Transcript Slide 1 - Garnet Valley School District

5-3
5-3 Solving
SolvingSystems
Systemsby
byElimination
Elimination
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Holt
McDougal
Algebra 1Algebra
Algebra11
Holt
McDougal
5-3 Solving Systems by Elimination
Warm Up
Simplify each expression.
1. 3x + 2y – 5x – 2y –2x
2. 5(x – y) + 2x + 5y 7x
3. 4y + 6x – 3(y + 2x) y
4. 2y – 4x – 2(4y – 2x) –6y
Write the least common multiple.
5. 3 and 6
6
6. 4 and 10 20
7. 6 and 8
24
8. 2 and 5
Holt McDougal Algebra 1
10
5-3 Solving Systems by Elimination
Objectives
Solve systems of linear equations in
two variables by elimination.
Compare and choose an appropriate
method for solving systems of linear
equations.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Another method for solving systems of
equations is elimination. Like substitution, the
goal of elimination is to get one equation that
has only one variable.
Remember that an equation stays balanced
if you add equal amounts to both sides.
Consider the system
. Since
5x + 2y = 1, you can add 5x + 2y to one
side of the first equation and 1 to the other
side and the balance is maintained.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Since –2y and 2y have opposite coefficients, you can
eliminate the y by adding the two equations. The
result is one equation that has only one variable:
6x = –18.
When you use the elimination method to solve a
system of linear equations, align all like terms in the
equations. Then determine whether any like terms
can be eliminated because they have opposite
coefficients.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Solving Systems of Equations by
Elimination
Step 1
Write the system so that like
terms are aligned.
Step 2
Eliminate one of the variables and
solve for the other variable.
Step 3
Substitute the value of the variable
into one of the original equations
and solve for the other variable.
Step 4
Write the answers from Steps 2 and 3
as an ordered pair, (x, y), and check.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Later in this lesson you will learn
how to multiply one or more
equations by a number in order to
produce opposites that can be
eliminated.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 1: Elimination Using Addition
Solve
3x – 4y = 10
by elimination.
x + 4y = –2
Step 1
Step 2
3x – 4y = 10
x + 4y = –2
4x + 0 = 8
4x = 8
4x = 8
4
4
x=2
Holt McDougal Algebra 1
Align like terms. −4y and
+4y are opposites.
Add the equations to
eliminate y.
Simplify and solve for x.
Divide both sides by 4.
5-3 Solving Systems by Elimination
Example 1 Continued
Step 3 x + 4y = –2
2 + 4y = –2
–2
–2
4y = –4
4y
–4
4
4
y = –1
Step 4 (2, –1)
Holt McDougal Algebra 1
Write one of the original
equations.
Substitute 2 for x.
Subtract 2 from both sides.
Divide both sides by 4.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
Check It Out! Example 1
Solve
y + 3x = –2
by elimination.
2y – 3x = 14
y + 3x = –2
2y – 3x = 14
Step 2 3y + 0 = 12
3y = 12
Step 1
Align like terms. 3x and
−3x are opposites.
Add the equations to
eliminate x.
Simplify and solve for y.
Divide both sides by 3.
y=4
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 1 Continued
Step 3 y + 3x = –2
4 + 3x = –2
–4
–4
3x = –6
3x = –6
3
3
x = –2
Step 4 (–2, 4)
Holt McDougal Algebra 1
Write one of the original
equations.
Substitute 4 for y.
Subtract 4 from both sides.
Divide both sides by 3.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
When two equations each contain
the same term, you can subtract
one equation from the other to
solve the system. To subtract an
equation, add the opposite of each
term.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 2: Elimination Using Subtraction
Solve
2x + y = –5
by elimination.
2x – 5y = 13
Step 1
2x + y = –5
–(2x – 5y = 13)
Step 2
2x + y = –5
–2x + 5y = –13
0 + 6y = –18
6y = –18
y = –3
Holt McDougal Algebra 1
Both equations contain
2x. Add the opposite
of each term in the
second equation.
Eliminate x.
Simplify and solve for y.
5-3 Solving Systems by Elimination
Example 2 Continued
Step 3 2x + y = –5
2x + (–3) = –5
2x – 3 = –5
+3
+3
Write one of the original
equations.
