4.2: Systems of Equations - The Substitution Method
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Solving a system by graphing has its limitations. First, it requires the graph to be precisely drawn. If the lines aren’t drawn correctly, it is possible to obtain the incorrect point of intersection. Second, graphing is not a sufficient method to use if the point of intersection is far away from the origin, like the ordered-pair (12567, 9832), or if the answer is a decimal (or irrational number) that the graph will not display well. For these reasons, we rarely use graphing to solve systems. Instead, we use an algebraic approach. There are two approaches and the first approach is called substitution. We build the concepts of substitution through several examples and then conclude with a general four-step process to solve problems using this method.
Introduction to Substitution
Solve the system of equations.
\[\left\{\begin{array}{l}x=5 \\ y=2x-3\end{array}\right.\nonumber\]
Solution
We are given \(x = 5\) in the first equation. Hence, we can plug-n-chug \(x = 5\) into the second equation to find \(y\) because we know any ordered-pair with \(x\) coordinate 5 will satisfy the equation:
\[\begin{array}{rl}y=2x-3&\text{Plug-n-chug }x=5 \\ y=2\color{blue}{(5)}\color{black}{}-3&\text{Evaluate} \\ y=10-3&\text{Subtract} \\ y=7&y\text{-coordinate of the solution}\end{array}\nonumber\]
Since it is given \(x = 5\) and we obtained \(y = 7\), then the solution to the system is the ordered-pair \((5, 7)\). Furthermore, if we were to graph these two lines, they would intersect at \((5, 7)\).
The Substitution Method
When we are given the solution to one of the two variables, we can easily plug-n-chug that value (or expression) in the other equation to obtain the value of the second variable. It is very important that when we substitute, we write parenthesis around the expression we are substituting; this will minimize errors with distribution.
Solve the system by substitution.
\[\left\{\begin{array}{l}2x-3y=7 \\ y=3x-7\end{array}\right.\nonumber\]
Solution
We can see that \(y\) is isolated in the second equation, \(y = 3x − 7\), and we can substitute the right side for \(y\) in the first equation.
\[\begin{array}{rl}2x-3y=7&\text{Plug-n-chug }y=3x-7&\text{ into the first equation} \\ 2x-3\color{blue}{(3x-7)}\color{black}{}=7&\text{Distribute} \\ 2x-9x+21=7&\text{Combine like terms} \\ -7x+21=7&\text{Isolate the variable term} \\ -7x=-14&\text{Multiply by the reciprocal of }-7 \\ x=2&x\text{-coordinate of the solution}\end{array}\nonumber\]
Since \(x = 2\), then we can plug-n-chug \(x = 2\) into one of the equations to obtain \(y\):
\[\begin{array}{rl}y=3x-7&\text{Plug-n-chug }x=2 \\ y=3\color{blue}{(2)}\color{black}{}-7&\text{Evaluate} \\ y=6-7&\text{Subtract} \\ y=-1&y\text{-coordinate of the solution}\end{array}\nonumber\]
The solution to the system is the ordered-pair \((2, −1)\). Furthermore, if we were to graph these two lines, we know they would intersect at \((2, −1)\). When we obtain a solution to a system, we call this system a consistent system. When we obtain one ordered-pair as the solution to the system, we call this solution an independent solution.
Given a system of equations,
- If there exists at least one solution to the system, then this is a consistent system.
- If the consistent system has one unique solution, i.e., the solution is an ordered-pair, then this is an independent system.
- If the consistent system has more than one solution, i.e., there are infinitely many solutions, then this is a dependent system.
- If a solution to the system doesn’t exist, i.e., there is no solution, then this is an inconsistent system.
Solve for a Variable
Solve the system by substitution:
\[\left\{\begin{array}{l}3x+2y=1 \\ x-5y=6\end{array}\right.\nonumber\]
Solution
Notice neither of the equations have \(y\) or \(x\) isolated. Hence, we will have to pick an equation and variable, and solve for that variable in that equation. We always want to work smarter, not harder, so let’s be clever in choosing the equation and variable. Looking at the first equation, there is a coefficient in front of each variable. Looking at the second equation, we see only the \(y\) has a coefficient other than 1, and \(x\)’s coefficient is one. Let’s pick this because solving for \(x\) is a one-step equation while the other is a two-step equation.
