Download Graphing Linear Systems

Graphing Linear Systems
Goal
Estimate the solution of a system of linear equations
by graphing.
VOCABULARY
System of linear equations
Solution of a linear system
Point of intersection
Find the Point of Intersection
Example 1
Use the graph at the right to estimate
the solution of the linear system. Then
check your solution algebraically.
x
x
+ 2y = -4
- 3y = 1
t--+-+--t--t=t--t--t--+-+--i
!-+--t--+-+ 3 f-----1f--t--+--+--1
Equation 1
h« +
Equation 2
I ,..................... .-­
y -
4
--.--~
x
Solution
31Y
1
I
3
.....
.-i-"'"
1
................
x
+
2(
+ 2y
=
-4
) J: -4
-4
for y in each equation.
x - 3y
- 3(
=
1
) J: 1
1
Answer Because (
,
) is a solution of each equation,
(
) is the solution of the system of linear equations.
142
Algebra 1. Concepts and Skills Notetaklng Guide . Chapter 7
--I
t-,
I
- - --
for x and
x
t-----1"""'.=I-,I'--...... j
The lines appear to intersect once at
(
,
).
Check Substitute
3
-......
SOLVING A LINEAR SYSTEM USING GRAPH·AND-CHECK
Step 1 Write each equation in a form that is
Step 2 Graph both equations in the
------
----------
Step 3 Estimate the coordinates 9f the
-------~-
Step 4 Check whether the coordinates give a solution by
them into each equation of the
::-;-------,------­
linear system.
Graph and Check a Linear System
Example 2
A line in slope­
intercept form,
y = mx + b, has a
slope of m and a
y-intercept of b.
Use the graph-and-check method to solve the linear system.
5x
+ 4y
-12
Equation 1
3x - 4y = -20
Equation 2
=
1. Write each equation in slope-intercept form.
Equation 1
5x
+
4y
=
4y
=
Equation 2
3x - 4y = -20
-12
- 12
-4y =
y=
- 20
y=
!
2. Graph both equations.
y
5r-­
3. Estimate from the graph that the
3 I-­
point of intersection is ( __ , _).
4. Check whether ( _ , _) is a
1 f-­
for
solution by substituting
for y in each of the
original equations.
-5
-3
-1
x and
Equation 1
5(
)
x
r 1 I-­
r 3 I-­
Equation 2
5x
+
4y = -12
+
4(
) J: -12
-12
3x - 4y = -20
3(
) - 4(
) J: - 20
-20
Answer Because ( __ , _) is a solution of each equation in the
linear system, ( __ , _) is a solution of the linear system.
Lesson 7.1 . Algebra 1 Concepts and Skills Notetaklng Guide
143
o Checkpoint Use the graph-and-check method to solve the
linear system.
2.5x + 2y = 4
9x + 2y = 12
1.3x - 4y = 4
x + 2y = 8
y I
-1-7 f-f---l--f-+-+-+-+---l
-r
~--+-1--15'f ---l-+-+-+-+--+--­ 1-1- 3
t--I-­
I
y
-I-II---I-t--+-+-+-f--I--l
--t--+---t---t--t---i
I
3
,
5
7 x
I.
-1-1 -+t--+-+-+-+--+--+--1
-I
- : - 3 f-t--+--+-+-+-+-.. .+---l
7 x
5
3
4. Y = 3x + 4
7x - 3y = -6
3. y = -2x - 3
2x + 5y = 25
}'
"""I
y
2
1-7
,.......... I-­
I
1
-4
I
5 t-- I-­
-2
2
4 x
2
3 -I-­
4
-7
144
-5
-3
I 1
i-I
._­
1-­
1
1 x
Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7
6
f-­
~
-
•
So ving Li ear Systems
by Subs itution
Goal
Solve a linear system by substitution.
