In the previous two sections, we’ve solved sets of two or even three linear equations. Here, we will do some examples of systems with four linear equations. The general strategy will be the same as before: we apply our method (either elimination or substitution) to our sets of equations in order to get rid of one equation. We repeat this until we’re left with just two or three equations, whereupon we know how to solve the problem. Also note that it takes n distinct equations to solve a system of n linear equations. (Two equations are distinct if you cannot obtain either one from the other).

Example (Elimination Method for 4 Equations)

    \[\begin{cases} 4a + 2b + 2c + d - e &= 8 \\ a - b + 3c + d &= 4 \\ 3a + 3b - c + d - e &= 6 \\ 3b - c + 2d - 3e &= -12 \\ a - b + c - d + e &= -1 \end{cases}\]

  1. Pick a variable we want to eliminate.
    Let’s pick a.
  2. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).
    Now, if we look at our equations, we see that equation 4 does not have any a’s in it. So no matter what constant we multiply the equation by, we will not get a term with a multiplied by some given coefficient. If we have an equation like this, we simply set it to the side and use it again later. This is because, recall, the point of this method is to eliminate all occurrences of a given variable, thereby simplifying our problem. Equation 4 has already eliminated all occurrences of a, so no further manipulation of the equation is needed.Now, we get:

        \[\begin{cases} 12a + 6b + 6c + 3d - 3e &= 24 \\ 12a - 12b + 36c + 12d &= 48 \\ 12a + 12b - 4c + 4d - 4e &= 24 \\ 12a - 12b + 12c - 12d + 12e &= -12 \end{cases}\]

    And we still have our equation 4:

        \[3b - c + 2d - 3e =& -12\]

  3. Subtract the equations from one another, thereby eliminating our variable.
    Note that this only applies to the equations that have a in them. And remember to use all of the equations that include a (but we only need to use each equation once).Then, we get:

        \[\begin{cases} 18b - 30c - 9d - 3e &= -24\\ -24b + 40c + 8d + 4e &= 24\\ -24b -16c + 16d - 16e &= 36 \\ 3b - c + 2d - 3e &= -12\end{cases}\]

  4. Now, we should have four equations with at most four different variables. Pick another variable to eliminate.
    Let’s eliminate b.
  5. Manipulate the equations so that the chosen variable has the same coefficient in all of them (or a coefficient of 0).

        \[\begin{cases} 18b - 30c - 9d - 3e &= -24\\ 18b - 30c - 6d - 3e &= -18\\ 18b +12c - 12d + 12e &= -27 \\ 18b - 6c + 12d - 18e &= -72\end{cases}\]

  6. Subtract the equations from one another, thereby eliminating our variable.

        \[\begin{cases} -3d &= -6\\ -42c + 6d -15e &= 9\\ 18c - 24d + 30e &= 45 \end{cases}\]

  7. Now, we should have three equations with at most three different variables.
    And we do.
  8. We can now solve this problem with the steps detailed in our previous post on the elimination method.

Example (Substitution Method with 4 Equations)

    \[\begin{cases} 4a + 2b + 2c + d - e &= 8 \\ a - b + 3c + d &= 4 \\ 3a + 3b - c + d - e &= 6 \\ 3b - c + 2d - 3e &= -12 \\ a - b + c - d + e &= -1 \end{cases}\]

  1. Pick a variable we want to substitute for.
    Let’s choose a.
  2. Use one of the lines of the equation to express that variable in terms of the other variables.
    Let’s use the second equation. Then, we get:

        \[a = 4 + b - 3c - d\]

  3. Substitute into the other equations.
    Since we used the second equation to get our equation for a, we now substitute for a in the first, third, fourth, and fifth equations. We then get (after simplifying):

        \[\begin{cases} 6b -10c -3d - e &= -8\\ 6b -10c -2d - e &= -6\\ 3b - c + 2d - 3e &= -12\\ -2c - 2d + e &= -5 \end{cases}\]

  4. Now, we should have four equations with at most four different variables.
    And this is true.
  5. Pick another variable to substitute for.
    Let’s pick e.
  6. Use one of the remaining lines of the equation to express that variable in terms of the other variables.
    Let’s use the first line. Then, we get

        \[e = 6b - 10c - 3d + 8\]

  7. Substitute into the other equations.
    Since we used the first equation, we need to plug in our formula for e into the second, third, and fourth equations. We then get (after simplifying):

        \[\begin{cases} d &= 2\\ -15b + 29c + 11d &= 12\\ 6b - 12c - 5d &= -13 \end{cases}\]

  8. Now, we should have three equations with at most three different variables.
    And we do.
  9. We can now solve this problem with the steps detailed in our previous post on the substitution method.

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