\[\begin{aligned}x-5y&=6 \\ x&=6+5y\end{aligned}\]
Now, we can substitute \(x\) into the first equation:
\[\begin{array}{rl}3x+2y=1&\text{Plug-n-chug }x=6+5y \\ 3\color{blue}{(6+5y)}\color{black}{}+2y=1&\text{Distribute} \\ 18+15y+2y=1&\text{Combine like terms} \\ 18+17y=1&\text{Isolate the variable term} \\ 17y=-17&\text{Multiply by the reciprocal of }17 \\ y=-1&y\text{-coordinate of the solution}\end{array}\nonumber\]
Since \(y = −1\), then we can plug-n-chug \(y = −1\) into one of the equations to obtain \(x\):
\[\begin{array}{rl}x=6+5y &\text{Plug-n-chug }y=-1 \\ x=6+5\color{blue}{(-1)}\color{black}{}&\text{Evaluate} \\ x=6-5&\text{Subtract} \\ x=1&x\text{-coordinate of the solution}\end{array}\nonumber\]
The solution to the system is the ordered-pair \((1, −1)\). Furthermore, if we were to graph these two lines, they would intersect at \((1, −1)\). Also, we know this system is a consistent system that is independent.
When choosing the equation and variable to use in substitution, we want to choose the one-step equation, i.e., one equation with one of the variables with a coefficient of one. Even though we are used to always solving for \(y\) when graphing lines, with this method, it doesn’t matter. We want to work smarter, not harder, and pick the equation that will give us the solution the quickest.
Given a system of two linear equations in two variables, we can use the following steps to solve by substitution.
- Step 1. Choose an equation and then solve for \(x\) or \(y\). (Choose the one-step equation when possible.)
- Step 2. Substitute the expression for \(x\) or \(y\) in the other equation.
- Step 3. Solve the equation.
- Step 4. Substitute the value into one of the original equations to find the remaining variable.
It is common practice to write your answer as an ordered-pair of the form \((x, y)\) since this is the point of intersection. Be sure to verify the solution.
Solve the system by substitution:
\[\left\{\begin{array}{l}4x-2y=2 \\ 2x+y=-5\end{array}\right.\nonumber\]
Solution
Notice none of the equations have \(y\) or \(x\) isolated. Hence, we will have to pick an equation and variable, and solve for that variable in that equation.
Step 1. We want to choose the one-step equation, if there is one. Looking at the second equation, we see only the \(y\) has a coefficient of one, and \(x\)’s coefficient is two. Let’s pick this because solving for \(y\) is a one-step equation while the other is a two-step equation. \[\begin{aligned}2x+y&=-5 \\ y&=-2x-5\end{aligned}\]
Step 2. Now, we can substitute \(y\) into the first equation: \[\begin{array}{rl}4x-2y=2&\text{Plug-n-chug }y=-2x-5 \\ 4x-2\color{blue}{(-2x-5)}\color{black}{}=2\end{array}\nonumber\]
Step 3. Solve for \(x\): \[\begin{array}{rl}4x-2\color{blue}{(-2x-5)}\color{black}{}=2&\text{Distribute} \\ 4x+4x+10=2&\text{Combine like terms} \\ 8x+10=2&\text{Isolate the variable term} \\ 8x=-8&\text{Multiply by the reciprocal of }8 \\ x=-1&x\text{-coordinate of the solution}\end{array}\nonumber\]
Step 4. Since \(x = −1\), then we can plug-n-chug \(x = −1\) into one of the equations to obtain \(y\): \[\begin{array}{rl}y=-2x-5&\text{Plug-n-chug }x=-1 \\ y=-2\color{blue}{(-1)}\color{black}{}-5&\text{Evaluate} \\ y=2-5&\text{Subtract} \\ y=-3&y\text{-coordinate of the solution}\end{array}\nonumber\] The solution to the system is the ordered-pair \((−1, −3)\). Furthermore, if we were to graph these two lines, they would intersect at \((−1, −3)\). Also, we know this system is a consistent system that is independent.
Substitution: Special Cases
Solve the system by substitution:
\[\left\{\begin{array}{l}y+4=3x \\ 2y-6x=-8\end{array}\right.\nonumber\]
Solution
Notice none of the equations have \(y\) or \(x\) isolated. Hence, we will have to pick an equation and variable, and solve for that variable in that equation.
