Worksheet 54 (10
Worksheet 54 (10.1)
Chapter 10 Systems of Equations
10.1 Systems of Two Linear Equations in Two Variables
Summary 1:
Systems of Equations
A system of two linear equations in two variables is two
equations
considered together. To solve a system is to find all the
ordered pairs that satisfy both equations.
When solving a system three situations can occur:
1. The lines intersect. There is one solution, the point where
they
intersect. The system is called a consistent system.
2. The lines are parallel. There is no solution, the lines do
not intersect.
The system is called an inconsistent system.
3. The lines are the same. There are infinitely many solutions,
the
lines coincide. The system is called dependent.
Systems of equations can be solved by various methods which will
be
discussed in this chapter.
Warm-up 1.a) Use the graphing approach to solve the system. Tell
whether the
system is consistent, inconsistent or dependent.
÷
÷
ø
ö
ç
ç
è
æ
4
=
2y
+
3x
5
=
y
-
2x
Graph the two lines using any method discussed in Chap. 7.
y
x
The solution to the system is the point of intersection: ( ,
).
The system is called .
Worksheet 54 (10.1)
Problem
1. Use the graphing approach to solve the system. Tell whether
the system is consistent, inconsistent or dependent.
÷
÷
ø
ö
ç
ç
è
æ
3
=
y
-
x
5
=
y
+
x
y
x
Summary 2:
The Substitution Method
To solve a system of equations by the substitution method:
1. Solve one of the equations for one of the variables.(Choose a
variable
with a coefficient of 1 or -1 if possible.)
2. Substitute this expression into the other equation to produce
an
equation with only one variable.
3. Solve the equation in Step 2 for the remaining variable.
4. Substitute this solution into the expression obtained in Step
1.
5. Solve for the second variable.
6. Write your solution set as an ordered pair, and check in each
equation.
Warm-up 2.a) Solve the system by the substitution method:
÷
÷
ø
ö
ç
ç
è
æ
2
=
y
+
3x
5
=
3y
-
2x
y= 2 - 3x 1) Solve the 2nd equation for y.
2x - 3( )= 5
2) Substitute 2 - 3x into the other
equation in place of y.
2x - + = 5 3) Solve for x.
11x=
x=
y= 2 - 3( ) 4) Substitute this value into step 1.
y= 5) Solve for y.
{( , )} 6) Write the solution set.
Check: 2( ) - 3( ) = 5 ? 7) Check the solution in both
equations.
3( ) + ( ) = 2 ?
Worksheet 54 (10.1)
Problem
2. Solve the system by the substitution method:
÷
÷
ø
ö
ç
ç
è
æ
2
=
9y
+
2x
12
-
=
2y
-
x
Warm-up 3.a) Solve by setting up a system of equations:
Brian invested $10,000, part of it at 6% interest and the
remainder at
4%. His yearly interest income from the two investments was
$540.
How much did he invest at each rate?
It is sometimes helpful to set up a chart.
rate
amount =
interest
investment 1
6% = 0.06
x
0.06x
investment 2
4% = 0.04
y
total
--------
10,000
Fill in the last two entries in the chart. Then use the last two
columns
to write your system of equations.
÷
÷
ø
ö
ç
ç
è
æ
540
=
y
0.04
+
x
0.06
10,000
=
y
+
x
x= Solve 1st equation for x.
0.06( ) + 0.04y= 540
Substitute this expression into other
equation in place of x.
- + 0.04y= 540
Solve for y.
-0.02y= -
y=
x= 10,000 - Sub. this value for y into 1st step.
x=
Solve for x
Brian invested at 6% and at 4%.
Worksheet 54 (10.1)
Problem - Solve by setting up a system of equations:
3. Two angles are complementary and the measure of one of them
is 30 more than the
measure of the other. Find the measure of each angle.
Worksheet 55 (10.2)
10.2 Elimination-by-Addition Method
Summary 1:
Elimination-by-Addition Method
In the elimination-by-addition method we use the following
operations to produce an equivalent system which can be easily
solved.
1. Any two equations can be interchanged.
2. Both sides of an equation can be multiplied by any nonzero
real
number.
3. Any equation can be replaced by the sum of that equation and
a
nonzero multiple of another equation.
