Question

How do you solve this system of equations \[3x-3y+5z=13,\] \[ 3x+y-3z=-5,\] \[19x-y-5z=0\]?

Answer 1

Hint: To solve the system of equations in three variables, we need to follow the steps given below in the same order:
Step 1: choose one of the equations to find the relationship between the three variables. this can be done by taking one of the variables to the other side of the equation and the remaining two to the other side of the equation.
Step 2: substitute this relationship in the other two equations to get two equations in two variables.
Step 3: solve this set of equations in two variables to find the value of the variable in the equation.
Step 4: substitute these values in any of the equations to find the value of the remaining variable.

Complete step by step solution:
We are given the two equations \[3x-3y+5z=13\], \[3x+y-3z=-5\] and \[19x-y-5z=0\]. We know the steps required to solve a system of equations in three variables. Let’s take the third equation, we get
\[\Rightarrow 19x-y-5z=0\]
Adding y to both sides of the equation, we get
\[\Rightarrow y=19x-5z\]
Substituting this in the remaining two equations, we get
The first equation as,
\[\begin{align}
  & \Rightarrow 3x-3\left( 19x-5z \right)+5z=13 \\
 & \Rightarrow 3x-57x+15z+5z=13 \\
 & \Rightarrow -54x+20z=13 \\
\end{align}\]
The second equation as,
\[\begin{align}
  & \Rightarrow 3x+19x-5z-3z=-5 \\
 & \Rightarrow 22x-8z=-5 \\
\end{align}\]
Now, we have two equations as \[-54x+20z=13\] and \[22x-8z=-5\].
We can solve this set of equations in two variables. On solving this set we get the solution as \[x=\dfrac{1}{2}\And z=2\]. Substituting these values in the equation \[y=19x-5z\], we get
\[\begin{align}
  & \Rightarrow y=\dfrac{19}{2}-5\left( 2 \right) \\
 & \Rightarrow y=-\dfrac{1}{2} \\
\end{align}\]
Thus, we get the solution of the given set of equations as \[x=\dfrac{1}{2},y=-\dfrac{1}{2}\And z=2\]

Note: We can check if the answer is correct or not by substituting these values in the equation. Substituting \[x=\dfrac{1}{2},y=-\dfrac{1}{2}\And z=2\] in the given three equation we get, For the first equation \[LHS=3\left( \dfrac{1}{2} \right)-3\left( -\dfrac{1}{2} \right)+5(2)=13=RHS\]. For the second equation \[LHS=3\left( \dfrac{1}{2} \right)+\left( -\dfrac{1}{2} \right)-3(2)=-5=RHS\]. For the third equation \[LHS=19\left( \dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)-5(2)=0=RHS\]. As the values satisfy all the equations, the solution is correct.