Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
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Sect.
7.1
Topic
Homework
HW √ On-Time
Introduction to
Page 431: 8-23
Systems
Solving Systems by
Graphing
Essential Question: How can we determine the solution of a linear system given a
graph?
7.3
Solving Systems Using
Page 447: 3-33 odd, 39-41
Elimination, part I
Essential Question: How can we determine how many solutions a linear system has?
7.4
Quiz #1 – Graphing &
Page 454: 3-20
Elimination
Solving Systems Using
Elimination, part II
Essential Question: How can we use the graphing calculator to verify algebraic
solutions?
7.2
Solving Systems Using
Page 439: 3-23 odd; 31,
Substitution
32, 35, 36
7.5
Solving Special Linear
Page 463: 8-23 odd
Systems
Essential Question: How do you determine which method of solving a linear system is
most efficient?
7.6
Quiz #2 – Elimination
Page 469: 3-23
& Substitution
Systems of Linear
Inequalities
Essential Question: How can we determine the solution of a system of linear
inequalities given a graph?
Unit Review
Page 479: #1-30 & Study
for Test
Essential Question: How do you determine which method of solving a linear system is
most efficient?
Unit 7 Test
Essential Question: What could I have done to better prepare for the exam?
Page 1 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Page 2 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Solving Systems by Graphing
Two or more linear equations form a system of linear equations. Any ordered pair that makes all
of the equations in a system true is a solution of a system of linear equations.
The solution is where the systems are equal.
The equations in a system of linear equations use the same variables.
The solution to a linear system is the point(s) that make both equations true.
The solution to a linear system is the intersection of the graphs of the linear equations.
Example:
The lines intersect at (1, 3).
Steps to Solve a Linear Equation by Graphing
1. Write each equation in slope-intercept form. (y = mx + b)
2. Carefully graph each line.
a. Plot the y-intercept (b).
b. Use the slope (m) to plot the second point.
c. Connect the dots to create the lines (don’t forget the arrows!).
3. Find the point where the lines intersect and write it as an ordered pair.
4. Check your solution by substitution (manually or with the graphing calculator).
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Examples: Solve by graphing.
𝒙+𝒚=𝟒
𝟐𝒙 − 𝒚 = 𝟓
𝒙+𝒚=𝟎
𝟑𝒙 − 𝟐𝒚 = 𝟏𝟎
𝟓
𝒚= − 𝒙+𝟑
𝟑
𝟏
𝒚 = 𝒙−𝟑
𝟑
Page 4 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Finding the Solution to a Linear System on a Graphing Calculator
1. Enter the equations in the
2. Graph the equations using the
3. Use the
CALC
Y=
screen.
GRAPH
feature.
feature. Choose INTERSECT to find the point where the lines intersect.
a. Note: Press ENTER 3 times to display the intersection.
Example: Find the intersection of the system using the graphing calculator.
𝒚 = −𝟓𝒙 + 𝟔
𝒚 = −𝒙 − 𝟐
Practice. Solve each system using the graphing calculator, identify the solution, and sketch the
graph.
𝟏
𝒚=− 𝒙−𝟏
𝟐
𝟏
𝒚 = 𝒙−𝟒
𝟒
−𝟒𝒙 + 𝒚 = 𝟑
𝒙 = −𝒚 − 𝟐
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Word Problem Practice. Create the linear system equations and graph to identify the solution.
Find the value of two numbers if their sum is 12 and their difference is 4.
THS is selling tickets to a choral performance. On the first day of ticket sales there were 3 senior
citizen tickets sold and 1 child ticket sold for a total of $19. On the second day, THS made $26 by
selling 3 senior citizen tickets and 2 child tickets. Find the price of each type of ticket – senior
citizen and child.
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Solving Systems Using Elimination
In the elimination method, you use the Addition and Subtraction Properties of Equality to add or
subtract equations in order to eliminate a variable in the system. The elimination method is also
known as linear combination.
We use linear combinations to eliminate one of the variables so that we can solve for the other
one.
Then, we can substitute the value that we found for the variable for which we solved into one of
the original equations and solve for the other variable.
Example. Find the solution to the system using elimination.
𝟐𝒙 + 𝟓𝒚 = 𝟏𝟕
𝟔𝒙 − 𝟓𝒚 = −𝟗
Step 1: Eliminate one variable. Since the sum of the y coefficients is 0, add the equations to
eliminate y.