Steps for elimination-by-addition method:
1. Write each equation in standard form if needed.
2. If necessary, multiply one or both equations by some
constant
which will make the x or y coefficients opposites.
3. Add the equations from step 2 together eliminating one of
the
variables.
4. Solve for the remaining variable.
5. Substitute this solution into either of the original
equations.
6. Solve for the second variable.
7. Write the solution set and check.
Warm-up 1.a) Solve the system by the elimination-by-addition
method:
÷
÷
ø
ö
ç
ç
è
æ
5
=
3y
-
2x
7
=
3y
+
x
The equations are in standard form.
3x= If we add the 2 equations the y's will cancel.
x= Solve for x.
+ 3y= 7
Substitute x = 4 into either equation.
3y= Solve for y.
y=
{( , )} Write the solution set and check.
check: 4 + 3(1) = 7 ?
2(4) - 3(1) = 5 ?
Worksheet 55 (10.2)
b) Solve the system by the elimination-by-addition method:
÷
÷
ø
ö
ç
ç
è
æ
1
-
=
3y
+
2x
21
-
=
2y
-
3x
Equations are in standard form.
9x - 6y= -63
Multiply the first equation by 3 and the
4x + 6y= -2
second equation by 2 so the coefficients
of the y terms become -6 and 6.
13x=
Add the two new equations together to
eliminate the y terms.
x=
Solve for x.
3( ) - 2y= -21
Substitute this solution for x into either
original equation.
- 2y= -21
Solve for y.
-2y=
y=
{ ( , ) }Write the solution set and check.
check: 3(-5) - 2(3) = -21 ?
2(-5) + 3(3) = -1 ?
Summary 2:
Which Method to Use?
1. If one equation is already solved for one of the variables,
substitution would probably be the easiest method.
2. If solving for either variable in either equation would
produce fractions to substitute, use the addition method.
3. Always clear fractions in both equations before deciding
which method to use.
Problems - Solve by either method:
1.
÷
÷
ø
ö
ç
ç
è
æ
0
=
2y
+
3x
13
=
y
-
5x
Worksheet 55 (10.2)
2.
÷
÷
ø
ö
ç
ç
è
æ
15
-
=
4y
+
5x
19
=
3y
-
4x
Warm-up 2.a) Solve the system:
÷
÷
ø
ö
ç
ç
è
æ
12
=
2y
-
4x
7
+
2x
=
y
4x - 2( )= 12
Since the first equation is already solved
for y, substitute 2x + 7 into the second
equation in place of y.
4x - - = 12
Solve for x
= 12
This is a false statement, which implies
The solution set is the system has no solution and is an
inconsistent system.
Problem
3. Solve the system:
÷
÷
ø
ö
ç
ç
è
æ
18
=
2y
+
4x
9
=
y
+
2x
Note: In Problem 3 the solution set can be expressed {(x,y)2x +
y = 9}. Obtaining an equation which is always true indicates that
the two lines are the same and any point that satisfies the
equations is a solution. The system is dependent.
Worksheet 56 (10.3)
10.3 Systems of Three Linear Equations in Three Variables
Summary 1:
Linear Equations in Three Variables
Linear equations in 3 variables such as x + y + z = 5 have
solutions which are ordered triples, (x,y,z).
The graph of a linear equation in 3 variables is a plane. To
graph you would need a 3-dimensional coordinate system.
To solve a system of 3 linear equations in 3 variables, means to
find all the ordered triples that satisfy all three equations.
Similar to a system of 2 equations:
1) There can be one solution. (All 3 planes intersect at one
point.)
2) There can be many solutions. (The 3 planes intersect in more
than one
point.)
3) There can be no solution. (All 3 planes never intersect.)
We will solve systems with 3 variables and 3 equations by the
elimination- by-addition method.
Warm-up 1.a) Solve the system by the elimination-by-addition
method:
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
1
-
=
2z
+
y
8
-
=
z
+
3y
5
-
=
z
-
2y
+
x
-6y - 2z= 16
Multiply equation 2 by -2.
y + 2z= -1
Add the equation obtained in the
-5y=
previous step to equation 3.
y=
Solve for y.
3( ) + z= -8
Substitute this solution into either
equation containing only y and z.
