Question

How do you solve the system of equations $4x+2y+z=7,2x+2y+4z=-4,x+3y-2z=-8$ ?

Answer 1

Hint: This is a system of equations. These are actually lines in the $xyz$- coordinate system. We know how to solve lines when they are in the $xy$ plane. We either use substitution or elimination. In the same way, we can either use substitution or elimination here as well. Solving two lines or three lines in any coordinate system is nothing but finding the point of intersection of these lines. These lines are concurrent lines as they pass through one single, common point.

Complete step by step solution:
Now we have to solve or find out the point of intersection of all these three lines.
Let us carry this out by using elimination process. So we have to eliminate two variables in order to find out the constant value of the third variable or we can eliminate just one variable by using two equations.
And then by using another two equations, we can eliminate the same variable. From the fresh two equations that we get, we can eliminate another variable and then find the constant value of the last standing variable.
Let us subtract $4x+2y+z=7,2x+2y+4z=-4$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow \left( 4x+2y+z-7 \right)-\left( 2x+2y+4z+4 \right)=0 \\
 & \Rightarrow 2x-3z-11=0 \\
 & \Rightarrow 2x-3z=11....eqn\left( 1 \right) \\
\end{align}$
Now let us subtract $2x+2y+4z=-4,x+3y-2z=-8$.
Before subtracting let us multiply the line $2x+2y+4z=-4$with $3$ and multiply the line $x+3y-2z=-8$ with $2$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow \left( 2x+2y+4z+4 \right)\times 3-\left( x+3y-2z+8 \right)\times 2=0 \\
 & \Rightarrow \left( 6x+6y+12z+12 \right)-\left( 2x+6y-4z+16 \right)=0 \\
 & \Rightarrow 4x+16z-4=0 \\
 & \Rightarrow x+4z=1......eqn\left( 2 \right) \\
\end{align}$
Let us subtract $eqn\left( 2 \right)$ and $eqn\left( 1 \right)$. Before doing that let us multiply $eqn\left( 2 \right)$with $2$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow \left( x+4z-1 \right)\times 2-\left( 2x-3z-11 \right)=0 \\
 & \Rightarrow \left( 2x+8z-2 \right)-\left( 2x-3z-11 \right)=0 \\
 & \Rightarrow 11z+9=0 \\
 & \Rightarrow z=\dfrac{-9}{11} \\
\end{align}$
Let us substitute the value of $z$ in $eqn\left( 2 \right)$to find the value of $x$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow x+4z=1 \\
 & \Rightarrow x+4\left( \dfrac{-9}{11} \right)=1 \\
 & \Rightarrow x-\dfrac{36}{11}=1 \\
 & \Rightarrow x=\dfrac{47}{11} \\
\end{align}$
Let us substitute the value of $x,z$ in any one of the line equations to find out the value of $y$.
Let us substitute them in $4x+2y+z=7$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow 4x+2y+z=7 \\
 & \Rightarrow 4\left( \dfrac{47}{11} \right)+2y-\dfrac{9}{11}=7 \\
 & \Rightarrow \dfrac{188-9}{11}+2y=7 \\
 & \Rightarrow \dfrac{179}{11}+2y=7 \\
 & \Rightarrow 2y=7-\dfrac{179}{11} \\
 & \Rightarrow y=\dfrac{-51}{11} \\
\end{align}$
$\therefore $ Hence, the point of intersection of the system of equations is $\left( \dfrac{47}{11},\dfrac{-51}{11},\dfrac{-9}{11} \right)$

Note: We should be very careful while solving as there is a huge scope for calculation errors. We need a lot of practice to carry out these calculations correctly and get the answer in the first shot or else it will become a huge mess of confusion. To check whether the point we got to be correct or not, we can substitute the point that we got in any of the line equations or in all the three. They will sum up to zero if our point is correct. If it is not, then at least one line equation will not sum up to zero.