Question

How do you solve the system $4x+4y+z=24,$ $2x-4y+z=0$ and $5x-4y-5z=12$?

Answer 1

Hint: To solve the given equations we will use the elimination method. First we will choose two equations such as by adding or subtracting we can eliminate one of the variables. Again we will choose two equations to eliminate a variable. Then we will solve the obtained two equations again by elimination method. Then we substitute the obtained values in the given equation.

Complete step by step solution:
We have been given the system $4x+4y+z=24,$ $2x-4y+z=0$ and $5x-4y-5z=12$.
We have to solve the given system of equations.
We have
$4x+4y+z=24..........(i)$
$2x-4y+z=0..........(ii)$
$5x-4y-5z=12........(iii)$
Now we can solve the given equations by using elimination methods. For this first we add equation (i) and equation (ii) then we will get
$\begin{align}
  & 4x+4y+z=24 \\
 & \underline{2x-4y+z=0} \\
 & 6x+2z=24..........(iv) \\
\end{align}$
Again we add equation (i) and equation (iii) then we will get
$\begin{align}
  & 4x+4y+z=24 \\
 & \underline{5x-4y-5z=12} \\
 & 9x-4z=36..........(v) \\
\end{align}$
Now, to solve the equation (iv) and equation (v), first we will multiply the equation (iv) by 2 then we will add both the equations. Then we will get
$\begin{align}
  & 12x+4z=48 \\
 & \underline{9x-4z=36} \\
 & 21x=84 \\
\end{align}$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow 21x=84 \\
 & \Rightarrow x=\dfrac{84}{21} \\
 & \Rightarrow x=4 \\
\end{align}$
Now, substituting the value of x in the equation (iv) we will get
$\begin{align}
  & \Rightarrow 6x+2z=24 \\
 & \Rightarrow 6\times 4+2z=24 \\
 & \Rightarrow 24+2z=24 \\
 & \Rightarrow 2z=24-24 \\
 & \Rightarrow z=0 \\
\end{align}$
Now, substituting the values of x and z in equation (i) we will get
$\begin{align}
  & \Rightarrow 4\times 4+4y+0=24 \\
 & \Rightarrow 16+4y=24 \\
 & \Rightarrow 4y=24-16 \\
 & \Rightarrow 4y=8 \\
 & \Rightarrow y=\dfrac{8}{4} \\
 & \Rightarrow y=2 \\
\end{align}$
Hence on solving the given equations we will get the values of x, y and z as 4, 2 and 0 respectively.

Note: We can verify the answer by putting the obtained values in other two given equations.
By putting the values in equation (ii) we will get
$\begin{align}
  & \Rightarrow 2\times 4-4\times 2+0=0 \\
 & \Rightarrow 8-8=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
Hence we get LHS=RHS
Now, by putting the values in equation (iii) we will get
$\begin{align}
  & \Rightarrow 5\times 4-4\times 2-5\times 0=12 \\
 & \Rightarrow 20-8-0=12 \\
 & \Rightarrow 12=12 \\
\end{align}$
Hence we get LHS=RHS
Therefore the obtained values are correct.