+ z= -8
Solve for z.
z=
x + 2( ) - ( )= -5
Substitute y and z into the first equation.
x - - = -5
Solve for x.
x - = -5
x=
{ ( , , ) }
Write the solution set as an ordered triple.
Checking this solution in each equation will
be left to you.
Worksheet 56 (10.3)
b) Solve the system by the elimination-by-addition method:
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
2
-
=
2z
+
y
-
3x
29
-
=
5z
-
2y
+
x
3
-
=
z
+
3y
-
2x
6x - 2y + 4z= -4
Multiply equation 3 by 2 and add to
x + 2y - 5z= -29
equation 2, eliminating the y terms.
7x - z=
-9x + 3y - 6z= 6
Multiply equation 3 by -3 and add to
2x - 3y + z= -3
equation 1 eliminating the y terms.
-7x =
7x - z= -33
Add the equations resulting from the first
-7x - 5z= 3
two steps, eliminating the x terms.
-6z=
Solve for z.
z=
7x - ( )= -33
Substitute this value for z into either
equation containing only x and z.
7x=
Solve for x.
x=
2( ) - 3y + ( )= -3
Substitute the solutions for x and z into
any of the original equations.
- 3y + 5= -3
Solve for y.
-3y=
y=
{ ( , , ) } Write the solution set as an ordered triple.
check: 2(-4) - 3(0) + 5= -3 ?
(-4) + 2(0) -5(5)= -29 ?
3(-4) - 0 + 2(5)= -2 ?
Worksheet 56 (10.3)
Problems - Solve the systems:
1.
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
10
-
=
3z
-
y
0
=
4z
+
2y
4
=
3z
-
y
-
x
2.
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
20
-
=
z
+
2y
+
3x
5
-
=
z
-
y
-
x
16
=
3z
-
5y
+
2x
Worksheet 57 (10.4)
10.4 Matrix Approach to Solving Systems
Summary 1:
Matrices
A matrix is an array of numbers arranged in horizontal rows and
vertical columns. If a matrix has 2 rows and 3 columns it is called
a 2 x 3 (two-by-three) matrix. A matrix can have any number of rows
or columns. In general a matrix with m rows and n columns is called
a matrix of dimension m x n.
Every system of linear equations has associated with it an
augmented matrix consisting of the coefficients and constant terms
of the system.
Summary 2:
Gaussian Elimination
A system of equations can be solved using the augmented matrix
and the following Elementary Row Operations:
1. Any two rows of an augmented matrix can be interchanged.
2. Any row can be multiplied by a nonzero constant.
3. Any row of the augmented matrix can be replaced by adding a
nonzero
multiple of another row to that row.
This method is called Gaussian Elimination.
Warm-up 1.Solve the systems using the augmented matrix of the
system:
a)
÷
÷
ø
ö
ç
ç
è
æ
7
-
=
2y
-
x
0
=
y
+
3x
ú
û
ù
ê
ë
é
7
-
2
-
1
0
1
3
Write the augmented matrix.
ú
û
ù
ê
ë
é
0
1
3
7
-
2
-
1
Interchange row 1 and row 2.
ú
û
ù
ê
ë
é
21
7
0
7
-
2
-
1
Multiply row 1 by -3 and add to row 2 to
produce a new row 2.
Worksheet 57 (10.4)
÷
÷
ø
ö
ç
ç
è
æ
21
=
7y
7
-
=
2y
-
x
The matrix can be rewritten as a new system.
7y= 21 Solve the last equation for y.
y=
x - 2( )= -7 Substitute the value for y into the first
equation
of the new system.
x - = -7 Solve for x.
x=
{( , )} Write the solution set.
b)
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
7
-
=
3z
-
y
+
x
4
-
=
2z
+
y
-
4x
15
=
5z
+
3y
-
2x
-
ú
ú
ú
û
ù
ê
ê
ê
ë
é
7
-
3
-
1
1
4
-
2
1
-
4
15
5
3
-
2
-
Write the augmented matrix.
ú
ú
ú
û
ù
ê
ê
ê
ë
é
15
5
3
-
2
-
4
-
2
1
-
4
7
-
3
-
1
1
Interchange row 3 and row 1.