SOLVING A LINEAR SYSTEM BY SUBSTITUTION
Step 1 Solve one of the equations for one"of its
----
Step 2 Substitute the expression from Step 1 into the other
equation and solve for the
------
I
Step 3 Substitute the value from
from
and sOllve.
into the revised equation
Step 4 Check the solution in each of the
equations.
Substitution Method: Solve for y First
Example 1
Solve the linear system.
4x
+Y=
-5
3x - y = 5
Equation 1
Equation 2
1. Solve for y in Equation 1.
4x
+Y=
Ori'ginal Equation 1
-5
Revised Equation 1
y =
2. Substitute
for y in Equation 2 and find the
value of x.
3x - y = 5
3x - (
- - - -)
x
When you use
I the substitution
method, you can
check the solution
by substituting it for
x and for y in each
of the original
equations. You can
also use a graph to
check your solution.
3. Substitute
of y.
y=
+
Write Equation 2.
=
5
Substitute
=
5
Simplify.
for y.
x=
Subtract
x =
Divide each side by
from each side.
for x in the revised Equation 1 and find the value
=
-----
4. Check that (_, __ ) is a solution by sUbstituting _
fory in each of the original equations.
for x and
Lesson 7.2 . Algebra 1 Concepts and Skills Notetaklng Guide
\­
145
When using
substitution, you will
get the same
solution whether you
solve for y first or x
first. You should
begin by solving for
the variable that is
easier to isolate.
Example 2
Substitution Method: Solve for x First
Solve the linear system.
2x - 5y = -13
Equation 1
x + 3y = -1
Equation 2
Solution
1. Solve for x in Equation 2.
x + 3y = -1
x=
Original Equation 2
Revised 'Equation 2
2. Substitute
for x in Equation 1 and find the
value of y.
2x - 5y
=
-13
) - 5y = -13
2(
_ _ _ _ - 5y
Write Equation 1.
Substitute
for x.
=
-13
Use the distributive property.
---- =
-13
Combine like terms.
Add
to each side.
Divide each side by
3. Substitute
for y in the revised Equation 2 and find the value
of x.
x=
Write revised Equation 2.
x=
x=
Substitute
for y.
Simplify.
4. Check that (
,
) is a solution by substit,uting
and
for y in eachof the original equations.
Answer The solution is (
146
,
---
Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7
).
for x
o Explain.
Checkpoint Name the variable that you would solve for first.
1.x - 2y
x - 8y
=
=
I
0
-5
2.4x + 2y = 10
7x - y = 12
Use substitution to solve the linear system.
,
1
3. y = x - 1
x - 5y = -15
4. Y = -5x + 3
3x + 2y = -8
Lesson 7.2 . A'igebra 1; Concepts and Skills Notetaking Guide
147
•
Solving Linear Systems
by Linear ombinat·ons
Goal
Solve a system of linear equations by linear combinations.
VOCABULARY
Linear combinations
SOLVING A LINEAR SYSTEM BY LINEAR COMBINAT,IONS
Step 1 Arrange the equations with
terms in columns.
Step 2 Multiply, if necessary, the equations by numbers to obtain
coefficients that are
for one of the variables.
Step 3
the equations from Step 2. Combining like terms
with opposite coefficents will
one variable.
Solve for the
--------
Step 4 Substitute the
obtained from Step 3 into
and
---------
----------
. Step 5 Check the solution in each of the
148
A'igebra 1 Concepts and Skills Notetaking Guide . Chapter 7
_
- - - equations.
Add the Equations
Example 1
Solve the linear system.
7x
+
2y = -6
Equation 1
5x - 2y = 6
Equation 2
Solution
Add the equations to get an equation in one variable.
7x
+ 2y
Write Equation 1.
= -6
5x - 2y = 6
Write Equation 2.
Add equations.
Solve for
Substitute
7(
)
+
for
in the first equation and solve for
2y = -6
Substitute
for
Solve for
Check that (_, __ ) is a solution by substituting
for y in each of the original equations.
for x and
Answer The solution is (_, __).
o
Checkpoint Use linear combinations to solve the system of
linear equations. Then check your solution.
