Question

How do you solve linear equations in three variables?

Answer 1

Hint: In this we are going to explain how to solve linear equations in three variables. An equation of the form \[ax + by + cz = 0\], where \[a,b\] and \[c\] are real numbers, such that \[a,b\] and \[c\] are not equal to zero, is called a linear equation in three variables. \[x,y\] and \[z\] are the three variables where \[a,b\] and \[c\] are their coefficients respectively. We are going to solve the linear equations in three variables by using the elimination method.

Complete answer:
To solve the linear equation in three variables, follow the given steps carefully.
Step 1: As linear equations in three variables have three variables there must be three equations. From that take any two equations.
Step 2: Solve those two equations for any one of the variables.
Step 3: Again take another pair of equations and solve them for the same variable which we chose previously.
Step 4: Now we will have two equations with two variables. Solve those two equations.
Step 5: Now we will get a value of any one of the variables, substituting them in the equation any one of the two variable equations we derive will get the value of another variable.
Step 6: Now substituting the found values in any one of the original equations we have found all the unknown variables.
Example: Solve the linear equation:
 \[x + y - 3z = - 10\] …………………………… (\[1\])
\[x - y + 2z = 3\] ………………………………………. (\[2\])
\[2x + y - z = - 6\] ……………………………………. (\[3\])
Let's take the equation \[1\] and \[2\].
\[x + y - 3z = - 10\]
\[x - y + 2z = 3\]
Now let's choose \[y\] to eliminate, we can just add those two equations,
Adding equation \[1\] and \[2\], \[ + y\] in equation \[1\] and \[ - y\] in equation \[2\] will get cancel,
\[
  \underline
  x + y - 3z = - 10 \\
  x - y + 2z = {3_{( + )}} \\
\\
  2x{\text{ }} - z = - 7 \\
\]
Now, \[2x - z = - 7\]……………………………… (\[4\]) a new equation in two variable
Now take another pair of equations, let take equation \[2\] and \[3\].
\[x - y + 2z = 3\]
\[2x + y - z = - 6\]
To eliminate the \[y\] in this pair we need to simply add them,
By adding equation \[2\] and \[3\], \[ + y\] in equation \[1\] and \[ - y\] in equation \[2\] will get cancel,
\[
  \underline
  x - y + 2z = 3 \\
  2x + y - z = - {6_{( + )}} \\
  \\
  3x{\text{ }} + z = - 3 \\
\]
Now we get another equation in two variable, \[3x + z = - 3\] ………………………………….. (\[5\])
Now we have two equations in two variables, let we solve them,
\[2x - z = - 7\]
\[3x + z = - 3\]
By adding them \[z\] will get eliminated,
\[
  \underline
  2x - z = - 7 \\
  3x + z = - {3_{( + )}} \\
\\
  5x{\text{ }} = - 10{\text{ }} \\
\]
Now we have an equation \[5x = - 10\] in terms of \[x\], by solving this we will get the value of \[x\].
\[5x = - 10\]
\[5\] in the L.H.S. will become into fraction when it moves to R.H.S.,
\[x = \dfrac{{ - 10}}{5}\]
\[x = - 2\]
Finally we found one variable to substitute this value in either equation \[4\] or in equation \[5\].
Let we substitute \[x = - 2\] in equation \[4\],
\[2( - 2) - z = - 7\]
\[ - 4 - z = - 7\]
Moving \[ - 4\] in L.H.S. to R.H.S. it will become \[ + 4\],
\[ - z = - 7 + 4\]
\[ - z = - 3\]
\[ - \] in L.H.S. and R.H.S. will get cancel,
\[z = 3\]
Now we have \[x = - 2\] and \[z = 3\], let substitute them in any of the one main equation(in \[1\] or \[2\] or \[3\]).
Substituting \[x = - 2\] and \[z = 3\], in equation \[2\],
\[ - 2 - y + 2(3) = 3\]
\[ - 2 - y + 6 = 3\]
Now keeping all the numerals in one side,
\[ - y = 3 + 2 - 6\]
By solving this,
\[ - y = 5 - 6\]
\[ - y = - 1\]
\[ - \] in L.H.S. and R.H.S. will get cancel,
\[y = 1\].
Thus \[x = - 2,y = 1\] and \[z = 3\] for the linear equations, \[x + y - 3z = - 10\], \[x - y + 2z = 3\] and \[2x + y - z = - 6\] We can check them by substituting the values in the equation and both the L.H.S. and R.H.S. would be the same for all the three equations.

Note:
If we were to graph a linear equation in three variables we would get a figure of a plane in a three dimensional coordinate system.
The possible outcomes for the linear equation in three variables are: One solution, no solution and infinite no solutions.
The linear system in three variables has one solution, if the values we found will satisfy all the three equations.
The linear system in three variables has no solution, they do not have points in common.
The linear system in three variables has infinite solutions, any solution that works in one equation will work in the other.