Question

How do you solve the system of equations with absolute value $\left| x+y \right|=5$ and $\left| x\cdot y \right|=2$?

Answer 1

Hint: The modulus function converts the positive or the negative value into the positive value. So on removing the modulus sign, we will get both the positive and the negative values and therefore we will obtain the given equations as $x+y=\pm 5$, and $xy=\pm 2$. We can divide our solution into two cases, one for $xy=2$ and the other for $xy=-2$. By solving the equations using the method of substitution, we will obtain four solutions in the respective cases, and hence total eight solutions of the given system of equations.

Complete step by step solution:
The given equations are
$\begin{align}
  & \Rightarrow \left| x+y \right|=5 \\
 & \Rightarrow \left| x\cdot y \right|=2 \\
\end{align}$
We know that the modulus of a negative or a positive value is positive. So we can write the first equation as
$\Rightarrow x+y=\pm 5$
From the above equation, we have the set of two equations given by
$\begin{align}
  & \Rightarrow x+y=5........\left( i \right) \\
 & \Rightarrow x+y=-5........\left( ii \right) \\
\end{align}$
Similarly from the second given equation we can write
$\begin{align}
  & \Rightarrow xy=\pm 2 \\
 & \Rightarrow xy=2........\left( iii \right) \\
 & \Rightarrow xy=-2........\left( iv \right) \\
\end{align}$
From the above two equations, we can form two cases.
Case I: $xy=2$
$\Rightarrow y=\dfrac{2}{x}........(v)$
Putting (v) in (i) we get
$\Rightarrow x+\dfrac{2}{x}=5$
Multiplying by $x$ both sides, we get
$\Rightarrow {{x}^{2}}+2=5x$
Subtracting $5x$ from both sides
\[\begin{align}
  & \Rightarrow {{x}^{2}}+2-5x=5x-5x \\
 & \Rightarrow {{x}^{2}}-5x+2=0.......\left( vi \right) \\
\end{align}\]
So we got a quadratic equation whose solution is given by
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Putting $a=1,b=-5,c=2$ from (vi) we get
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 2 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25-8}}{2} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{17}}{2} \\
 & \Rightarrow x=\dfrac{5+\sqrt{17}}{2},x=\dfrac{5-\sqrt{17}}{2} \\
\end{align}$
Putting these in (v), we get the respective values of $y$ as
$\Rightarrow y=\dfrac{4}{5+\sqrt{17}},y=\dfrac{4}{5-\sqrt{17}}$
Similarly putting (v) in (ii) we get
$\begin{align}
  & \Rightarrow x+\dfrac{2}{x}=-5 \\
 & \Rightarrow {{x}^{2}}+5x+2=0 \\
\end{align}$
Solving the above equation by using the quadratic formula we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-5\pm \sqrt{17}}{2} \\
 & \Rightarrow x=\dfrac{-5+\sqrt{17}}{2},x=\dfrac{-5-\sqrt{17}}{2} \\
\end{align}\]
Putting these in (v), we get the respective values of $y$ as
$\Rightarrow y=\dfrac{4}{-5+\sqrt{17}},y=\dfrac{4}{-5-\sqrt{17}}$
Case II: $xy=-2$
$\Rightarrow y=-\dfrac{2}{x}........\left( vii \right)$
Putting (vii) in (i) we get
$\Rightarrow x-\dfrac{2}{x}=5$
Multiplying by $x$ both sides, we get
$\Rightarrow {{x}^{2}}-2=5x$
Subtracting $5x$ from both sides
\[\begin{align}
  & \Rightarrow {{x}^{2}}-2-5x=5x-5x \\
 & \Rightarrow {{x}^{2}}-5x-2=0.......(viii) \\
\end{align}\]
So we got a quadratic equation whose solution is given by
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Putting $a=1,b=-5,c=-2$ from (viii) we get
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( -2 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25+8}}{2} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{33}}{2} \\
 & \Rightarrow x=\dfrac{5+\sqrt{33}}{2},x=\dfrac{5-\sqrt{33}}{2} \\
\end{align}$
Putting these in (viii), we get the respective values of $y$ as
$\Rightarrow y=\dfrac{4}{5+\sqrt{33}},y=\dfrac{4}{5+\sqrt{33}}$
Similarly putting (viii) in (ii) we get
$\begin{align}
  & \Rightarrow x-\dfrac{2}{x}=-5 \\
 & \Rightarrow {{x}^{2}}+5x-2=0 \\
\end{align}$
Solving the above equation by using the quadratic formula we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-5\pm \sqrt{33}}{2} \\
 & \Rightarrow x=\dfrac{-5+\sqrt{33}}{2},x=\dfrac{-5-\sqrt{33}}{2} \\
\end{align}\]
Putting these in (viii), we get the respective values of $y$ as
$\Rightarrow y=\dfrac{4}{-5+\sqrt{33}},y=\dfrac{4}{-5-\sqrt{33}}$
Hence, the given pair of equations is solved.

Note: We can also take the square of the first equation to get ${{\left( x+y \right)}^{2}}=25$. We know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$ so we will obtain ${{x}^{2}}+2xy+{{y}^{2}}=25$. Then putting the two value of the product $xy$ from the equation $\left| xy \right|=2$, we will get four equations in terms of ${{x}^{2}}$ and ${{y}^{2}}$ on solving which wec will get the eight solutions as obtained above.