𝟐𝒙 + 𝟓𝒚 = 𝟏𝟕
𝟔𝒙 − 𝟓𝒚 = −𝟗
--------------------𝟖𝒙 + 𝟎 = 𝟖
Add the two equations.
𝒙=𝟏
Solve for x
Step 2: Substitute 1 for x to solve for the eliminated variable, y.
𝟐𝒙 + 𝟓𝒚 = 𝟏𝟕
You can use either equation. We are using the first.
𝟐(𝟏) + 𝟓𝒚 = 𝟏𝟕
Substitute 1 for x.
𝟐 + 𝟓𝒚 = 𝟏𝟕
Simplify
𝒚=𝟑
Solve for y.
Since x = 1 and y = 3, the solution is (1, 3).
Page 7 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Steps for Solving Systems Using Elimination/Linear Combination
1. Arrange the equations with like terms in columns.
2. Look for coefficients that are opposites for one of the variables. For example, 2x and -2x
are opposites (they eliminate each other when added together).
a. Note: If you have no opposites, you may have to multiply one, or both, of the
equations to create an opposite.
For example: If one equation has 2x and the other has 4x, multiple the equation
with 2x by -2 to create the opposite (-4x). Remember to multiply the entire
equation to keep the equation true.
3. Add the two equations together to eliminate one of the variables.
4. Solve for the variable and substitute it into one of the original equations.
5. Solve for the second variable.
6. Check your solutions (through substitution in the second original equation) and then write
your answer as an ordered pair.
Examples. Solve the linear systems using elimination/linear combination.
−𝑥 + 2𝑦 = −8
𝑥 + 6𝑦 = −16
11𝑥 + 3𝑦 = 1
−5𝑥 − 3𝑦 = −7
Sometimes you have to rearrange the equations to get the terms in the same order.
−3𝑥 − 2𝑦 = −8
2𝑦 = 12 − 5𝑥
11𝑥 = −3𝑦 + 1
−5𝑥 − 3𝑦 = −7
Page 8 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Sometimes you have to multiply one of the equations to get opposite coefficients.
3𝑥 + 6𝑦 = 12
−𝑥 + 3𝑦 = 6
5𝑥 − 2𝑦 = 3
5𝑥 + 𝑦 = −9
Practice. Solve the linear systems using elimination/linear combination.
−4𝑥 − 2𝑦 = −12
4𝑥 + 8𝑦 = −24
𝑥 − 𝑦 = 11
2𝑥 + 𝑦 = 19
−6𝑥 + 6𝑦 = 6
−6𝑥 + 3𝑦 = −12
−7𝑥 + 𝑦 = −19
−2𝑥 + 3𝑦 = −19
The senior classes at High School A and High School B planned separate trips to New York City.
High School A rented and filled 1 van and 6 busses with 372 students. High School B rented and
filled 4 vans and 12 busses with 780 students. Each van and bus carried the same number of
students. How many students can a van carry? How many students can a bus carry?
Page 9 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Elimination/Linear Combination – Modifying BOTH Equations
Sometimes, you need to multiply BOTH equations in order to get coefficients that are opposite.
Example. Solve the linear system.
2𝑥 + 2𝑦 = −8
3𝑥 − 3𝑦 = 18
There is no opposite to eliminate. Need to multiply
both equations to create an opposite.
(3)(2𝑥 + 2𝑦) = −8 (𝟑)
(2)(3𝑥 − 3𝑦) = 18 (𝟐)
6/-6 can be created as an “opposite” for y by multiplying
the top equation by 3 and the bottom by 2.
6𝑥 + 6𝑦 = −24
6𝑥 − 6𝑦 = 36
-------------------12𝑥 = 12
𝑥=1
2𝑥 + 2𝑦 = −8
2(1) + 2𝑦 = −8
2 + 2𝑦 = −8
2𝑦 = −10
𝑦 = −5
The solution set is (1, -5).
Practice.
2𝑥 − 5𝑦 = −19
3𝑥 + 2𝑦 = 0
5𝑥 + 4𝑦 = −30
3𝑥 − 9𝑦 = −18
Page 10 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Solving Systems by Substitution
You can solve linear systems by solving one of the equations for one of the variables. Then,
substitute the expression for the variable into the other equation. This is called the substitution
method.