1.4x + Y = -4
-4x + 2y = 16
2.4x + 3y = 10
12x- 3y = 6
Lesson 7.3 . Algebra 1 Concepts and Skills Notetaking Guide
.149
Linear Systems and
Problem Solving
Goal
Use linear systems to solve real-Ufe problems.
Example 1
Choosing a Solution Method
Health Food A health food store mixes granola and raisins to malke
20 pounds of raisin granola. Granola costs $4 per pound and raisins
cost $5 per pound. How many pounds of each should be included for
the mixture to cost a total of $85?
Solution
Verbal
Model
Pounds of
granola
Price of
granola
Pounds of
granola
Labels
Algebraic
Model
Pounds of
raisins
+
Price of
raisins
+
•
Total
pounds
Pounds
of raisins
=
Total
cost
Pounds of granola = _
(pounds)
Pounds of raisins =
(pounds)
Total pounds =
(pounds)
Price of granola = _
(dollars per pound)
Price of raisins =
(dollars per pound)
Total cost ~
(dollars)
-
++
Equation 1
=
Equation 2
Because the coefficients of x and yare 1 in Equation 1,
for
and
- - - - - - is most convenient. Solve Equation
. Simplify to obtain y =
the result in Equation
----Substitute
for y in Equation 1 and solve for x.
Answer The solution is
granola.
pounds of raisins and
pounds of
Lesson 7.4 . Algebra 1 Concepts and Skills Notetaking Guide
151
Multiply then Add
Example 2
Solve the linear system.
3x - 5y = 15
Equation 1
2x + 4y = -1
Equation 2
Solution
You can get the coefficients of x to be opposites by multiplying the
first equation by
and the second equation by
3x - 5y = 15
Multiply by
x-
y=
+ 4y = -1
Multiply by
x-
y=
2x
Add the equations and solve for
Substitute - - for
2x
2x
+-
+ 4y
4(
=
in the second equation and solve for
-1
) = -1
2x -
= -1
Write Equation 2.
Substitute
for
Simplify.
Solve for
-
Answer The solution is (
--
).
o linear equations.
Use linear combinations to solve the system of
Then check your solution.
Checkpoint
3. x - 3y = 8
3x + 4y = 11
150
Algebra 1 Concepts and Skills Notetaklng Guide . Chapter 7
.. _--_.~----
4.6x
+ 5y = 23
9x - 2y = -32
Linear Systems and
Problem Solving
Goal
Use linear systems to solve real-life problems.
Choosing a Solution Method
Example 1
Health Food A health food store mixes granola and raisins to make
20 pounds of raisin granola. Granola costs $4 per pound and raisins
cost $5 per pound. How many pounds of each should be included for
the mixture to cost a total of $85?
Solution
Verbal
Model
Price of
granola
Labels
Algebraic
Model
Pounds of
granola
•
Pounds of
granola
+
Total
pounds
Pounds of
raisins
+
Price of
raisins
•
Pounds
of raisins
i
'
-
Total
cost
Pounds of granola = _
(pounds)
Pounds of raisins =
(pounds)
Total pounds =
(pounds)
Price of granola = _
(dollars per pound)
Price of raisins =
(dollars per pound)
Total cost ~
(dollars)
-
++
Equation 1
Equation 2
Because the coefficients of x and yare 1 in Equation 1,
for
and
- - - - - - is most convenient. Solve Equation
the result in Equation
. Simplify to obtain y =
----Substitute
for y in Equation 1 and solve for x.
Answer The solution is
granola.
pounds of raisins and
pounds of
Lesson 7.4 . Algebra 1 Concepts and Skills Notetaking Guide
151
WAYS TO SOLVE A SYSTEM OF LINEAR EQUATIONS
Substitution requires that one of the variables be
on
one side of the equation. It is especially convenient when one of
the variables has a coefficient of
or
Linear Combinations can be applied to any system, but it is
especially convenient when a
appears in different
equations with
that are
-----
Graphing can provide a useful method for
o
a solution.
