Question

How do you solve the system of linear equations $x-y+2z=9$, $-2x+y+z=-1$ and $2x+2y-z=-5$?

Answer 1

Hint: To solve the system of linear equations we will first consider the equation $2x+2y-z=-5$ and write z in terms of x and y. Now we will substitute the value of z in the other two equations. Hence we have obtained two new equations which are linear equations in two variables. We will solve the two linear equations by substitution method and hence find the value of x and y. Now we will substitute the value of x and y in any given equation and find the value of z. Hence we have the solution to the equation.

Complete step by step solution:
Note to solve the linear equations we will first consider the equation $2x+2y-z=-5$
Now transposing 5 on LHS and z on RHS we get,
$z=2x+2y+5................\left( 1 \right)$
Now let us substitute the value of z in the two equation $x-y+2z=9$ we get
$\begin{align}
  & \Rightarrow x-y+2\left( 2x+2y+5 \right)=9 \\
 & \Rightarrow x-y+4x+4y+10=9 \\
 & \Rightarrow 5x+3y=-1...................\left( 2 \right) \\
\end{align}$
Now substituting the value of z in the equation $-2x+y+z=-1$ we get,
$\begin{align}
  & \Rightarrow -2x+y+\left( 2x+2y+5 \right)=-1 \\
 & \Rightarrow 3y+5=-1 \\
 & \Rightarrow 3y=-6 \\
 & \Rightarrow y=-3....................\left( 3 \right) \\
\end{align}$
Now we have the value of y.
Substituting the value of y in equation (2) we get,
$\begin{align}
  & \Rightarrow 5x+3\left( -3 \right)=-1 \\
 & \Rightarrow 5x-9=-1 \\
 & \Rightarrow 5x=8 \\
 & \Rightarrow x=\dfrac{8}{5} \\
\end{align}$
Now substituting the value of x and value of y in the equation (1) we get,
$\begin{align}
  & \Rightarrow z=2\times \dfrac{8}{5}+2\left( -3 \right)+5 \\
 & \Rightarrow z=\dfrac{16}{5}-6+5 \\
 & \Rightarrow z=\dfrac{16}{5}-1 \\
\end{align}$
$\begin{align}
  & \Rightarrow z=\dfrac{16-5}{5} \\
 & \Rightarrow z=\dfrac{11}{5} \\
\end{align}$
Hence we have the value of $z=\dfrac{11}{5}$

Hence the solution of the given equation is $x=\dfrac{8}{5},y=-3$ and $z=\dfrac{11}{5}$

Note: Now note that we can also solve this equation by using the concept of matrices. First we write the equations in the form of AX = B. Now we know where A is a 3 × 3 matrix with each row representing the coefficient of equations. Now to solve the equation using matrices we will have to first find inverse matrix of A. we can do this by row transformation method or by using the formula ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$ . Now once we have an inverse of A, we will find the matrix ${{A}^{-1}}B$ . Now we have that $X={{A}^{-1}}B$ and hence we have the solution of the equation.