Substitute –3 for y.
2x
Simplify and solve for x.
= –2
x = –1
Step 4 (–1, –3)
Holt McDougal Algebra 1
Add 3 to both sides.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
Remember!
Remember to check by substituting your answer
into both original equations.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 2
Solve
Step 1
Step 2
3x + 3y = 15
by elimination.
–2x + 3y = –5
3x + 3y = 15
–(–2x + 3y = –5)
3x + 3y = 15
+ 2x – 3y = +5
5x + 0 = 20
5x = 20
x=4
Holt McDougal Algebra 1
Both equations contain
3y. Add the opposite
of each term in the
second equation.
Eliminate y.
Simplify and solve for x.
5-3 Solving Systems by Elimination
Check It Out! Example 2 Continued
Step 3
3x + 3y = 15
3(4) + 3y = 15
12 + 3y = 15
–12
–12
3y = 3
y=1
Step 4
Holt McDougal Algebra 1
(4, 1)
Write one of the original
equations.
Substitute 4 for x.
Subtract 12 from both sides.
Simplify and solve for y.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
In some cases, you will first need to
multiply one or both of the equations by
a number so that one variable has
opposite coefficients.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 3A: Elimination Using Multiplication First
Solve the system by elimination.
x + 2y = 11
–3x + y = –5
Step 1
Step 2
x + 2y = 11
–2(–3x + y = –5)
x + 2y = 11
+(6x –2y = +10)
7x + 0 = 21
7x = 21
x=3
Holt McDougal Algebra 1
Multiply each term in the
second equation by –2 to
get opposite y-coefficients.
Add the new equation to
the first equation to
eliminate y.
Solve for x.
5-3 Solving Systems by Elimination
Example 3A Continued
Step 3 x + 2y = 11
3 + 2y = 11
–3
–3
2y = 8
y=4
Step 4
Holt McDougal Algebra 1
(3, 4)
Write one of the original
equations.
Substitute 3 for x.
Subtract 3 from both sides.
Solve for y.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
Example 3B: Elimination Using Multiplication First
Solve the system by elimination.
–5x + 2y = 32
2x + 3y = 10
Step 1
2(–5x + 2y = 32)
5(2x + 3y = 10)
–10x + 4y = 64
+(10x + 15y = 50)
Step 2
Holt McDougal Algebra 1
19y = 114
y=6
Multiply the first equation
by 2 and the second
equation by 5 to get
opposite x-coefficients
Add the new equations to
eliminate x.
Solve for y.
5-3 Solving Systems by Elimination
Example 3B Continued
Step 3
2x + 3y = 10
2x + 3(6) = 10
2x + 18 = 10
–18 –18
Step 4
Holt McDougal Algebra 1
2x = –8
x = –4
(–4, 6)
Write one of the original
equations.
Substitute 6 for y.
Subtract 18 from both sides.
Solve for x.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
Check It Out! Example 3a
Solve the system by elimination.
3x + 2y = 6
–x + y = –2
Step 1
3x + 2y = 6
3(–x + y = –2)
3x + 2y = 6
+(–3x + 3y = –6)
0
Step 2
Holt McDougal Algebra 1
+ 5y = 0
5y = 0
y=0
Multiply each term in the
second equation by 3 to get
opposite x-coefficients.
Add the new equation to
the first equation.
Simplify and solve for y.
5-3 Solving Systems by Elimination
Check It Out! Example 3a Continued
Step 3
–x + y = –2
–x + 3(0) = –2
–x + 0 = –2
–x = –2
x=2
Step 4
(2, 0)
Holt McDougal Algebra 1
Write one of the original
equations.
Substitute 0 for y.
Solve for x.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
Check It Out! Example 3b
Solve the system by elimination.
2x + 5y = 26
–3x – 4y = –25
Step 1
Step 2
3(2x + 5y = 26)
+(2)(–3x – 4y = –25)
Multiply the first equation
by 3 and the second
equation by 2 to get
opposite x-coefficients
6x + 15y = 78
+(–6x – 8y = –50) Add the new equations to
eliminate x.