Checkpoint Choose a method to solve the linear system.
Explain your choice, and then solve the system.
1. In Example 1, suppose the health food store wants to make
30 pounds of raisin granola that will cost a total of $125. How
many pounds of granola and raisins do they need? Use the
prices given in Example 1.
152
Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7
--_.--_._-------~-----
-­
•
Special Types of Linear Systems
al
Identify how many solutions a linear system has.
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
If the two sollutions have
- - - - slopes, then the system has
one solution.
Lines intersect:
- - - - - solution.
If the two solutions have the
slope but
y-intercepts, then the system has no solution.
_
Lines are parallel:
solution.
If the two equa.tions have the
slope and the
_
y-intercepts, then the system has infinitely many solutions.
Lines coincide:
solutions.
x
Lesson 7.5 . Algebra 1 Concepts and Skills Notetaking Guide
153
A Linear System with No Solution
Example 1
Show that the linear system has no solution.
-x + y = -3
-x + Y = 2
Equation 1
Equation 2
Solution
Method 1: Graphing Rewrite each equation in slope-intercept form.
Then graph the linear system.
y =
Revised Equation 1
y =
Revised Equation 2
y
3
I
I
[
1
-5
-3
-[
[
[
I
3 x
I
I
3
I
I
1­
I
5
Answer Because the lines have the same slope but different
y-intercepts, they are
lines do not
, so the system has
----'
-----­
Method 2: Substitution Because Equation 2 can be rewritten as
y =
, you can substitute
for y in
Equation 1.
-x + Y = -3
= -3
-x +
Write Equation 1.
Substitute
for y.
Combine like terms.
Answer The variables are
and you are left with a
statement that is
regardless of the values of x and y. This
tells you that the system has
-----
154
Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7
A Linear System with Infinitely Many Solutions
Example 2
Show that the linear system has many solutions.
3x
+y
- 6x -' 2y
=
-1
Equation 1
= 2
Equation 2
Solution
Method 1: Graphing Rewrite each equation
in slope-intercept form. Then
graph the linear system.
y =
Revised Equation 1
y
Revised Equation 2
=
I
5
IY
I
1-. -
f-­
-
3
1
I
-4
, 2
-2
I
Answer From these equations you can see
that the equations represent the same line.
point on the line is a solution.
x
1
.:- 3
I-­
-
.-
­
---
Method 2: Linear Combinations You can multiply Equation 1 by
Ox + Oy = 0
- 6x -
2y
o
=
=
2
0
Multiply Equation 1 by
.
Write Equation 2.
Add equations.
statement
Answer The var'iables are
and you are left with a
statement that is
regardless of the values of x and y. This
tells you that the system has
-----------
o
Checkpoint Solve the linear system and tell how many
solutions the system has.
1. x - 2y = 3
-5x + 10y
=
-15
2. -2x + 3y = 4
-4x + 6y = 10
Lesson 7.5 . Algebra 1 Concepts and Skills Notetaking Guide
155
Example 2
I
A Linear System with Infinitely Many Solutions
Show that the linear system has many solutions.
3x
+y
Equation 1
= -1
- 6x - 2y = 2
Equation 2
Solution
Method 1: Graphing Rewrite each equation
in slope-intercept form. Then
graph the linear system.
y =
Revised Equation 1
y
Revised Equation 2
=
y
I
5
i
I
I
3
1
-4
I
Answer From these equations you can see
that the equations represent the same line.
point on the line is a solution.
II
I
-2
2
x
1
.~ 3
---
Method 2: Linear Combinations You can multiply Equation 1 by
Ox + Oy 0
=
- 6x -
2y =
o
=
2
0
Multiply Equation 1 by
Write Equation 2.
