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Precalculus for Team-Based Inquiry Learning: 2024 Development Edition — Instructor Version

TBIL Fellows

Section 3.6 Systems of Linear Equations (LF6)

Subsection 3.6.1 Activities

Observation 3.6.1.

Often times when solving a real-world application, more than one equation is necessary to describe the information. We’ll investigate some of those in this section.

Activity 3.6.2.

Admission into a carnival for \(4\) children and \(2\) adults is \(\$128.50\text{.}\) For \(6\) children and \(4\) adults, the admission is \(\$208\text{.}\) Assuming a different price for children and adults, what is the price of the child’s admission and the price of the adult admission?
(a)
Let \(c\) represent the cost of a child’s admission and \(a\) the cost of an adult admission. Write an equation to represent the total cost for 4 children and 2 adults.
  1. \(\displaystyle 4a+2c=128.50\)
  2. \(\displaystyle a+c=128.50\)
  3. \(\displaystyle 4c+2a=128.50\)
  4. \(\displaystyle a+c=336.50\)
Answer.
C: \(4c+2a=128.50\)
(b)
Now write an equation to represent the total cost for 6 children and 4 adults.
  1. \(\displaystyle 6c+4a=208\)
  2. \(\displaystyle a+c=208\)
  3. \(\displaystyle 4a+6c=208\)
  4. \(\displaystyle a+c=336.50\)
Answer.
A: \(6c+4a=208\)
(c)
Using the above equations, check by substitution which admission prices would satisfy both equations?
  1. \(c=\$20\) and \(a=\$24.25\)
  2. \(c=\$24.50\) and \(a=\$15.25\)
  3. \(c=\$20\) and \(a=\$22\)
  4. \(c=\$14.50\) and \(a=\$30.25\)
Answer.
B: \(c=\$24.50\) and \(a=\$15.25\)

Definition 3.6.3.

A system of linear equations consists of two or more linear equations made up of two or more variables. A solution to a system of equations is a value for each of the variables that satisfies all the equations at the same time.

Activity 3.6.4.

Consider the following system of equations.
\begin{equation*} \begin{cases} y=2x+4\\ 3x+2y=1 \end{cases} \end{equation*}
(a)
Which of the ordered pairs is a solution to the system?
  1. \(\displaystyle (3,10)\)
  2. \(\displaystyle (0,4)\)
  3. \(\displaystyle (1,-1)\)
  4. \(\displaystyle (-1,2)\)
Answer.
D: \((-1,2) \)

Remark 3.6.5.

While we can test points to determine if they are solutions, it is not feasible to test every possible point to find a solution. We need a method to solve a system.

Activity 3.6.6.

Consider the following system of equations.
\begin{equation*} \begin{cases} 3x-y=2\\ x+4y=5 \end{cases} \end{equation*}
(a)
Rewrite the first equation in terms of \(y\text{.}\)
(b)
Rewrite the second equation in terms of \(y\text{.}\)
(c)
Graph the two equations on the same set of axes. Where do the lines intersect?
(d)
Check that the point of intersection of the two lines is a solution to the system of equations.

Remark 3.6.7.

Sometimes it is difficult to determine the exact intersection point of two lines using a graph. Let’s explore another possible method for solving a system of equations.

Activity 3.6.8.

Consider the following system of equations.
\begin{equation*} \begin{cases} 3x+y=4\\ x+3y=10 \end{cases} \end{equation*}
(a)
Graph the two equations on the same set of axes. Is it possible to determine exactly where the lines intersect?
(b)
Solve the first equation for \(y\) and substitute into the second equation. What is the resulting equation?
  1. \(\displaystyle x+4-3x=10\)
  2. \(\displaystyle x+3(4-3x)=10\)
  3. \(\displaystyle 4-3x+3y=10\)
  4. \(\displaystyle 3x+(4-3x)=4\)
Answer.
B: \(x+3(4-3x)=10\)
(c)
Solve the resulting equation from part (b) for \(x\text{.}\)
  1. \(\displaystyle x=-3\)
  2. \(\displaystyle x=\frac{1}{4}\)
  3. \(\displaystyle x=\frac{7}{3}\)
  4. \(\displaystyle x=0\)
Answer.
B: \(x=\frac{1}{4}\)
(d)
Substitute the value of \(x\) into the first equation to find the value of \(y\text{.}\)
  1. \(\displaystyle y=-5\)
  2. \(\displaystyle y=\frac{13}{4}\)
  3. \(\displaystyle y=-3\)
  4. \(\displaystyle y=\frac{4}{3}\)
Answer.
B: \(y=\frac{13}{4}\)
(e)
Write the solution to the system of equations (the found values of \(x\) and \(y\)) as an ordered pair.