0 + 7y = 28
Solve for y.
y =4
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 3b Continued
Step 3
2x + 5y = 26
2x + 5(4) = 26
2x + 20 = 26
–20 –20
2X
= 6
x=3
Step 4
(3, 4)
Holt McDougal Algebra 1
Write one of the original
equations.
Substitute 4 for y.
Subtract 20 from both
sides.
Solve for x.
Write the solution as an
ordered pair.
5-3 Solving Systems by Elimination
Example 4: Application
Paige has $7.75 to buy 12 sheets of felt and
card stock for her scrapbook. The felt costs
$0.50 per sheet, and the card stock costs
$0.75 per sheet. How many sheets of each
can Paige buy?
Write a system. Use f for the number of felt
sheets and c for the number of card stock sheets.
0.50f + 0.75c = 7.75
f + c = 12
Holt McDougal Algebra 1
The cost of felt and card
stock totals $7.75.
The total number of sheets
is 12.
5-3 Solving Systems by Elimination
Example 4 Continued
Step 1
0.50f + 0.75c = 7.75 Multiply the second
equation by –0.50 to get
+ (–0.50)(f + c) = 12
opposite f-coefficients.
0.50f + 0.75c = 7.75
Add this equation to the
+ (–0.50f – 0.50c = –6)
first equation to
0.25c = 1.75
Step 2
eliminate f.
Solve for c.
c=7
Step 3
f + c = 12
f + 7 = 12
–7 –7
f
= 5
Holt McDougal Algebra 1
Write one of the original
equations.
Substitute 7 for c.
Subtract 7 from both sides.
5-3 Solving Systems by Elimination
Example 4 Continued
Step 4
(7, 5)
Write the solution as an
ordered pair.
Paige can buy 7 sheets of card stock and 5
sheets of felt.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 4
What if…? Sally spent $14.85 to buy 13
flowers. She bought lilies, which cost $1.25
each, and tulips, which cost $0.90 each. How
many of each flower did Sally buy?
Write a system. Use l for the number of lilies
and t for the number of tulips.
1.25l + 0.90t = 14.85
l + t = 13
Holt McDougal Algebra 1
The cost of lilies and tulips
totals $14.85.
The total number of flowers
is 13.
5-3 Solving Systems by Elimination
Check It Out! Example 4 Continued
Step 1
1.25l + .90t = 14.85
+ (–.90)(l + t) = 13
Multiply the second
equation by –0.90 to get
opposite t-coefficients.
1.25l + 0.90t = 14.85
+ (–0.90l – 0.90t = –11.70) Add this equation to the
0.35l = 3.15
first equation to
eliminate t.
Step 2
Solve for l.
l=9
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 4 Continued
Step 3
Step 4
l + t = 13
9 + t = 13
–9
–9
t = 4
(9, 4)
Write one of the original
equations.
Substitute 9 for l.
Subtract 9 from both
sides.
Write the solution as
an ordered pair.
Sally bought 9 lilies and 4 tulips.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
All systems can be solved in more than
one way. For some systems, some
methods may be better than others.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Lesson Quiz
Solve each system by elimination.
1.
2x + y = 25
3y = 2x – 13
2.
–3x + 4y = –18
x = –2y – 4
(2, –3)
3.
–2x + 3y = –15
3x + 2y = –23
(–3, –7)
(11, 3)
4. Harlan has $44 to buy 7 pairs of socks. Athletic
socks cost $5 per pair. Dress socks cost $8 per
pair. How many pairs of each can Harlan buy?
4 pairs of athletic socks and 3 pairs of dress socks
Holt McDougal Algebra 1
5-3
Elimination
5-4 Solving
SolvingSystems
Special by
Systems
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Holt
McDougal
Algebra 1Algebra
Algebra11
Holt
McDougal
5-3 Solving Systems by Elimination
Warm Up
Solve each equation.
1. 2x + 3 = 2x + 4
no solution
2. 2(x + 1) = 2x + 2 infinitely many solutions
3. Solve 2y – 6x = 10 for y
y =3x + 5
Solve by using any method.