Add equations.
statement
Answer The variables are
and you are left with a
statement that is
regardless of the values of x and y. This
tells you that the system has
-----------
o solutions theSolve
the linear system and tell how many
system has.
Checkpoint
1.x - 2y = 3
-5x + 10y = -15
+ 3y = 4
-4x + 6y = 10
2. -2x
Lesson 7.5 . Algebra 1 Concepts and Skills Notetaking Guide
155
•
Systems of Linear Inequalities
Goal
Graph a system of linear i'nequalities.
VOCABULARY
System of linear inequalities
Solution of a system of linear inequalities
GRAPHING A SYSTEM OF LINEAR INEQUALITIES
Step 1
Step 2
Step 3
156
the boundary lines of each inequality. Use a
line if
- - ­ line if the inequality is < or > and a
the inequality is :::; or ·z.
the appropriate half-plane for each inequality.
the solution of the system of inequalities as the
intersection of the half-planes from Step 2.
Algebra 1 Concepts and Skills Notetaking Guide . Chapter 7
-------------------~--------_ ..
_-_..._.. _ - - ­
1\
I
i
I
I
Example 1
Graph a System of Two Linear Inequalities
Graph the system of linear inequalities.
To check your
graph, choose a
point in the overlap
of the half-planes.
Then substitute the
coordinates into
each inequality. If
each inequality is
true, then the point
is a solution.
x
y - x ::::: -1
Inequality 1
+ 2y < 1
Inequality 2
Solution
Graph both inequalities in the same
coordinate plane. The graph of the
system is the overlap, or - - - - - of the two half-planes.
I
r~;--l
y
i
I
I3
1
1
~i
J
+1
1
-1
-3
1
T
1
I
i
I
I­
i
!I
I
xl
3
JI
3
I
I
I
Example 2
I
I
Graph a System of Three Linear Inequalities
Graph the system of linear inequalities.
y::::: - 3
Inequality 1
x<2
Inequality 2
y <x
+1
Inequality 3
Solution
,.
The graph of y ::::: -3 is the half-plane
and
the
line
y
I
i
3
I
I
1
The graph of x < 2 is the half-plane to
the
of the
line
The graph of y < x
the
---
+
1 is the half-plane
line
-3
-1
I
i
1
3
,
X
1
1_- t3
-i
i
I
-----
Finally, the graph of the system is the
, of the
, or
:-:----:---:-:
three half-planes.
Lesson 7.6 . Algebra 1 Concepts and Skills Notetaking Guide
157
Example 3
Write a System of Linear Inequalities
Write a system of inequalities that
defines the shaded region at the right.
I
-il
Y
i
3
r---
Solution
1
The graph of one inequality is the
. half-plane to the left of
-3
-1
-t- ~~
I
II
---
!
The graph of the other inequality is
_
the half-plane to the right of
The shaded region of the graph is the vertical band that lies
and
, but not
- - - - the two vertical lines,
the Hnes.
Answer The system of linear inequalities below defines the
shaded region.
Inequality 1
Inequality 2
o
Checkpoint Complete the following exercises.
1. Graph the system of linear
inequalities.
y
3
I
y<2x+2
1
2
1
y> --x - 1
-3
,-I
-1
3
1
x
+3
I
2. Write a system of linear inequalities
that defines the shaded region.
I
y
I
I
.
I
I
I
I
I
5
3
I
I
-3
-1
i
158
Algebra 1 Concepts and Skills Notetaklng Guide . Chapter 7
1
1
3
x
Words to Review
Give an example of the vocabulary word.
System of linear equations
Solution of a linear system
Point of intersection
Linear combination
System of linear inequalities
Solution of a system of Unear
inequalities
Review your notes and Chapter 7 by using the Chapter Review on
pages 431-434 of your textbook.
Words to Review . Algebra 1 Concepts and Skills Notetaking Guide
159