Step 1. We want to choose the one-step equation, if there is one. Looking at the first equation, we see only the \(y\) has a coefficient of one, and \(x\)’s coefficient is three. Let’s pick this because solving for \(y\) is a one-step equation while the other is a two-step equation. \[\begin{aligned}y+4&=3x \\ y&=3x-4\end{aligned}\]
Step 2. Now, we can substitute \(y\) into the second equation: \[\begin{array}{rl}2y-6x=-8&\text{Plug-n-chug }y=3x-4 \\ 2\color{blue}{(3x-4)}\color{black}{}-6x=-8\end{array}\nonumber\]
Step 3. Solve for \(x\): \[\begin{array}{rl}2\color{blue}{(3x-4)}\color{black}{}-6x=-8&\text{Distribute} \\ 6x-8-6x=-8&\text{Combine like terms} \\ -8=-8\end{array}\nonumber\] Since all the variables cancel and we are left with a statement without variables, we ask,“ Is this statement true?” \[\begin{array}{rl}-8\stackrel{?}{=}-8&\text{Is this true?} \\ -8=-8&\checkmark\text{ True}\end{array}\nonumber\] Since this statement is true, then there are infinitely many solutions on the line \(y + 4 = 3x\). Furthermore, if we were to graph these two lines, we know they would be the same line and intersect at every point on the line. Also, we know this system is a consistent system that is dependent.
Solve the system by substitution:
\[\left\{\begin{array}{l}6x-3y=-9 \\ -2x+y=5\end{array}\right.\nonumber\]
Solution
Notice none of the equations have \(y\) or \(x\) isolated. Hence, we will have to pick an equation and variable, and solve for that variable in that equation.
Step 1. We want to choose the one-step equation, if there is one. Looking at the second equation, we see only the \(y\) has a coefficient of one, and \(x\)’s coefficient is \(−2\). Let’s pick this because solving for \(y\) is a one-step equation while the other is a two-step equation. \[\begin{aligned}-2x+y&=5 \\ y&=2x+5\end{aligned}\]
Step 2. Now, we can substitute \(y\) into the first equation: \[\begin{array}{rl}6x-3y=-9&\text{Plug-n-chug }y=2x+5 \\ 6x-3\color{blue}{(2x+5)}\color{black}{}=-9\end{array}\nonumber\]
Step 3. Solve for \(x\): \[\begin{array}{rl}6x-3\color{blue}{(2x+5)}\color{black}{}=-9&\text{Distribute} \\ 6x-6x-15=-9&\text{Combine like terms} \\ -15=-9\end{array}\nonumber\] Since all the variables cancel and we are left with a statement without variables, we ask,“ Is this statement true?” \[\begin{array}{rl}-15\stackrel{?}{=}-9&\text{Is this true?} \\ -15\neq -9&X \text{ False}\end{array}\nonumber\] Since this statement is false, then there is no solution to this system. Furthermore, if we were to graph these two lines, we know they would be parallel. Hence, this system is an inconsistent system.
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 the reason we are often solving for the variables \(x,\: y,\) and \(z\).
Solve the system by substitution:
\[\left\{\begin{array}{l}5x-6y=-14 \\ -2x+4y=12\end{array}\right.\nonumber\]
Solution
Notice none of the equations have \(y\) or \(x\) isolated. Hence, we will have to pick an equation and variable, and solve for that variable in that equation.
Step 1. We want to choose the one-step equation, if there is one. Looking at both equations, we see none of these are one-step equations. Hence, we can just pick one equation and solve for a variable. Notice in the second equation, all coefficients are divisible by 2. Let’s pick this because solving for \(x\) would avoid fractions. \[\begin{aligned}-2x+4y&=12 \\ -2x&=-4y+12 \\ x&=2y-6\end{aligned}\]
Step 2. Now, we can substitute \(x\) into the first equation: \[\begin{array}{rl}5x-6y=-14 &\text{Plug-n-chug }x=2y-6 \\ 5\color{blue}{(2y-6)}\color{black}{}-6y=-14\end{array}\nonumber\]
Step 3. Solve for \(y\): \[\begin{array}{rl}5\color{blue}{(2y-6)}\color{black}{}-6y=-14&\text{Distribute} \\ 10y-20-6y=-14&\text{Combine like terms} \\ 4y-30=-14&\text{Isolate the variable term} \\ 4y=16&\text{Multiply by the reciprocal of }4 \\ y=4&y\text{-coordinate of the solution}\end{array}\nonumber\]
Step 4. Since \(y = 4\), then we can plug-n-chug \(y = 4\) into one of the equations to obtain \(x\): \[\begin{array}{rl}x=2y-6&\text{Plug-n-chug }y=4 \\ x=2\color{blue}{(4)}\color{black}{}-6&\text{Evaluate} \\ x=8-6&\text{Subtract} \\ x=2&x\text{-coordinate of the solution}\end{array}\nonumber\] The solution to the system is the ordered-pair \((2, 4)\). Furthermore, if we were to graph these two lines, we know they would intersect at \((2, 4)\). Also, we know this system is a consistent system that is independent.
