Question

How do you solve the system \[w-2x+3y+z=3\],\[2w-x-y+z=4\],\[w+2x-3y-z=1\] and \[3w-x+y-2z=-4\]?

Answer 1

Hint: To solve this system of equations we have to observe all the equations and then we can observe that the first equation and third equation have opposite signed same terms except ‘z’. By adding them we can get the value of ‘z’. By performing operations to all the equations we will get all values.

Complete step by step solution:
For the given problem we are given to solve the system of equations. Let us consider the given equations as equation (1), equation (2), equation (3) and equation (4).
\[w-2x+3y+z=3.......\left( 1 \right)\]
\[2w-x-y+z=4.........\left( 2 \right)\]
\[w+2x-3y-z=1........\left( 3 \right)\]
\[3w-x+y-2z=-4.......\left( 4 \right)\]
By comparing equation (1) and equation (3) we can see that they are three terms with opposite signs. So, by adding them we can get the value of w.
By adding equation (1) and equation (3), we get
\[\begin{align}
  & \Rightarrow 2w=4 \\
 & \Rightarrow w=2 \\
\end{align}\]
Let us consider the equation as equation (5), we get
\[\Rightarrow w=2.........\left( 5 \right)\]
Now by substituting the equation (5) in system of equations, we get
\[\Rightarrow \left( \begin{align}
  & -2x+3y+z=1 \\
 & -x-y+z=0 \\
 & 2x-3y-z=-1 \\
 & -x+y-2z=-10 \\
\end{align} \right)\]
From the second equation, we get
\[\Rightarrow z=x+y\]
Let us substitute this equation in third and fourth equations, we get
\[\Rightarrow \left( \begin{align}
  & x-4y=-1 \\
 & -3x-y=-10 \\
\end{align} \right)\]
Now we have to solve the above equations to get the values of ‘x’ and ‘y’.
Let us multiply the second equation by -4 then above equations will become:
\[\Rightarrow \left( \begin{align}
  & x-4y=-1 \\
 & 12x+4y=40 \\
\end{align} \right)\]
Now by adding the above equations, we get
\[\begin{align}
  & \Rightarrow 13x=39 \\
 & \Rightarrow x=3 \\
\end{align}\]
Let us consider of as equation (6), we get
\[\Rightarrow x=3........................\left( 6 \right)\]
Substitute the ‘x’ value in first equation, we get
\[\begin{align}
  & \Rightarrow 3-4y=-1 \\
 & \Rightarrow y=1 \\
\end{align}\]
Let us consider the above equation as equation (7), we get
\[\Rightarrow y=1..............\left( 7 \right)\]
By substituting equation (5), equation (6) and equation (7) in equation (1), we get
\[\begin{align}
  & \Rightarrow 2-6+3+z=3 \\
 & \Rightarrow z=4 \\
\end{align}\]
Let us consider the above equation as equation (8), we get
\[\Rightarrow z=4..................\left( 8 \right)\]
Therefore equation (5),(6),(7) and equation (8) are the solutions of the system of equations.

Note: If the given system of equations contains only 3 variables then we can solve that problem by using the matrix method. We can do this problem by another method i.e. we have to find the value of ‘z’ by the first equation and then by substituting the equation into other equations, we can solve the solution.