# Linear Equations in Two Variables

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A system of equations is a set of equations that involve the same variables. A solution of a system is an assignment of values for the variables that make each equation in the system true.
1. College Algebra
Sixth Edition
James Stewart  Lothar Redlin  Saleem Watson
2. Systems of
5 Equations and
Inequalities
3. Chapter Overview
We have already seen how a real-world
situation can be modeled by an equation.
However, Many real-world situations have
too many variables to be modeled by
a single equation.
• For example, weather depends on
many variables, including temperature, wind speed,
air pressure, and humidity.
4. Chapter Overview
So, to model (and forecast) the weather,
scientists use many equations, each
having many variables.
• Such systems of equations work together
to describe the weather.
5. Chapter Overview
Systems of equations with hundreds,
or even thousands, of variables are also
used extensively in the air travel and
telecommunications industries to:
• Establish consistent airline schedules.
• Find efficient routing for telephone calls.
6. Systems of
Linear Equations
5.1 in Two Variables
7. Systems of Linear Equations
and Their Solutions
8. Systems of Linear Equations and Their Solutions
A system of equations is a set of equations
that involve the same variables.
A solution of a system is an assignment
of values for the variables that makes each
equation in the system true.
• To solve a system means to find
all solutions of the system.
9. Systems of Linear Equations and Their Solutions
Here is an example of a system of
linear equations in two variables:
2 x  y 5 Equation 1

 x  4 y 7 Equation 2
10. Systems of Linear Equations and Their Solutions
We can check that x = 3 and y = 1
is a solution of this system.
Equation 1 Equation 2
2x – y = 5 x + 4y = 7
2(3) – 1 = 5 3 + 4(1) = 7
• The solution can also be written as
the ordered pair (3, 1).
11. Systems of Linear Equations and Their Solutions
Note that the graphs of Equations 1
and 2 are lines.
• As the solution (3, 1)
satisfies each equation,
the point (3, 1) lies on
each line.
• So, it is the point of
intersection of the two
lines.
12. Substitution Method
13. Substitution Method
To solve a system using the
equation in the system and solve for one
variable in terms of the other variable.
1. Solve for one variable.
2. Substitute.
3. Back-substitute.
14. Step 1
Solve for one variable.
• Choose one equation, and solve
for one variable in terms of the other
variable.
15. Step 2
• Substitute the expression you found in step 1
into the other equation to get an equation
in one variable.
• Then, solve for that variable.
16. Step 3
• Substitute the value you found in step 2
back into the expression found in step 1
to solve for the remaining variable.
17. E.g. 1—Substitution Method
Find all solutions of the system.
2 x  y  1 Equation 1

3 x  4 y 14 Equation 2
• We solve for y in the first equation.
y = 1 – 2x
18. E.g. 1—Substitution Method
Now, we substitute for y in the second
equation and solve for x:
3x + 4(1 – 2x) = 14
3x + 4 – 8x = 14
–5x + 4 = 14
–5x = 10
x = –2
19. E.g. 1—Substitution Method
Next, we back-substitute x = –2
into the equation y = 1 – 2x:
y = 1 – 2(–2) = 5
• Thus, x = –2 and y = 5.
• So, the solution is the ordered pair (–2, 5).
20. E.g. 1—Substitution Method
The figure shows that the graphs
of the two equations intersect at
the point (–2, 5).
21. Elimination Method
22. Elimination Method
To solve a system using the elimination
method, we try to combine the equations
using sums or differences so as to eliminate
one of the variables.
3. Back-substitute.
23. Step 1
• Multiply one or more of the equations by
appropriate numbers so that the coefficient
of one variable in one equation is the negative
of its coefficient in the other equation.
24. Step 2
• Add the two equations to eliminate
one variable.
• Then, solve for the remaining variable.
25. Step 3
• Substitute the value you found in step 2
back into one of the original equations.
• Then, solve for the remaining variable.
26. E.g. 2—Elimination Method
Find all solutions of the system.
3 x  2y 14 Equation 1

2 x  4 y  4 Equation 2
• Multiply Equation 2 by ½ so that the coefficients of
the y-terms are negatives of each other.
• Now, we can add the equations to eliminate y.
27. E.g. 2—Elimination Method
3 x  2y 14
 x  2y 2
4x 16
x 4
• Now, we back-substitute x = 4 into one of
the original equations and solve for y.
• Let’s choose the second equation because
it looks simpler.
28. E.g. 2—Elimination Method
x – 2y = 2
4 – 2y = 2
–2y = –2
y=1
• The solution is (4, 1).
29. E.g. 2—Elimination Method
The figure shows
that the graphs of
the equations in
the system intersect
at the point (4, 1).
30. Graphical Method
31. Graphical Method
In the graphical method, we use
a graphing device to solve the system
of equations.
32. Graphical Method
1. Graph each equation.
2. Find the intersection points.
33. Step 1
Graph each equation.
• Express each equation in a form suitable
for the graphing calculator by solving for y
as a function of x.
• Graph the equations on the same screen.
34. Step 2
Find the intersection points.
• The solutions are the x- and y-coordinates
of the points of intersection.
35. E.g. 3—Graphical Method
Find all solutions of the system.
1.35x  2.13 y  2.36

