$
As we know expression $\left( a+1 \right)\left( b+1 \right)$ can be given as
$\left( a+1 \right)\left( b+1 \right)=ab+a+b+1...............\left( iv \right)$
Now, by comparing the R.H.S of above equation by L.H.S of all three equations given in the problem. We get that if we add 1 to L.H.S. of all of three equation (i), (ii) and (iii), we can write them in form of expression given in equation (iv).
So, adding 1 to both sides of the equation (i), we get
$\begin{align}
  & xy+x+y+1=23+1 \\
 & \Rightarrow xy+x+y+1=24 \\
\end{align}$
Taking $x$ as common from first two terms and 1 as common from last two terms of the expression given in the L.H.S of the above equation. So, we get
$\begin{align}
  & x\left( y+1 \right)+1\left( y+1 \right)=24 \\
 & \Rightarrow \left( x+1 \right)\left( y+1 \right)=24..............\left( v \right) \\
\end{align}$
Now, similarly we can add 1 to both sides of equation (ii) as well. So, we can re-write the equation (ii) as
$\left( x+1 \right)\left( z+1 \right)=42..............\left( vi \right)$
And hence, equation (iii) can be re-written as
$\left( y+1 \right)\left( z+1 \right)=28...............\left( vii \right)$
On multiplying equations (v), (vi) and (vii), we can get relation as
\[\begin{align}
  & \left( x+1 \right)\left( y+1 \right)\left( x+1 \right)\left( z+1 \right)\left( y+1 \right)\left( z+1 \right)=24\times 42\times 28 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}{{\left( y+1 \right)}^{2}}{{\left( z+1 \right)}^{2}}=24\times 42\times 28 \\
 & \Rightarrow {{\left[ \left( x+1 \right)\left( y+1 \right)\left( z+1 \right) \right]}^{2}}=4\times 6\times 6\times 7\times 4\times 7 \\
 & \Rightarrow {{\left[ \left( x+1 \right)\left( y+1 \right)\left( z+1 \right) \right]}^{2}}={{\left( 4\times 6\times 7 \right)}^{2}} \\
\end{align}\]
On taking square roots to both sides of the above equation, weget
$\left( x+1 \right)\left( y+1 \right)\left( z+1 \right)=\pm 168...............\left( viii \right)$
Now, putting $\left( x+1 \right)\left( y+1 \right)=24$ from the equation (v) to the above equation, we get
$24\left( z+1 \right)=\pm 168$
On dividing the whole equation by 24, we get
$\begin{align}
  & z+1=\pm \dfrac{168}{24}=\pm 7 \\
 & z+1=\pm 7 \\
 & \Rightarrow z+1=7,z+1=-7 \\
 & \Rightarrow z=7-1,z=-7-1 \\
 & \Rightarrow z=6,z=-8 \\
\end{align}$
So, we get values of z as
$z=6,-8$
Similarly, putting value of $\left( x+1 \right)\left( z+1 \right)=42$ from equation (vi) to the equation (viii), we get
     \[\begin{align}
  & 42\left( y+1 \right)=\pm 168 \\
 & y+1=\pm \dfrac{168}{42}=\pm 4 \\
 & \Rightarrow y+1=4,y+1=-4 \\
 & y=4-1,y=-4-1 \\
 & y=3,y=-5 \\
\end{align}\]
Hence, we get values of ‘y’ as
$y=3,-5$
Similarly, we can put $\left( y+1 \right)\left( z+1 \right)=28$ from the equation (vii) to the equation (viii). We get
$28\left( x+1 \right)=\pm 168$
On dividing the whole equation by 28, we get
$\begin{align}
  & x+1=\pm \dfrac{168}{28}=\pm 6 \\
 & \Rightarrow x+1=6,x+1=-6 \\
 & x=6-1,x=-6-1 \\
 & x=5,x=-7 \\
\end{align}$
Hence, we get values of $x$ as
$x=5,-7$
Hence, we get values of $x,y,z$ as $x=5,-7,y=3,-5,z=6,-8$
So, option (c) is the correct answer of the problem.

Note: One may try to eliminate any two variables using all three equations. It will be a longer approach and would be complex as well. So, try to observe the given equations to proceed further to get solutions for them. Answer remains the same but it would not be flexible than given in solution.
One may go wrong with the step while taking the square root of the equation
${{\left( \left( x+1 \right)\left( y+1 \right)\left( z+1 \right) \right)}^{2}}={{\left( 4\times 6\times 7 \right)}^{2}}$
One may consider only positive signs after taking the square root to both sides, which is wrong. So, take care of it that square of any number with the positive sign will be the same as the square of the same number with a negative sign. So, take care of it.