College Algebra Lecture Notes
Section 5.1
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Section 5.1: Linear Systems in Two Variables with Applications
Big Idea: Many times, when we want to solve a real-world problem that involves two or more
variable quantities, we can write two or more equations that describe the problem, and then solve
those multiple equations to get the solution to the problem that we want.
Big Skill: You should be able to solve a system of two linear equations in two unknowns by
graphing, using substitution, and using elimination.
A. Solutions to a System of Equations
A system of linear equations is a grouping of two or more linear equations, each of which
contains one or more variables.
A solution of a system of linear equations consists of values for the variables that are solutions
to ALL of the equations in the system.
Practice:
1. Algebraically determine which of the following ordered pairs is a solution of the
following system of linear equations
 2x  y  2
.

4 x  3 y  9
a. (-2, 3)
b. (1.5, -1)
c. Graph the lines below and see how the solution relates to the lines:
College Algebra Lecture Notes
Section 5.1
B. Solving Systems Graphically

Graph both the lines.

Read the coordinates of the intersection point off the graph.

Check to see if those coordinates are the solution.
Practice:
 x  2 y  6
2. Solve the system 
by graphing.
2 x  3 y  16
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College Algebra Lecture Notes
Section 5.1
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C. Solving Systems by Substitution

Solve one of the equations for one of the unknowns.

Substitute the expression for that unknown into the other equation.

Solve the resulting equation in one unknown.

Substitute that solution into the first equation to solve for the original variable.

Check your answer.
Practice:
2 x  3 y  5
3. Solve the system 
using substitution.
4 x  9 y  10
College Algebra Lecture Notes
Section 5.1
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D. Solving Systems Using Elimination

Multiply one or both equations by nonzero constants so that the coefficients of one of the
variables are additive inverses.

Add the two equations to obtain a new equation in one unknown.

Solve the resulting equation in one unknown.

Substitute that solution into either of the original equations and solve for the remaining
variable.

Check your answer.
Operations that produce an equivalent system of equations:

Changing the order of the equations.

Replacing an equation by a nonzero constant multiple of that equation.

Replacing an equation with the sum of two equations from the system.
Practice:
17 x  3 y  24
4. Solve the system 
using elimination.
 6 x  3 y   9
College Algebra Lecture Notes
Section 5.1
1

 3x  2 y  3
5. Solve the system 
using elimination.
 1 x  3 y  11
 2
2
3
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College Algebra Lecture Notes
Section 5.1
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E. Inconsistent and Dependent Systems
The number of solutions to a system of linear equations in two variables falls under one of
three cases:
 Zero solutions: The lines never intersect (i.e., they are parallel to one another), and thus
the system has no solutions. This type of system is called inconsistent.
o Note: m1 = m2; b1  b2

One solution: The lines intersect at one point, and thus the system has exactly one
solution. This type of system is called consistent and the equations are called
independent.
o Note: m1  m2

Infinitely many solutions: The lines lie on top of each other (i.e., the equations are
equivalent), and thus the system has infinitely many solutions. This type of system is
called consistent and the equations are called dependent.
o Note: m1 = m2; b1 = b2
Vocabulary:
Consistent: The system has one or more solutions.
Inconsistent: The system has no solutions.
Dependent: The equations produce the same line when graphed.
Independent: The equations produce different lines when graphed.
College Algebra Lecture Notes
Section 5.1
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Practice:
6. Graph the following systems and describe them as consistent, inconsistent, dependent, or
independent
 x  y 1

2 x  y  7
 x y 5

2 x  2 y  8
 3x  2 y  10

6 x  4 y  20
College Algebra Lecture Notes
Section 5.1
The Solution to any System of Two Linear Equations in Two Variables:
ax  by  k1

 cx  dy  k2
 (c)(eqn. #1)
acx  bcy  ck1

 dy  k2
 cx
 (a)(eqn. #2)
acx  bcy  ck1

 acx  ady  ak2
 (eqn. #1)  (eqn. #2)is placed in row #2
bcy

ck1
acx 

 ad  bc  y  ak2  ck1

  ad  bc  (eqn. #1)
ac  ad  bc  x  bc  ad  bc  y  c  ad  bc  k1

ak2  ck1
 ad  bc  y 

  bc  (eqn. #2)
ac  ad  bc  x  bc  ad  bc  y  c  ad  bc  k1

bc  ad  bc  y  bc  ak2  ck1 

 (eqn. #1)  (eqn. #2)is placed in row #1
ac  ad  bc  x


bc  ad  bc  y 
So,
bc  ad  bc  y  bc  ak2  ck1 
y
bc  ak2  ck1 
bc  ad  bc  y
y
ak2  ck1
ad  bc
And
 bc  ak2  ck1   c  ad  bc  k1
bc  ak2  ck1 
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College Algebra Lecture Notes
x
bc  ak2  ck1   c  ad  bc  k1
ac  ad  bc 
x
b  ak2  ck1    ad  bc  k1
a  ad  bc 
x
abk2  bck1  adk1  bck1
a  ad  bc 
x
abk2  adk1
a  ad  bc 
x
a  dk1  bk2 
a  ad  bc 
x
dk1  bk2
ad  bc
Section 5.1
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