4.
y = 3x + 2 (1, 5)
2x + y = 7
Holt McDougal Algebra 1
5.
x – y = 8 (6, –2)
x+y=4
5-3 Solving Systems by Elimination
Objectives
Solve special systems of linear
equations in two variables.
Classify systems of linear equations and
determine the number of solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Vocabulary
inconsistent system
consistent system
independent system
dependent system
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
In Lesson 6-1, you saw that when two lines intersect
at a point, there is exactly one solution to the
system. Systems with at least one solution are called
consistent.
When the two lines in a system do not intersect
they are parallel lines. There are no ordered pairs
that satisfy both equations, so there is no solution.
A system that has no solution is an inconsistent
system.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 1: Systems with No Solution
Show that
y=x–4
has no solution.
–x + y = 3
Method 1 Compare slopes and y-intercepts.
y=x–4
–x + y = 3
y = 1x – 4 Write both equations in slopeintercept form.
y = 1x + 3
The lines are parallel because
they have the same slope and
different y-intercepts.
This system has no solution.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 1 Continued
Show that
y=x–4
has no solution.
–x + y = 3
Method 2 Solve the system algebraically. Use the
substitution method because the first
equation is solved for y.
Substitute x – 4 for y in the
second equation, and solve.
–4 = 3  False.
–x + (x – 4) = 3
This system has no solution.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 1 Continued
Show that
y=x–4
has no solution.
–x + y = 3
Check Graph the system.
–x+y=3
The lines appear are
parallel.
y=x– 4
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 1
y = –2x + 5
Show that
has no solution.
2x + y = 1
Method 1 Compare slopes and y-intercepts.
y = –2x + 5
2x + y = 1
y = –2x + 5
y = –2x + 1
Write both equations in
slope-intercept form.
The lines are parallel because
they have the same slope
and different y-intercepts.
This system has no solution.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 1 Continued
Show that
y = –2x + 5
has no solution.
2x + y = 1
Method 2 Solve the system algebraically. Use the
substitution method because the first
equation is solved for y.
2x + (–2x + 5) = 1
5 = 1
Substitute –2x + 5 for y in the
second equation, and solve.
False.
This system has no solution.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 1 Continued
Show that
y = –2x + 5
has no solution.
2x + y = 1
Check Graph the system.
y = –2x + 5
y = – 2x + 1
The lines are parallel.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
If two linear equations in a system
have the same graph, the graphs are
coincident lines, or the same line.
There are infinitely many solutions of
the system because every point on the
line represents a solution of both
equations.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 2A: Systems with Infinitely Many Solutions
Show that
y = 3x + 2
has infinitely
3x – y + 2= 0 many solutions.
Method 1 Compare slopes and y-intercepts.
y = 3x + 2
3x – y + 2= 0
y = 3x + 2 Write both equations in slopeintercept form. The lines
y = 3x + 2
have the same slope and
the same y-intercept.
If this system were graphed, the graphs
would be the same line. There are infinitely
many solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 2A Continued
Show that
y = 3x + 2
has infinitely
3x – y + 2= 0 many solutions.
Method 2 Solve the system algebraically. Use
the elimination method.
y = 3x + 2
3x − y + 2= 0
y − 3x =
2
−y + 3x = −2
0 = 0
Write equations to line up
like terms.
Add the equations.
True. The equation is an
identity.
There are infinitely many solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Caution!
0 = 0 is a true statement. It does not mean the
system has zero solutions or no solution.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 2
Show that
Method 1
y=x–3
has infinitely
x – y – 3 = 0 many solutions.
Compare slopes and y-intercepts.
y=x–3
y = 1x – 3
x–y–3=0
y = 1x – 3
Write both equations in slopeintercept form. The lines
have the same slope and
the same y-intercept.
If this system were graphed, the graphs
would be the same line. There are infinitely
many solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 2 Continued
Show that
y=x–3
has infinitely
x – y – 3 = 0 many solutions.
Method 2 Solve the system algebraically. Use
the elimination method.
y=x–3
x–y–3=0
y= x–3
–y = –x + 3
0 = 0
Write equations to line up
like terms.
Add the equations.
True. The equation is an
identity.
There are infinitely many solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Consistent systems can either be independent
or dependent.
