In a previous post, we talked about solving systems of linear equations with the method of elimination. Here, we talk about the method of substitution.

Essentially, for this method we want to get an equation that expresses one variable in terms of the others. For example, suppose we start with the following system:

Example

    \[\begin{cases} x + 2y &= 3 \\ 2x - 3y &= 4 \end{cases}\]

We can re-write the first equation as:

    \[x = -2y + 3\]

Thereby giving us x in terms of y and some constants. We can now plug this into the second equation, giving us:

    \[2(-2y + 3) - 3y = 4\]

And simplifying, we get:

    \[-4y + 6 - 3y = -7y + 6 = 4\]

    \[\implies y = \frac{2}{7}\]

Now that we have the value of y, we can plug it into one of our equations to get the value of x:

    \[x + 2(\frac{2}{7}) = 3\]

    \[\implies x = \frac{17}{7}\]

In general, the steps of our method are as follows. Suppose we have some system of equations:

    \[\begin{cases} ax + by &= A \\ cx + dy &= B\end{cases}\]

where A, B are constants.

Substitution Method

  1. Pick one variable to substitute.

For our example, let’s pick x to substitute.

  1. Define that variable in terms of the other variables.

We can use either equation to define the value of x. Let’s use the first. Then, we get:

    \[x = \frac{A-by}{a}\]

  1. Substitute that variable into the remaining equation to solve for one of the variables.

Since we used the first equation to define x, we need to plug it into the second equation. Then, we get:

    \[cx + dy = \frac{A-by}{a}c + dy = B\]

    \[\implies y = \frac{B - \frac{A-by}{a}c}{d}\]

  1. Now that you have one of the variables, use the equations to find the other.

Just plug in your value for y into ax + by = A to find the value of x.

Now, as before, we can also use this method for systems of 3 equations. There, we can turn that problem into a system of 2 equations as follows.

Example

Suppose we have the following system of linear equations:

    \[\begin{cases} 3x + 4y - z &= 8 \\ x + 2y + 2z &= 4 \\ 2x - 3z &= 2 \end{cases}\]

Now, we pick a variable that we want to substitute. Suppose we want to substitute x first. The key is that we can pick any of the three equations to define x in terms of the other variables, y and z. Then, we use that equation to substitute for x in the other two equations.

Suppose we want to use equation 2 to define x. Then, we get:

    \[x = 4 - 2y - 2z\]

Then, since we used equation 2 to define x, we need to substitute for x in equations 1 and 3 (i.e. the other equations). We then get:

    \[\begin{cases} 3(4-2y-2z) + 4y - z &= 8 \\ 2(4-2y-2z) - 3z &= 2 \end{cases}\]

Simplifying, we get:

    \[\begin{cases}-2y - 7z &= -4 \\ -4y - 7z &= -6 \end{cases}\]

Which we can solve via our standard substitution methods.

Practice Problems

Solve the following systems of equations using the method of substitution.

  1.  

        \[\begin{cases} 7x - 2y &= x + 3 \\ x + 2y &= 0 \end{cases}\]

    Answer
  2.  

        \[\begin{cases} 7x - 2y &= 26\\ 2x + 2y &= 15 \end{cases}\]

    Answer

  3.  

        \[\begin{cases} x + 3y &= 15\\ x &= -y + 5 \end{cases}\]

     

    Answer
  4.  

        \[\begin{cases} 4x + 2y &= 3x - 3y + 3\\ 2x - y + 2 &= x - 2y \end{cases}\]

     

    Answer
  5.  

        \[\begin{cases} x + 4y - 3z &= 8 \\ x + 2y + 2z &= 4 \\ 2x - 3z = 7 \end{cases}\]

     

    Answer

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