Question

How do you solve x, y, z in \[x + 3y - z = 4, - x - 2y + 3z = 4,2x - y - 2z = - 6\]?

Answer 1

Hint: We simply solve the system of linear equations by solving two equations at a time such that we eradicate one of the variables, write the value of each variable in terms of the same variable and then substitute the values in one of the equations.

Complete step by step answer:
We are given three equations:
\[x + 3y - z = 4\] … (1)
\[ - x - 2y + 3z = 4\] … (2)
\[2x - y - 2z = - 6\] … (3)
Now we will solve two equations at a time to remove one variable i.e. possible.
Add equations (1) and (2)
$
  x + 3y - z = 4 \\
  \underline { - x - 2y + 3z = 4} \\
  0x + y + 2z = 8 \\
$
So, we get the equation as \[y + 2z = 8\] … (4)
From equation (4); shift 2z to right hand side of the equation
\[ \Rightarrow y = 8 - 2z\] … (5)
Multiply equation (3) by 2
\[4x - 2y - 4z = - 12\] … (6)
Subtract equation (6) from equation (2)
$
- x - 2y + 3z = 4 \\
\underline {4x - 2y - 4z = - 12} \\
- 5x + 0y + 7z = 16 \\
$
So, we get the equation as \[ - 5x + 7z = 16\] … (7)
From equation (7); shift 7z to right hand side of the equation
\[ \Rightarrow - 5x = 16 - 7z\]
Divide both sides by -5
\[ \Rightarrow x = \dfrac{{7z - 16}}{5}\] … (8)
Substitute the values of y and x from equations (5) and (8) respectively in equation (1), i.e. \[x + 3y - z = 4\]
\[ \Rightarrow \dfrac{{7z - 16}}{5} + 3(8 - 2z) - z = 4\]
Open the bracket and multiply terms inside the bracket with terms outside the bracket
\[ \Rightarrow \dfrac{{7z - 16}}{5} + 24 - 6z - z = 4\]
\[ \Rightarrow \dfrac{{7z - 16}}{5} + 24 - 7z = 4\]
Shift 24 to right hand side of the equation
\[ \Rightarrow \dfrac{{7z - 16}}{5} - 7z = 4 - 24\]
\[ \Rightarrow \dfrac{{7z - 16}}{5} - 7z = - 20\]
Take LCM on left hand side of the equation
\[ \Rightarrow \dfrac{{7z - 16 - 35z}}{5} = - 20\]
\[ \Rightarrow \dfrac{{ - 28z - 16}}{5} = - 20\]
Cross multiply 5 from left hand side to right hand side of the equation
\[ \Rightarrow - 28z - 16 = - 100\]
Cancel -1 from both sides of the equation
\[ \Rightarrow 28z + 16 = 100\]
Shift constant value to right hand side of the equation
\[ \Rightarrow 28z = 100 - 16\]
\[ \Rightarrow 28z = 84\]
Divide both sides by 28
\[ \Rightarrow \dfrac{{28z}}{{28}} = \dfrac{{84}}{{28}}\]
Cancel same factors from numerator and denominator on both sides of the equation
\[ \Rightarrow z = 3\]
Substitute the value of z as 3 in equation (5) and (8)
\[ \Rightarrow y = 8 - 2 \times 3\]
\[ \Rightarrow y = 8 - 6\]
\[ \Rightarrow y = 2\]
Now from equation (8)
\[ \Rightarrow x = \dfrac{{7z - 16}}{5}\]
\[ \Rightarrow x = \dfrac{{7 \times 3 - 16}}{5}\]
\[ \Rightarrow x = \dfrac{{21 - 16}}{5}\]
\[ \Rightarrow x = \dfrac{5}{5}\]
Cancel same term from numerator and denominator
\[ \Rightarrow x = 1\]

The values of x, y and z are 1, 2 and 3 respectively.

Note: Many students make the mistake of solving the pair of equations as they just add or subtract equations randomly and end up with different variables in both obtained equations. Keep in mind we have to have one common variable in the obtained equations such that we can write other two variables in terms of this variable. When subtracting equations always remember to change signs of variables in the second equation.