Using substitution. What is the solution to the system (use substitution).
𝑦 = 3𝑥
𝑥 + 𝑦 = −32
Because y = 3x, you can substitute 3x for y in the second equation.
𝑥 + 3𝑥 = −32
4𝑥 = −32
𝑥 = −8
𝑦 = 3(−𝟖)
𝑦 = −24
The solution is (-8, -24).
Oftentimes, you will need to solve for a variable in order to use the substitution
method. You may not be given a solved equation like the one above, y = 3x. You will
need to use your properties of equality to solve for a variable in one equation and
then substitute the expression into the second equation.
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Example. What is the solution to the system? Use substitution to solve.
2𝑦 + 4𝑥 = 24
−2𝑥 + 𝑦 = −4
Solve for a variable in one equation:
−2𝑥 + 𝑦 = −4
+2𝑥
+ 2𝑥
------------------------𝑦 = 2𝑥 − 4
Substitute the solution into the other equation.
2(𝟐𝒙 − 𝟒) + 4𝑥 = 24
4𝑥 ∓ 8 + 4𝑥 = 24
8𝑥 + 8 = 24
−8
−8
--------------------8𝑥 = 16
𝑥=2
Substitute the solution into the first equation to solve for the other variable.
−2(2) + 𝑦 = −4
−4 + 𝑦 = −4
+4
+4
---------------------𝑦=0
The solution set is 2, 0).
𝑦 = 6𝑥 − 11
−2𝑥 − 3𝑦 = −7
−7𝑥 − 2𝑦 = −13
𝑥 − 2𝑦 = 11
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Practice. Use substitution to solve the system.
2𝑥 − 3𝑦 = −1
𝑦 =𝑥−1
−3𝑥 − 3𝑦 = 3
𝑦 = −5𝑥 − 17
𝑦 = 5𝑥 − 7
−3𝑥 − 2𝑦 = −12
−4𝑥 + 𝑦 = 6
−5𝑥 − 𝑦 = 21
−7𝑥 − 2𝑦 = −13
𝑥 − 2𝑦 = 11
−5𝑥 + 𝑦 = −2
3𝑥 − 8𝑦 = 24
Page 13 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Practical Applications of Systems (Real World Scenarios)
You can solve systems of linear equations using a graph, substitution, or elimination/linear
combination. The best method depends on the forms of the given equations.
Choosing a Method for Solving Linear Systems
Method
When to use
Graphing
When you want a visual display or when you want to estimate a solution.
Substitution
When one equation is already solved for one of the variables, or when it is
easy to solve for one of the variables.
Elimination/Linear When the coefficients of one variable are the same or opposites, or when it is
Combination
not convenient to use graphing or substitution.
Examples.
Finding a Break-Even Point (i.e. when income = expenses)
A fashion designer makes and sells hats. The material for each hat costs $5.50. The hats sell for
$12.50 each. The designer spends $1,400 on advertising. How many hats must the designer sell to
break even?
Step 1. Write a system of equations. Let x = the number of hats sold, and let y = the dollars of
expense or income.
Expense: y = 5.5x + 1400
Income: y = 12.5x
Step 2. Choose a method to solve. Substitution is recommended because both equations are
already solved for y.
𝒚 = 5.5𝑥 + 1400
Write the first equation
𝟏𝟐. 𝟓𝒙 = 5.5𝑥 + 1400
Substitute
7𝑥 = 1400
Subtract 5.5x from both sides
𝑥 = 200
Divide both sides by 7.
Since x is the number of hats, the designer must sell 200 hats to break even.
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
A puzzle expert wrote a new Sudoku puzzle book. His initial costs are $864. Binding and packaging
each book costs $0.80. The price of the book is $2.00. How many copies must be sold to break
even?
Page 15 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Solving a Mixture Problem
A dairy owner produces low fat milt containing 1% fat and whole milk containing 3.5% fat. How
many gallons of each type should be combined to make 100 gallons of milk that is 2% fat?
Let x = the number of gallons of low fat milk, and let y = the number of gallons of whole milk.
Total gallons: x+ y = 100
Fat content: 0.01x + 0.035y = 0.02(100)
The first equation is easy to solve for x or y so use substitution.