2.16 x  0.32y  1.06
• To graph each equation we solve for y in terms
of x, and we get the equivalent system where
we have rounded the coefficients to
two decimals.
 y  0.63 x 1.11

 y  6.75 x  3.31
36. E.g. 3—Graphical Method
The figure shows that the two lines
• Zooming in,
we see that
the solution is
approximately
(0.30, 1.30).
37. The Number of Solutions of a
Linear System in Two Variables
38. The Number of Solutions of a Linear System in Two Variables
The graph of a linear system in two
variables is a pair of lines.
• So, to solve the system graphically,
we must find the intersection point(s)
of the lines.
39. The Number of Solutions of a Linear System in Two Variables
Two lines may intersect in a single point,
they may be parallel, or they may coincide.
• So, there are three possible outcomes
when solving such a system.
40. The Number of Solutions of a Linear System in Two Variables
For a system of linear equations in two
variables, exactly one of the following is true.
• The system has exactly one solution.
• The system has no solution.
• The system has infinitely many solutions.
41. Inconsistent and Dependent Systems
A system that has no solution is said
to be inconsistent.
A system with infinitely many solutions
is called dependent.
42. E.g. 4—A Linear System with One Solution
Solve the system and graph
the lines.
3 x  y 0 Equation 1

5 x  2y 22 Equation 2
43. E.g. 4—A Linear System with One Solution
We eliminate y from the equations and
solve for x.
6 x  2y 0 2  Equation1

5 x  2 y  22
11x 22
x 2
44. E.g. 4—A Linear System with One Solution
Now, we back-substitute into the first
equation and solve for y:
6(2) – 2y = 0
–2y = –12
y =6
• The solution of the system is the ordered pair
(2, 6).
• That is, x = 2, y = 6
45. E.g. 4—A Linear System with One Solution
The graph shows that
the lines in the system
intersect at the point
(2, 6).
46. E.g. 5—A Linear System with No Solution
Solve the system.
 8 x  2y 5 Equation 1

 12 x  3 y 7 Equation 2
• This time, we try to find a suitable combination
of the two equations to eliminate the variable y.
47. E.g. 5—A Linear System with No Solution
Multiplying the first equation by 3 and
the second by 2 gives:
 24 x  6 y 15

 24 x  6 y 14
0 29
• Adding the two equations eliminates
both x and y in this case.
• So, we end up with 0 = 29, which is obviously
false.
48. E.g. 5—A Linear System with No Solution
No matter what values we assign to x
and y, we cannot make this statement
• So, the system has no solution.
49. E.g. 5—A Linear System with No Solution
The figure shows that the lines in
the system are parallel and do not
• The system is
inconsistent.
50. E.g. 6—A Linear System with Infinitely Many Solutions
Solve the system.
3 x  6 y 12 Equation 1

4 x  8 y 16 Equation 2
• We multiply the first equation by 4 and
the second by 3 to prepare for subtracting
the equations to eliminate x.
51. E.g. 6—A Linear System with Infinitely Many Solutions
The new equations are:
12 x  24 y 48

