Section 4.2 Substitution
When solving a system by graphing has several limitations. First, it requires the graph to be perfectly drawn, if the lines are not straight we may arrive at the wrong answer. Second, graphing is not a great method to use if the answer is really large, over \(100\) for example, or if the answer is a decimal the that graph will not help us find, \(3.2134\) for example. For these reasons we will rarely use graphing to solve our systems. Instead, an algebraic approach will be used.
The first algebraic approach is called substitution. We will build the concepts of substitution through several example, then end with a five-step process to solve problems using this method.
Example 4.2.1.
When we know what one variable equals we can plug that value (or expression) in for the variable in the other equation. It is very important that when we substitute, the substituted value goes in parenthesis. The reason for this is shown in the next example.
Example 4.2.2.
By using the entire expression \(3x - 7\) to replace \(y\) in the other equation we were able to reduce the system to a single linear equation which we can easily solve for our first variable. However, the lone variable (a variable without a coefficient) is not always alone on one side of the equation. If this happens we can isolate it by solving for the lone variable.
Example 4.2.3.
The process in the previous example is how we will solve problems using substitution. This process is described and illustrated in the following table which lists the five steps to solving by substitution.
Problem | \(\begin{matrix} 4x-2y=2 \\ 2x+y=-5\end{matrix}\) |
1. Find the lone variable | \(\begin{matrix} \text{Second Equation, } y \\2x+{\mathbf y} =-5\end{matrix}\) |
2. Solve for the lone variable | \(\begin{matrix} -2x \qquad -2x \\ y=-5-2x \end{matrix} \) |
3. Substitute into the untouched equation | |
4. Solve | \(\begin{matrix} 4x+10+4x=2 \\ 8x+10=2 \\ -10\ -10 \\ \frac{8x}{8}=\frac{-8}{8} \\ x=-1\end{matrix} \) |
5. Plug into lone variable equation and evaluate | \(\begin{matrix} y=-5-2(-1) \\ y=-5+2\\ y=-3 \end{matrix} \) |
Solution | (-1,-3) |
Sometimes we have several lone variables in a problem. In this case we will have the choice on which lone variable we wish to solve for, either will give the same final result.
Example 4.2.5.
Just as with graphing it is possible to have no solution \(\emptyset\) (parallel lines) or infinite solutions (same line) with the substitution method. While we won’t have a parallel line or the same line to look at and conclude if it is one or the other, the process takes an interesting turn as shown in the following example.
Example 4.2.6.
Because we had a true statement, and no variables, we know that anything that works in the first equation, will also work in the second equation. However, we do not always end up with a true statement.
Example 4.2.7.
Because we had a false statement, and no variables, we know that nothing will work in both equations.
World View Note: French mathematician Rene Descartes wrote a book which included an appendix on geometry. It was in this book that he suggested using letters from the end of the alphabet for unknown values. This is why often we are solving for the variables \(x\text{,}\) \(y\text{,}\) and \(z\text{.}\)
One more question needs to be considered, what if there is no lone variable? If there is no lone variable substitution can still work to solve, we will just have to select one variable to solve for and use fractions as we solve.
Example 4.2.8.
Using the fractions does make the problem a bit more tricky. This is why we have another method for solving systems of equations that will be discussed in another lesson.
Exercises Exercises - Substitution
Exercise Group.
Solve each system by substitution.
1.
\(\begin{matrix} y=-3x\\ y=6x-9 \end{matrix} \)
2.
\(\begin{matrix} y=x+5\\ y=-2x-4 \end{matrix} \)
3.
\(\begin{matrix} y=-2x-9 \\ y=2x-1 \end{matrix} \)
4.
\(\begin{matrix} y=-6x+3 \\y=6x+3 \end{matrix} \)
5.
\(\begin{matrix} y=6x+4\\ y=-3x-5 \end{matrix} \)
6.
\(\begin{matrix} y=3x+13\\ y=-2x-22 \end{matrix} \)
7.
\(\begin{matrix} y=3x+2\\ y=-3x+8 \end{matrix} \)
8.
\(\begin{matrix} y=-2x-9\\ y=-5x-21 \end{matrix} \)
9.
\(\begin{matrix} y=2x-3\\ y=-2x+9 \end{matrix} \)
10.
\(\begin{matrix} y=7x-24 \\ y=-3x+16 \end{matrix} \)
11.
\(\begin{matrix} y=6x-6 \\ -3x-3y=-24 \end{matrix} \)
12.
\(\begin{matrix} -x+3y=12 \\ y=6x+21 \end{matrix} \)
13.
\(\begin{matrix} y=-6\\ 3x-6y=30 \end{matrix} \)
14.
\(\begin{matrix} 6x-4y=-8 \\ y=-6x+2 \end{matrix} \)
15.
\(\begin{matrix} y=-5 \\ 3x+4y=-17 \end{matrix} \)
16.
\(\begin{matrix} 7x+2y=-7 \\y=5x+5 \end{matrix} \)
17.
\(\begin{matrix} -2x+2y=18 \\ y=7x+15 \end{matrix} \)
18.
\(\begin{matrix} y=x+4 \\3x-4y=-19 \end{matrix} \)
19.
\(\begin{matrix} y=-8x+19 \\-x+6y=16 \end{matrix} \)
20.
\(\begin{matrix} y=-2x+8\\-7x-6y=-8 \end{matrix} \)
21.
\(\begin{matrix} 7x-2y=-7 \\ y=7 \end{matrix} \)
22.
\(\begin{matrix} x-2y=-13 \\ 4x+2y=18 \end{matrix} \)
23.
\(\begin{matrix} x-5y=7\\2x+7y=-20 \end{matrix} \)
24.
\(\begin{matrix}3x-4y=15 \\7x+y=4 \end{matrix} \)
25.
\(\begin{matrix} - 2x - y = - 5\\ x-8y=-23 \end{matrix} \)
26.
\(\begin{matrix} 6x+4y=16\\ -2x+y=-3 \end{matrix} \)
27.
\(\begin{matrix} -6x+y=20\\-3x-3y=-18 \end{matrix} \)
28.
\(\begin{matrix} 7x+5y=-13\\x-4y=-16 \end{matrix} \)
29.
\(\begin{matrix} 3x + y = 9\\2x+8y=-16 \end{matrix} \)
30.
\(\begin{matrix}-5x-5y=-20 \\ -2x+y=7 \end{matrix} \)
31.
\(\begin{matrix} 2x+y=2\\3x+7y=14 \end{matrix} \)
32.
\(\begin{matrix}2x+y=-7 \\ 5x+3y=-21 \end{matrix} \)
33.
\(\begin{matrix} x+5y=15 \\ -3x+2y=6 \end{matrix} \)
34.
\(\begin{matrix} 2x+3y=-10\\ 7x+y=3 \end{matrix} \)
35.
\(\begin{matrix} -2x+4y=-16 \\y=-2 \end{matrix} \)
36.
\(\begin{matrix} -2x+2y=-22\\ -5x-7y=-19 \end{matrix} \)
37.
\(\begin{matrix} -6x+6y=-12\\ 8x-3y=16 \end{matrix} \)
38.
\(\begin{matrix} -8x+2y=-6\\-2x+3y=11 \end{matrix} \)
39.
\(\begin{matrix} 2x+3y=16 \\ -7x-y=20 \end{matrix} \)
40.
\(\begin{matrix}-x-4y=-14 \\ -6x+8y=12 \end{matrix} \)