Solve the following equations:
$\begin{align}
  & 3x-2y=7 \\
 & xy=20 \\
\end{align}$

Hint: We will first start by finding the value of y from the linear equation in x and y and then substitute it in the other equation to form a quadratic equation in x and then solve it to find the value of x and using these values we will find the value of y.

Complete step-by-step answer:
Now, we have been given two equations as,
$\begin{align}
  & 3x-2y=7........\left( 1 \right) \\
 & xy=20.........\left( 2 \right) \\
\end{align}$
Now, we have the value of y from (1) as $y=\dfrac{3x-7}{2}$ using this value we will substitute in equation (2) to form a quadratic equation in x. So, we have,
$\begin{align}
  & x\left( \dfrac{3x-7}{2} \right)=20 \\
 & x\left( 3x-7 \right)=40 \\
 & 3{{x}^{2}}-7x-40=0 \\
\end{align}$
Now, using factorization method, we can write it as,
$\begin{align}
  & 3{{x}^{2}}-15x+8x-40=0 \\
 & 3x\left( x-5 \right)+8\left( x-5 \right)=0 \\
 & \left( 3x+8 \right)\left( x-5 \right)=0 \\
 & either\ 3x+8=0\ or\ x-5=0 \\
 & x=\dfrac{-8}{3}\ or\ x=5 \\
\end{align}$
Now, substituting these values of x in (2) we have,
$\begin{align}
  & y\left( \dfrac{-8}{3} \right)=20 \\
 & y=\dfrac{-15}{2} \\
 & y\left( 5 \right)=20 \\
 & y=4 \\
\end{align}$
So, the value of x and y are,
\[\left( \dfrac{-8}{3},\dfrac{-15}{2} \right),\left( 5,4 \right)\]

Note: To solve these types of questions it is important to note that we have used the factorization method of solving quadratic equations of the form $a{{x}^{2}}+bx+c$ in this method. We write the term with x as the sum of factors of the product of the constant term and coefficient of ${{x}^{2}}$.