Solve the following equations by reduction method: 2x + y = 5, 3x + 5y = –3

Hint: We had to only write the given equations in the form of AX = B, where A, X and B are matrices of different orders. A is the matrix of the coefficients of the variables X in the given equations. B is the column matrix of the variable Y. And B is also a column matrix of constant terms.

Complete step-by-step answer:
As we know that according to the reduction method to solve linear equations if we are given two linear equations like \[ax + by = c\] and \[dx + ey = f\]. Then we have to write the equations in form of AX = B, where A = \[\left[ {\begin{array}{*{20}{c}}
  a&b \\
  d&e
\end{array}} \right]\], X = \[\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\] and B = \[\left[ {\begin{array}{*{20}{c}}
  c \\
  f
\end{array}} \right]\]. And after that we had to apply row or column operations of the matrix A to change matrix A to identity matrix i.e. I = \[\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\].
As it is given that two equations are,
\[ \Rightarrow 2x + y = 5\] (1)
\[ \Rightarrow 3x + 5y = - 3\] (2)
Now we had to write equations 1 and 2 in the form of AX = B.
So, A = \[\left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&5
\end{array}} \right]\], X = \[\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]\] and B = \[\left[ {\begin{array}{*{20}{c}}
  5 \\
  { - 3}
\end{array}} \right]\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  2&1 \\
  3&5
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5 \\
  { - 3}
\end{array}} \right]\] (3)
Now we had to apply different row operations to change matrix A to identity matrix.
So, applying \[{R_2} \to 2{R_2}\]
 \[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  2&1 \\
  6&{10}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5 \\
  { - 6}
\end{array}} \right]\]
Now applying \[{R_2} \to {R_2} - 3{R_1}\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  2&1 \\
  0&7
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  5 \\
  { - 21}
\end{array}} \right]\]
Now applying \[{R_1} \to 7{R_1} - {R_2}\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {14}&0 \\
  0&7
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {56} \\
  { - 21}
\end{array}} \right]\]
Applying \[{R_1} \to \dfrac{{{R_1}}}{{14}}\] and \[{R_2} \to \dfrac{{{R_2}}}{7}\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  4 \\
  { - 3}
\end{array}} \right]\]
Now as we know that when we multiply identity matrix to any other matrix then the result does not change So, IX = X
So, we can write \[\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  4 \\
  { - 3}
\end{array}} \right]\]
Now comparing elements of the matrix of both the sides of the above equation.
\[ \Rightarrow x = 4,y = - 3\]
Hence, the solution of equation 2x + y = 5 and 3x + 5y = –3 will be x = 4 and y = –3.

Note: - Whenever we come up with this type of problem, if a particular method to solve the linear equations is given then we have to use only that method. Otherwise there are easier methods than reduction methods to solve linear equations. And one of them is a substitution method for that we had to take one of the equations and from that find the value of variable x in terms of y. And then put that value of x in another equation to find the value of y. Now we will get the value of x by putting the value of y in the previous equation of x. This will be the easiest and efficient method to find the solution of the problem if the method to be used is not given.