ú
ú
ú
û
ù
ê
ê
ê
ë
é
1
1
-
1
-
0
24
14
5
-
0
7
-
3
-
1
1
Multiply row 1 by -4 and add to row 2.
ú
ú
ú
û
ù
ê
ê
ê
ë
é
24
14
5
-
0
1
1
-
1
-
0
7
-
3
-
1
1
Interchange row 2 and row 3.
ú
ú
ú
û
ù
ê
ê
ê
ë
é
19
19
0
0
1
1
-
1
-
0
7
-
3
-
1
1
Multiply row 2 by -5 and add to row 3.
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
19
=
19z
1
=
z
-
y
-
7
-
=
3z
-
y
+
x
Rewrite as a new system.
Note: This matrix represents a system that is in triangular
form.
Worksheet 57 (10.4)
19z= 19
Solve the 3rd equation for z.
z=
-y - ( )= 1
Substitute this value for z into the 2nd
equation.
-y=
Solve for y.
y=
x + ( ) - 3( )= -7
Substitute z and y into the 1st equation.
x - 2 - = -7 Solve for x.
x - = -7
x=
{( , , )}Write the solution set.
Problems - Solve using the augmented matrix of the system.
1.
÷
÷
ø
ö
ç
ç
è
æ
19
=
5y
+
2x
18
-
=
3y
-
x
Worksheet 55 (10.4)
2.
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
1
-
=
3z
-
6y
-
x
2
-
=
z
-
3y
+
2x
9
=
2z
+
y
-
x
Worksheet 58 (10.5)
10.5 Determinants
Summary 1:
Determinants
A square matrix has the same number of rows as columns. For
example, a
2 x 2 matrix or a 3 x 3 matrix.
The determinant of a square matrix is a real number.
For the square matrix
ú
ú
û
ù
ê
ê
ë
é
b
a
b
a
2
2
1
1
the determinant is defined by:
b
a
b
a
2
2
1
1
To evaluate the determinant, multiply the element in row 1,
column 1
(a1) times the element in row 2, column 2 (b2). Subtract from
this
quantity the product of the element in row 2, column 1 (a2)
and
the element in row 1, column 2 (b1):
b
a
-
b
a
=
b
a
b
a
1
2
2
1
2
2
1
1
The vertical bars used above are the algebraic symbols
indicating to evaluate the determinant.
Warm-up 1.a) Find the determinant of the matrix:
ú
û
ù
ê
ë
é
1
-
2
3
5
-
determinant =
)
(
)
(
-
(-1)
)
(-5
=
1
-
2
3
5
-
= -
=
b) Evaluate:
)
(
)
(
-
(-6)
3
=
6
-
5
4
-
3
= -
=
Problems - Evaluate each determinant:
1.
4
-
3
2
7
-
2.
12
-
5
5
1
2
1
Worksheet 58 (10.5)
Summary 2:
Cramer's Rule
Cramer's Rule is a method of using determinants to solve a
system of 2 linear equations in 2 variables. Given the system:
0
_
b
a
-
b
a
with
c
=
y
b
+
x
a
c
=
y
b
+
x
a
2
2
1
1
2
2
2
1
1
1
÷
÷
ø
ö
ç
ç
è
æ
,
to use Cramer's Rule we need to calculate three
determinants.
D = determinant of the coefficients of x and y =
b
a
b
a
2
2
1
1
Dx = determinant formed by replacing the x coefficients with
the
constants =
b
c
b
c
2
2
1
1
Dy = determinant formed by replacing the y coefficients with
the
constants =
c
a
c
a
2
2
1
1
;
then
D
D
=
y
and
D
D
=
x
y
x
Warm-up 2.Use Cramer's Rule to solve the system:
a)
÷
÷
ø
ö
ç
ç
è
æ
8
-
=
2y
+
3x
4
-
=
6y
-
5x
)
)(
(
-
5(2)
=
2
3
6
-
5
=
D
= +
=
)
(
)
(
-
4(2)
-
=
2
8
-
6
-
4
-
=
D
x
= -
=
Worksheet 58 (10.5)
)
(
)
(
-
)
(
)
(
=
8
-
3
4
-
5
=
D
y
= +
=
=
)
(
)
(
=
D
D
=
y
=
)
(
)
(
=
D
D
=
x
y
x
Solution Set = { ( , ) }
b)
÷
÷
ø
ö
ç
ç
è
æ
1
-
=
4y
-
x
9
=
4y
-
3x
)
(
)
(
-
3(-4)
=
4
-
1
4
-
3
=
D
= +
=
)
(
)
(
-
)
(
)
(
=
4
-
4
-
=
D
x
= -
=
)
1(
-
)
3(
=
1
3
=
D
y
= -
=
=
)
(
)
(
=
D
D
=
y
=
)
(
)
(
=
D
D
=
x
y
x
Solution Set = { ( , )}
Problems - Use Cramer's Rule to solve the systems:
3.