𝑥 + 𝑦 = 100
𝑥 = 100 − 𝑦
Write the equation
Subtract y from both sides to solve for x.
Substitute 100 – y for x in the second equation and solve for y.
0.01𝒙 + 0.035𝑦 = 0.02(100)
Write the first equation.
0.01(𝟏𝟎𝟎 − 𝒚) + 0.035𝑦 = 0.02(100)
Substitute
1 − .01𝑦 + 0.035𝑦 = 2
Distribute
1 + 0.025𝑦 = 2
Simplify
0.025𝑦 = 1
Subtract 1 from both sides
𝑦 = 40
Divide both sides by 0.025
Substitute y = 40 in either equation and solve for x.
𝑥 + 𝟒𝟎 = 100
Substitute
𝑥 = 60
Subtract 40 from both sides
The owner should mix 60 gallons of low-fat milk with 40 gallons of whole milk.
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
One antifreeze solution is 20% alcohol. Another antifreeze solution is 12% alcohol. How many
liters of each solution should be combined to make 15 liters of antifreeze solution that is 18%?
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Practice.
Gym A charges a membership fee of $100 plus $2.00 per visit. Gym B charges a membership fee of
$60, plus $4.00 per visit. At how many visits, x, is the total cost, y, the same for both gyms?
A ticket booth sold 226 tickets and collected 843 in ticket sales. Adult tickets are $5.50 and child
tickets are $1.50. How many tickets of each type were sold?
Page 18 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Solving Systems of Inequalities
A system of inequalities is made up of two or more linear inequalities. A solution of a system of
linear inequalities is an ordered pair that makes all the inequalities in the system true. The graph
of a system of linear inequalities is the set of points that represent all of the solutions to the
system.
y>x
y < -x
Each point on a dashed line is not
a solution. A dashed line is used
for inequalities with > or <.
Each point on a solid line is not a
solution. A solid line is used for
inequalities with > or <.
The overlap of the shaded
areas (including the solid
line) represents the
solution set.
Page 19 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Examples. Graph each system.
𝑦≤1
𝑥≥2
𝑦 > 2𝑥 − 4
1
𝑦 > −3𝑥 + 2
𝑦 ≤ 3𝑥 + 1
𝑦 > 3𝑥 − 3
𝑥+𝑦 >2
𝑦 ≤ 3𝑥 + 1
2
2
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Practice. Graph each system.
𝑦 ≤ −𝑥 − 2
𝑦 ≥ −5𝑥 + 2
𝑦 ≤ −3
5
𝑦 < 3𝑥 + 2
1
𝑦 ≤ 2𝑥 + 2
𝑦 < −2𝑥 − 3
Page 21 of 23
Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Identifying Solutions to Linear Inequalities
Use substitution to determine if an ordered pair is a solution to a given system. If the ordered pair
makes both inequalities true, then it is a solution. Otherwise, it is not a solution.
Practice. Determine whether the ordered pair is a solution of the given system.
(-2, -4)
5
𝑦 ≤ −2𝑥 − 2
(1, 1)
3𝑥 + 2𝑦 ≥ −2
𝑦 < −2𝑥 + 2
𝑥 + 2𝑦 ≤ 2
1
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Briggs – Algebra 1 – Unit 6: Systems of Equations and Inequalities -- Notes
Graphing Linear Inequalities with a Graphing Calculator
A graphing calculator can show the solutions of an inequality or a system of inequalities.
1) To enter an inequality, press APPS and scroll down to select Inequalz.
2) Move the cursor over the = for one of the equations. Notice the inequality symbols at the
bottom of the screen. They correspond to the buttons below labeled F2, F3, F4, and F5 in
green.
3) Change the = symbol to the appropriate inequality symbol by pressing ALPHA followed by
the corresponding F2-F5 button.
4) Move the cursor to the right, using your arrow, and enter the expression for Y1.
5) Press enter.
6) Move the cursor to the left and change the = symbol to the appropriate inequality symbol
by pressing ALPHA followed by the corresponding F2-F5 button.
7) Move the cursor to the right, using your arrow, and enter the expression for Y1.
8) Press enter.
9) Press GRAPH to graph the system of inequalities.
Example. Use a graphing calculator to graph the system of inequalities. Sketch the graph.
𝑦 < −2𝑥 − 3
𝑦 ≥𝑥+4
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