Question

Find the values of x and y by solving the following two equations $x+2+y+3+\sqrt{\left( x+2 \right)\left( y+3 \right)}=39$ and ${{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+\left( x+2 \right)\left( y+3 \right)=741$.

Answer 1

Hint: Since the second equation contains the square of the terms of the first equation, we will rearrange terms in the first equation and then we will square this rearranged equation. Then, we will be able to substitute the second equation in the first equation and we will get a linear equation in $x$ and $y$.

Complete step-by-step answer:
In the question, we are given two equations $x+2+y+3+\sqrt{\left( x+2 \right)\left( y+3 \right)}=39$ and ${{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+\left( x+2 \right)\left( y+3 \right)=741$. Let us number $x+2+y+3+\sqrt{\left( x+2 \right)\left( y+3 \right)}=39$ as equation $\left( 1 \right)$ and ${{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+\left( x+2 \right)\left( y+3 \right)=741$ equation $\left( 2 \right)$ respectively.
If we carefully notice both the equations i.e. equation $\left( 1 \right)$ and equation $\left( 2 \right)$, we will see that the equation $\left( 2 \right)$ contains the squared terms of the equation $\left( 1 \right)$. So let us first rearrange some terms in equation $\left( 1 \right)$ and then we will square the terms on the both sides of the equality in equation $\left( 1 \right)$.
Form equation $\left( 1 \right)$, we have,
$x+2+y+3+\sqrt{\left( x+2 \right)\left( y+3 \right)}=39$
We can also write this equation as,
$\sqrt{\left( x+2 \right)\left( y+3 \right)}=39-\left( x+2 \right)-\left( y+3 \right)$
Squaring both the sides in the above rearranged equation, we get,
${{\left( \sqrt{\left( x+2 \right)\left( y+3 \right)} \right)}^{2}}={{\left( 39-\left( x+2 \right)-\left( y+3 \right) \right)}^{2}}................\left( 3 \right)$
To solve the right side of the above equation, we have to use a formula,
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac.............\left( 4 \right)$
Solving right side of equation $\left( 3 \right)$ by substituting $a=39,b=-\left( x+2 \right),c=-\left( y+3 \right)$ in equation $\left( 4 \right)$, we get,
$\begin{align}
  & \left( x+2 \right)\left( y+3 \right)={{39}^{2}}+{{\left( -\left( x+2 \right) \right)}^{2}}+{{\left( -\left( y+3 \right) \right)}^{2}}+2\times 39\left( -\left( x+2 \right) \right)+2\times 39\left( -\left( y+3 \right) \right)+2\left( -\left( x+2 \right) \right)\left( -\left( y+3 \right) \right) \\
 & \Rightarrow \left( x+2 \right)\left( y+3 \right)={{39}^{2}}+{{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}-2\left( 39 \right)\left( x+2 \right)-2\left( 39 \right)\left( y+3 \right)+2\left( x+2 \right)\left( y+3 \right) \\
 & \Rightarrow {{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+\left( x+2 \right)\left( y+3 \right)=78\left( x+2+y+3 \right)-1521 \\
 & \Rightarrow {{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+\left( x+2 \right)\left( y+3 \right)=78\left( x+y+5 \right)-1521...........\left( 5 \right) \\
\end{align}$
Substituting ${{\left( x+2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+\left( x+2 \right)\left( y+3 \right)=741$ from equation $\left( 2 \right)$ in equation $\left( 5 \right)$, we get,
 $\begin{align}
  & 741=78\left( x+y+5 \right)-1521 \\
 & \Rightarrow 78\left( x+y+5 \right)=2262 \\
 & \Rightarrow x+y+5=29 \\
 & \Rightarrow x+y=24 \\
 & \Rightarrow y=24-x...........\left( 6 \right) \\
\end{align}$
Substituting $y=24-x$ from equation $\left( 6 \right)$ in equation $\left( 2 \right)$, we get,
$\begin{align}
  & {{\left( x+2 \right)}^{2}}+{{\left( 24-x+3 \right)}^{2}}+\left( x+2 \right)\left( 24-x+3 \right)=741 \\
 & \Rightarrow {{\left( x+2 \right)}^{2}}+{{\left( 27-x \right)}^{2}}+\left( x+2 \right)\left( 27-x \right)=741 \\
\end{align}$
We have a formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Using this formula in the above equation, we get,
$\begin{align}
  & {{x}^{2}}+4+4x+729+{{x}^{2}}-54x+27x-{{x}^{2}}+54-2x=741 \\
 & \Rightarrow 787-25x+{{x}^{2}}=741 \\
 & \Rightarrow {{x}^{2}}-25x+46=0 \\
 & \Rightarrow {{x}^{2}}-2x-23x+46=0 \\
 & \Rightarrow x\left( x-2 \right)-23\left( x-2 \right)=0 \\
 & \Rightarrow \left( x-23 \right)\left( x-2 \right)=0 \\
 & \Rightarrow x=2,x=23...........\left( 7 \right) \\
\end{align}$
Since we got $x=2,x=23$, substituting both $x=2,x=23$ from equation $\left( 7 \right)$ in equation $\left( 6 \right)$, we get,
$\begin{align}
  & y=24-2,y=24-23 \\
 & \Rightarrow y=22,y=1 \\
\end{align}$
Hence, there are two possible pairs of $\left( x,y \right)$. They are $\left( 2,22 \right)$ and $\left( 23,1 \right)$.

Note: If someone wants to check whether his/her answer is correct or not, he/she can check this by substituting the obtained $\left( x,y \right)$ pairs in the equations given in the question. If those $\left( x,y \right)$ pairs satisfy the equations given in the question, this means that we have got a correct answer.