÷
÷
ø
ö
ç
ç
è
æ
3
-
=
2y
-
3x
11
=
3y
+
2x
Worksheet 58 (10.5)
4.
÷
÷
ø
ö
ç
ç
è
æ
2
=
9y
+
2x
12
-
=
2y
-
x
Worksheet 59 (10.6)
10.6 3 x 3 Determinants and Systems of Three Linear
Equations
Summary 1:
The determinant of a 3 x 3 matrix is defined by:
b
a
c
-
a
c
b
-
c
b
a
-
b
a
c
+
a
c
b
+
c
b
a
=
c
b
a
c
b
a
c
b
a
1
2
3
1
2
3
1
2
3
3
2
1
3
2
1
3
2
1
3
3
3
2
2
2
1
1
1
An easier way to calculate the determinant of a 3 x 3 matrix is
a method called expansion of a determinant by minors.
A minor of an element in a determinant is the determinant that
remains after deleting the row and column in which the element
appears.
Warm-up 1.Use expansion by minors to evaluate the
determinant:
a)
2
-
6
1
4
-
7
5
3
1
-
2
We will expand by minors on the first column.
1) Multiply the number in row 1, column 1 (2) times the
determinant of the four
numbers remaining if you cross out the row and column containing
the 2.
2
-
6
4
-
7
2
2) From this subtract the product of the number in row 2, column
1 (5) and the
determinant of the four numbers remaining if you cross out the
row and column
containing the 5:
2
-
6
3
1
-
5
-
3) To this add the product of the number in row 3, column 1 (1)
and the determinant
of the four numbers remaining if you cross out the row and
column containing the 1.
4
-
7
3
1
-
1
+
Worksheet 59 (10.6)
2
-
6
1
4
-
7
5
3
1
-
2
=
2
-
6
4
-
7
2
2
-
6
3
1
-
5
-
EMBED Equation.3
4
-
7
3
1
-
1
+
step 1 step 2 step 3
= 2(-14 - (-24)) - 5(2 - 18) + 1( - )
= 2( ) - 5( ) + 1( )
= + +
=
Problem - Use expansion by minors to evaluate the
determinant:
1.
3
1
-
6
4
-
5
0
2
3
-
1
Summary 2:
Cramer's Rule for 3 x 3 Systems
Given the system:
÷
÷
÷
÷
ø
ö
ç
ç
ç
ç
è
æ
d
=
z
c
+
y
b
+
x
a
d
=
z
c
+
y
b
+
x
a
d
=
z
c
+
y
b
+
x
a
3
3
3
3
2
2
2
2
1
1
1
1
with
,
c
b
d
c
b
d
c
b
d
=
D
0,
_
c
b
a
c
b
a
c
b
a
=
D
3
3
3
2
2
2
1
1
1
x
3
3
3
2
2
2
1
1
1
,
d
b
a
d
b
a
d
b
a
=
D
,
c
d
a
c
d
a
c
d
a
=
D
3
3
3
2
2
2
1
1
1
z
3
3
3
2
2
2
1
1
1
y
then
.
D
D
=
z
and
,
D
D
=
y
,
D
D
=
x
z
y
x
Worksheet 59 (10.6)
Warm-up 2.Use Cramer's Rule to find the solutions of the
system:
a)
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
7
-
=
5z
+
2y
+
3x
-
17
=
4z
+
3y
-
2x
2
-
=
z
-
y
+
x
We will find the determinant D, Dx, Dy and Dz by expanding by
minors on
the first column.