Remark 3.6.9.

This method of solving a system of equations is referred to as the Substitution Method.
  1. Solve one of the equations for one variable.
  2. Substitute the expression into the other equation to solve for the remaining variable.
  3. Substitute that value into either equation to find the value of the first variable.

Activity 3.6.10.

Solve the following system of equations using the substitution method.
\begin{equation*} \begin{cases} x+2y=-1\\ -x+y=3 \end{cases} \end{equation*}
Answer.
\(\left( -\dfrac{7}{3}, \dfrac{2}{3} \right)\)

Remark 3.6.11.

While the substitution method will always work, sometimes the resulting equations will be difficult to solve. Let’s explore a third method for solving a system of two linear equations with two variables.

Activity 3.6.12.

Consider the following system of equations.
\begin{equation*} \begin{cases} 5x+7y=12\\ 3x-7y=37 \end{cases} \end{equation*}
(a)
Add the two equations together. What is the resulting equation?
  1. \(\displaystyle 2x=-15\)
  2. \(\displaystyle 14y=49\)
  3. \(\displaystyle 8x+14y=49\)
  4. \(\displaystyle 8x=49\)
Answer.
D: \(8x=49\)
(b)
Use the resulting equation after addition, to solve for the variable.
  1. \(\displaystyle x=-\frac{15}{2}\)
  2. \(\displaystyle y=\frac{49}{14}\)
  3. \(\displaystyle x=\frac{49}{22}\)
  4. \(\displaystyle x=\frac{49}{8}\)
Answer.
D: \(x=\frac{49}{8}\)
(c)
Use the value to find the solution to the system of equations.
Answer.
\(\left( \dfrac{49}{8}, -\dfrac{149}{56}\right)\)

Remark 3.6.13.

This method of solving a system of equations is referred to as the Elimination Method.
  1. Combine the two equations using addition or subtraction to eliminate one of the variables.
  2. Solve the resulting equation.
  3. Substitute that value into either equation to find the value of the other variable.

Activity 3.6.14.

Solve the following system of equations using the elimination method.
\begin{equation*} \begin{cases} 7x-4y=3\\ 3y-7x=8 \end{cases} \end{equation*}
Answer.
\(\left(-\frac{41}{7} ,-11 \right)\)

Activity 3.6.15.

Consider the following system of equations.
\begin{equation*} \begin{cases} 5x-9y=6\\ -10x+4y=2 \end{cases} \end{equation*}
Notice that if you simply add the two equations together, it will not eliminate a variable. Substitution will also be difficult since it involves fractions.
(a)
What value can you multiply the first equation by so that when you add the result to the second equation one variable cancels?
  1. \(-1\) and the \(x\) will cancel
  2. \(2\) and the \(x\) will cancel
  3. \(3\) and the \(y\) will cancel
  4. \(-2\) and the \(y\) will cancel
Answer.
B: Multiply the second equation by \(2\) and the \(x\) will cancel
(b)
Perform the multiplication and add the two equations. What is the resulting equation?
  1. \(\displaystyle -5y=8\)
  2. \(\displaystyle -14y=14\)
  3. \(\displaystyle -14y=8\)
  4. \(\displaystyle -5x=4\)
Answer.
B: \(-14y=14\)
(c)
What is the solution to the system of equations?
  1. \(\displaystyle (-1,-1.6)\)
  2. \(\displaystyle (-1,-0.6)\)
  3. \(\displaystyle (-0.6,-1)\)
  4. \(\displaystyle (-1.6,-1)\)
Answer.
C: \((3,1)\)

Activity 3.6.16.

For each system of equations, determine which method (graphical, substitution, or elimination) might be best for solving.
(a)
\begin{equation*} \begin{cases} 5x+9y=16\\ x+2y=4 \end{cases} \end{equation*}
  1. Graphical
  2. Substitution
  3. Elimination
Answer.
B: Substitution
(b)
\begin{equation*} \begin{cases} y=4x-6\\ y=-5x+21 \end{cases} \end{equation*}
  1. Graphical
  2. Substitution
  3. Elimination
Answer.
A: Graphical
(c)
\begin{equation*} \begin{cases} x+y=10\\ x-y=12 \end{cases} \end{equation*}
  1. Graphical
  2. Substitution
  3. Elimination
Answer.
C: Elimination

Activity 3.6.17.

Solve each of the systems of equations from Activity 3.6.16 using the method you chose.
Answer.
(a) \((-4,4)\) (b) \((3,6)\) (c) \((11,-1)\)

Subsection 3.6.2 Videos

It would be great to include videos down here, like in the Calculus book!
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