Systems of Equations: The Substitution Method Homework
Solve each system by substitution. Determine if each system is consistent, independent or dependent, or inconsistent.
\(\begin{array}{l}y=-3x \\ y=6x-9\end{array}\)
\(\begin{array}{l}y=-2x-9 \\ y=2x-1\end{array}\)
\(\begin{array}{l}y=6x+4 \\ y=-3x-5\end{array}\)
\(\begin{array}{l}y=3x+2 \\ y=-3x+8\end{array}\)
\(\begin{array}{l}y=2x-3 \\ y=-2x+9\end{array}\)
\(\begin{array}{l}y=6x-6 \\ -3x-3y=-24\end{array}\)
\(\begin{array}{l}y=-6 \\ 3x-6y=30\end{array}\)
\(\begin{array}{l}y=-5 \\ 3x+4y=-17\end{array}\)
\(\begin{array}{l}-2x+2y=18 \\ y=7x+15\end{array}\)
\(\begin{array}{l}y=-8x+19 \\ -x+6y=16\end{array}\)
\(\begin{array}{l}7x-2y=-7 \\ y=7\end{array}\)
\(\begin{array}{l}x-5y=7 \\ 2x+7y=-20\end{array}\)
\(\begin{array}{l}-2x-y=-5 \\ x-8y=-23\end{array}\)
\(\begin{array}{l}-6x+y=20 \\ -3x-3y=-18\end{array}\)
\(\begin{array}{l}3x+y=9 \\ 2x+8y=-16\end{array}\)
\(\begin{array}{l}y=x+5 \\ y=-2x-4\end{array}\)
\(\begin{array}{l}y=-6x+3 \\ y=6x+3 \end{array}\)
\(\begin{array}{l}y=3x+13\\y=-2x-22 \end{array}\)
\(\begin{array}{l}y=-2x-9\\y=-5x-21 \end{array}\)
\(\begin{array}{l}y=7x-24 \\ y=-3x+16 \end{array}\)
\(\begin{array}{l}-x+3y=12 \\ y=6x+21 \end{array}\)
\(\begin{array}{l}6x-4y=-8 \\ y=-6x+2 \end{array}\)
\(\begin{array}{l}7x+2y=-7 \\ y=5x+5 \end{array}\)
\(\begin{array}{l}y=x+4 \\ 3x-4y=-19 \end{array}\)
\(\begin{array}{l}y=-2x+8 \\ -7x-6y=-8 \end{array}\)
\(\begin{array}{l}x-2y=-13 \\ 4x+2y=18 \end{array}\)
\(\begin{array}{l}3x-4y=15 \\ 7x+y=4 \end{array}\)
\(\begin{array}{l}6x+4y=16 \\ -2x+y=-3 \end{array}\)
\(\begin{array}{l}7x+5y=-13 \\ x-4y=-16 \end{array}\)
\(\begin{array}{l}-5x-5y=-20 \\ -2x+y=7 \end{array}\)
\(\begin{array}{l}2x+y=2 \\ 3x+7y=14 \end{array}\)
\(\begin{array}{l}x+5y=15 \\ -3x+2y=6 \end{array}\)
\(\begin{array}{l}-2x+4y=-16 \\ y=-2 \end{array}\)
\(\begin{array}{l}-6x+6y=-12 \\ 8x-3y=16 \end{array}\)
\(\begin{array}{l}2x+3y=16 \\ -7x-y=20 \end{array}\)
\(\begin{array}{l}2x+y=-7 \\ 5x+3y=-21 \end{array}\)
\(\begin{array}{l}2x+3y=-10 \\ 7x+y=3 \end{array}\)
\(\begin{array}{l}-2x+2y=-22 \\ -5x-7y=-19 \end{array}\)
\(\begin{array}{l}-8x+2y=-6 \\ -2x+3y=11 \end{array}\)
\(\begin{array}{l}-x-4y=-14 \\ -6x+8y=12 \end{array}\)