4
3
-
1
-
1
(-3)
+
5
2
1
-
1
2
-
5
2
4
3
-
1
=
5
2
3
-
4
3
-
2
1
-
1
1
=
D
= 1(-15 - 8) - 2(5 - ( )) + (-3)( - )
= 1( ) - 2( ) + (-3)( )
= - -
=
4
3
-
1
-
1
(-7)
+
5
2
1
-
1
17
-
5
2
4
3
-
2
-
=
5
2
7
-
4
3
-
17
1
-
1
2
-
=
D
x
= -2(-15 - 8) - 17(5 - ( )) - 7(4 - )
= -2( ) - 17( ) - 7( )
= - -
=
)
(
)
(
)
(
)
(
(-3)
+
5
7
-
)
(
)
(
2
-
5
7
-
4
17
1
=
5
7
-
3
-
4
17
2
1
-
2
-
1
=
D
y
= 1(85 - ( )) - 2(-10 - ) - 3(-8 - ( ))
= 1( ) - 2( ) - 3( )
= + -
=
)
(
)
(
)
(
)
(
(-3)
+
)
(
)
(
2
-
1
2
-
7
-
2
17
3
-
1
=
7
-
2
3
-
17
3
-
2
2
-
1
1
=
D
z
= 1(21 - ) - 2( - ( )) - 3( - )
= 1( ) - 2( ) - 3( )
= + -
=
=
)
(
)
(
=
D
D
=
z
;
=
)
(
)
(
=
D
D
=
y
;
=
)
(
)
(
=
D
D
=
x
z
y
x
Solution Set = {( , , )}
Worksheet 59 (10.6)
Problem - Use Cramer's Rule to find the solution of the
system:
2.
÷
÷
÷
ø
ö
ç
ç
ç
è
æ
14
-
=
3z
-
2y
+
5x
10
=
4z
+
3y
-
x
0
=
z
+
y
-
2x
Worksheet 60 (10.7)
10.7 Systems Involving Nonlinear Equations and Systems of
Inequalities
Summary 1:
Systems of Nonlinear Equations
Most nonlinear systems involve two quadratic equations or a
quadratic
equation and a linear equation. The graphs of the quadratic
equations will be one of the four conic sections. Systems of this
type are solved by the substitution method or the elimination
method.
Warm-up 1.a) Graph the following system to approximate the
solutions, then
solve by substitution or elimination method.
÷
÷
ø
ö
ç
ç
è
æ
1
-
2x
=
y
4
-
x
=
y
2
y
x
The solutions should be a point in Quadrant I and a point in
Quadrant III.
By inspection, points of intersection could be (-1, -3) and (3,
5).
Solve the system by substitution.
1
-
2x
=
y
4
-
x
=
y
2
= x2 - 4
Substitute 2x - 1 into equation
1 in place of y.
0= x2 - -
Set equal to zero.
0= (x - 3)( )
Factor.
x - 3= 0 or = 0
Solve for x.
x= or x =
y = ( )2 - 4 or y = ( )2 - 4
Substitute each of these 2 values
for x into either equation.
y = y =
Solve for the corresponding y's.
Solution Set: {( , ), ( , )}
Worksheet 60 (10.7)
Problems - Graph the system to approximate the solutions then
solve the system.
1.
÷
÷
ø
ö
ç
ç
è
æ
4
-
x
=
y
4
=
y
+
x
2
2
2
y
x
Summary 2:
Systems of Linear Inequalities
To find the solution set for a system of linear inequalities,
graph each
inequality on the same coordinate system. The intersection of
the two graphs is the solution set for the system.
Warm-up 2.a) Solve the system of inequalities by graphing:
÷
÷
ø
ö
ç
ç
è
æ
³
6
3y
-
x
3
-
2x
>
y
y
1) Graph the first line, using a
dashed line, the line will be
shaded above theline.
2) Graph the second line, using
a solid line, and shade below X
the line.
3) The solution to the system is
the region which is above the lst
line and below the 2nd line.
Worksheet 60 (10.7)
Problems Solve the system of inequalities by graphing:
2.
÷
÷
ø
ö
ç
ç
è
æ
£
3
<
y
-
2x
3
y
y
x