12 x  24 y 48
• We see that the two equations in the original
system are simply different ways of expressing
the equation of one single line.
• The coordinates of any point on this line give
a solution of the system.
52. E.g. 6—A Linear System with Infinitely Many Solutions
Writing the equation in slope-intercept
form, we have: y = ½x – 2
• So, if we let t represent any real number,
we can write the solution as:
x=t
y = ½t – 2
• We can also write the solution
in ordered-pair form as:
(t, ½t – 2)
where t is any real number.
53. E.g. 6—A Linear System with Infinitely Many Solutions
The system has infinitely many
54. A Linear System with Infinitely Many
Solutions
In Example 3, to get specific solutions,
we have to assign values to t.
• If t = 1, we get
3
the solution (1, – 2 ).
• If t = 4, we get
the solution (4, 0).
• For every value of t,
we get a different
solution.
55. Modeling with Linear Systems
56. Modeling with Linear Systems
Frequently, when we use equations to
solve problems in the sciences or in other
areas, we obtain systems like the ones
we’ve been considering.
57. Guidelines for Modeling with Systems of Equations
When modeling with systems of equations,
we use these guidelines—similar to
those in Section 1.5.
1. Identify the variables.
2. Express all unknown quantities in terms of
the variables.
3. Set up a system of equations.
4. Solve the system and interpret the results.
58. Guideline 1 for Modeling with Systems of Equations
Identify the variables.
• Identify the quantities the problem asks you to find.
• These are usually determined by a careful reading
of the question posed at the end of the problem.
• Introduce notation for the variables.
(Call them x and y or some other letters).
59. Guideline 2 for Modeling with Systems of Equations
Express all unknown quantities
in terms of the variables.
• Read the problem again, and express all
the quantities mentioned in the problem
in terms of the variables you defined in step 1.
60. Guideline 3 for Modeling with Systems of Equations
Set up a system of equations.
• Find the crucial facts in the problem that give
the relationships between the expressions
you found in Step 2.
• Set up a system of equations (or a model)
that expresses these relationships.
61. Guideline 4 for Modeling with Systems of Equations
Solve the system and interpret
the results.
• Solve the system you found in Step 3.
answers the question posed in the problem.
62. E.g. 7—A Distance-Speed-Time Problem
A woman rows a boat upstream from
one point on a river to another point 4 mi
away in 1½ hours.
• The return trip,
traveling with
the current,
takes only
45 min.
63. E.g. 7—A Distance-Speed-Time Problem
How fast does she row relative to the water?
At what speed is the current flowing?
64. E.g. 7—A Distance-Speed-Time Problem
We are asked to find the rowing speed
and the speed of the current.
• So, we let:
x = rowing speed (mi/h)
y = current speed (mi/h)
65. E.g. 7—A Distance-Speed-Time Problem
The woman’s speed when she rows
upstream is:
• Her rowing speed minus the speed of the current
Her speed downstream is:
• Her rowing speed plus the speed of the current.
66. E.g. 7—A Distance-Speed-Time Problem
Now, we translate this information into
the language of algebra.
67. E.g. 7—A Distance-Speed-Time Problem
The distance upstream and downstream
is 4 mi.
• So, using the fact that
speed  time = distance for both legs of the trip
we get:
speed upstream  time upstream distance traveled
speed downstream  time downstream distance traveled
68. E.g. 7—A Distance-Speed-Time Problem
In algebraic notation, that translates into
these equations.
( x  y ) 32 4 Equation 1
( x  y ) 34 4 Equation 2
• The times have been converted to hours, since
we are expressing the speeds in miles per hour.
69. E.g. 7—A Distance-Speed-Time Problem
We multiply the equations by 2 and 4,
respectively, to clear the denominators.
3 x  3 y 8

3 x  3 y 16
6x 24
x 4
70. E.g. 7—A Distance-Speed-Time Problem
Back-substituting this value of x into
the first equation (the second works
just as well) and solving for y gives:
3(4)  3 y 8
 3 y 8  12
y  34
• The woman rows at 4 mi/h.
• The current flows at 1⅓ mi/h.
71. E.g. 8—A Mixture Problem
A vintner fortifies wine that contains 10%
alcohol by adding 70% alcohol solution to it.
The resulting mixture has an alcoholic
strength of 16% and fills 1000 one-liter
• How many liters (L) of the wine and
of the alcohol solution does vintner use?
72. E.g. 8—A Mixture Problem
Since we are asked for the amounts
of wine and alcohol, we let:
x = amount of wine used (L)
y = amount of alcohol solution used (L)
73. E.g. 8—A Mixture Problem
From the fact that the wine contains 10%
alcohol and the solution contains 70%
alcohol, we get the following.
74. E.g. 8—A Mixture Problem
The volume of the mixture must be
the total of the two volumes the vintner
• Thus,
x + y = 1000
75. E.g. 8—A Mixture Problem
Also, the amount of alcohol in the mixture
must be the total of the alcohol contributed
by the wine and by the alcohol solution.
• That is,
0.10x + 0.70y = (0.16)1000
0.10x + 0.70y = 160
x + 7y = 1600
76. E.g. 8—A Mixture Problem
Thus, we get the system
 x  y 1000 Equation1

 x  7 y 1600 Equation 2
• Subtracting the first equation from the second
eliminates the variable x, and we get:
6y = 600
y = 100
77. E.g. 8—A Mixture Problem
We now back-substitute y = 100 into
the first equation and solve for x:
x + 100 = 1000
x = 900
• The vintner uses 900 L of wine
and 100 L of the alcohol solution.