An independent system has exactly one
solution. The graph of an independent system
consists of two intersecting lines.
A dependent system has infinitely many
solutions. The graph of a dependent system
consists of two coincident lines.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 3A: Classifying Systems of Linear Equations
Classify the system. Give the number of solutions.
Solve
3y = x + 3
3y = x + 3
x+y=1
x+y=1
y=
x+1
Write both equations in
slope-intercept form.
y=
x+1
The lines have the same slope
and the same y-intercepts.
They are the same.
The system is consistent and dependent. It has
infinitely many solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 3B: Classifying Systems of Linear equations
Classify the system. Give the number of solutions.
Solve
x+y=5
4 + y = –x
x+y=5
y = –1x + 5
4 + y = –x
y = –1x – 4
Write both equations in
slope-intercept form.
The lines have the same
slope and different yintercepts. They are
parallel.
The system is inconsistent. It has no solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 3C: Classifying Systems of Linear equations
Classify the system. Give the number of solutions.
Solve
y = 4(x + 1)
y–3=x
y = 4(x + 1)
y–3=x
y = 4x + 4
y = 1x + 3
Write both equations in
slope-intercept form.
The lines have different
slopes. They intersect.
The system is consistent and independent. It has
one solution.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 3a
Classify the system. Give the number of solutions.
Solve
x + 2y = –4
–2(y + 2) = x
x + 2y = –4
y=
x–2
–2(y + 2) = x
y=
x–2
Write both equations in
slope-intercept form.
The lines have the same
slope and the same yintercepts. They are the
same.
The system is consistent and dependent. It has
infinitely many solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 3b
Classify the system. Give the number of solutions.
Solve
y = –2(x – 1)
y = –x + 3
y = –2(x – 1)
y = –2x + 2
y = –x + 3
y = –1x + 3
Write both equations in
slope-intercept form.
The lines have different
slopes. They intersect.
The system is consistent and independent. It has
one solution.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 3c
Classify the system. Give the number of solutions.
2x – 3y = 6
Solve
y=
x
2x – 3y = 6
y=
x–2
Write both equations in
slope-intercept form.
y=
y=
x
The lines have the same
slope and different yintercepts. They are
parallel.
x
The system is inconsistent. It has no solutions.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 4: Application
Jared and David both started a savings account
in January. If the pattern of savings in the
table continues, when will the amount in
Jared’s account equal the amount in David’s
account?
Use the table to write a system of linear
equations. Let y represent the savings total
and x represent the number of months.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Example 4 Continued
Jared
David
Total
saved
is
y
=
$25
+
$5
x
y
=
y = 5x + 25
y = 5x + 40
$40
+
$5
x
y = 5x + 25
y = 5x + 40
start
amount plus
amount
saved
for each
month.
Both equations are in the slopeintercept form.
The lines have the same slope
but different y-intercepts.
The graphs of the two equations are parallel lines, so
there is no solution. If the patterns continue, the
amount in Jared’s account will never be equal to the
amount in David’s account.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Check It Out! Example 4
Matt has $100 in a checking account and deposits
$20 per month. Ben has $80 in a checking account
and deposits $30 per month. Will the accounts
ever have the same balance? Explain.
Write a system of linear equations. Let y represent the
account total and x represent the number of months.
y = 20x + 100
y = 30x + 80
Both equations are in slope-intercept
form.
y = 20x + 100 The lines have different slopes..
y = 30x + 80
The accounts will have the same balance. The graphs
of the two equations have different slopes so they
intersect.
Holt McDougal Algebra 1
5-3 Solving Systems by Elimination
Lesson Quiz: Part I
Solve and classify each system.
1.
y = 5x – 1
5x – y – 1 = 0
infinitely many solutions;
consistent, dependent
2.
y=4+x
–x + y = 1
no solution; inconsistent
3.
y = 3(x + 1)
y=x–2
Holt McDougal Algebra 1
consistent,
independent
5-3 Solving Systems by Elimination
Lesson Quiz: Part II
4. If the pattern in the table continues, when will
the sales for Hats Off equal sales for Tops?
never
Holt McDougal Algebra 1