Download 3.1 Solving Equations Using Addition and Subtraction

5-5B Linear Systems and
Problems Solving
Algebra 1
Glencoe McGraw-Hill
Linda Stamper
Ways to Solve a System of Linear Equations
Graphing – can provide a useful method for estimating a solution and to
provide a visual model of the problem.
Substitution – requires that one of the variables be isolated on one
side of the equation. It is especially convenient when one of the
variables has a coefficient of 1 or –1.
Elimination Using Addition –convenient when a variable appears in
different equations with coefficients that are opposites.
Elimination Using Subtraction –convenient if one of the variables has
the same coefficient in the two equations.
Elimination Using Multiplication –can be applied to create opposites in
any system.
Solving Word Problems Using A Linear System
1) Write two sets of labels, if necessary (one set for number,
one set for value, weight etc.)
2) Write two verbal models. (Given in problem.)
3) Write two algebraic models - equations. (Translate from sentences.)
4) Solve the linear system.
5) Write a sentence and check your solution in the word problem.
Meg’s age is 5 times Jose’s age. The sum of their ages is 18. How
=
old is each person?
Assign Labels. Choose a different variable for each person.
Let m = Meg’s age
Let j = Jose’s age
Write an equation for each of the first two sentences.
m = 5j
5j
m + j = 18
Solve the system of equations.
How old is Meg? m  5 j
 53
Sentence.
 15
(
)  j  18
6j  18
j3
Jose is 3 and Meg is 15.
The length of a rectangle is 1 m more than twice its width. If the
=
perimeter is 110 m, find the dimensions.
let l = length
let w = width
length
width
width
length
Formula
2l  2w 
l  2w  1
 218  1
 36  1
 37
2(
)  2w  110
4w  2  2w  110
The width is 18 m and the
length is 37 m.
6w  2  110
6w  108
w  18
Example 1 A class has a total of 25 students. Twice the number of
girls is equal to 3 times the number of boys. How many boys and girls
are there in the class?
Assign Labels. Choose a different variable for each type of person.
Let g = # of girls
Let b = # of boys
Write an equation for each of the first two sentences.
g + b = 25
g   b  25
2g = 3b
  3b
2
 2b  50  3b
50  5b
10  b
g  b  25
g  10  25
g  15
There are 15 girls and 10 boys in the class.
Example 2 The length of a rectangle is 4 m more than twice its width.
=
If the perimeter is 38 m, find the dimensions.
1. Labels.
let w = width
2. Translate first sentence.
let l = length
length
width
width
length
l  2w  4
3. Use perimeter formula. 2l  2w 
4. Solve the system.
w
l  22w
w444
and
2(
5. Sentence.
 25   4
 10  4
 14
The width is 5 m and the length is 14 m.
2 l  2 w  38
)  2w  38
4w  8  2w
6w  8
6w
w
 38
 38
 30
5
Example 3 Admission to the play was $2 for an adult and $1.50 for a
student. Total income from the sale of tickets was $550. The number
of adult tickets sold was 100 less than 3 times the number of student
tickets. How many tickets of each type were sold?
Number Labels. let a = # of adult tickets let s = # of student tickets
value of
value of
Value Labels.
let 2a =
let
1.50s
=
adult tickets
student tickets
Example 3 Admission to the play was $2 for an adult and $1.50 for a
student. Total income from the sale of tickets was $550. The number
of adult tickets sold was 100 less than 3 times the number of student
= of each type were sold?
tickets. How many tickets
Number Labels. let a = # of adult tickets let s = # of student tickets
value of
value of
Value Labels.
let 2a =
let
1.50s
=
adult tickets
student tickets
a =3
3s
– 100
s
2a + 1.50s = 550
200a  150s  55000
  150s  55000
200
600 s  20,000  150 s  55,000
Clear the decimals.
Multiply both sides by 100.
a  3100   100
750 s  20,000  55,000
 300  100
750 s  75,000
 200
s  100
The school sold 200 adult tickets and 100 student tickets.
Example 4 The number of quarters that Tom has is 3 times the number
of nickels. He has $1.60 in all. How many coins of each type does he
have?
let n = # of nickels
Number Labels. let q = # of quarters
Value Labels.
let .25q = value of quarters let .05n = value of nickels
Example 4 The number of quarters that Tom has is 3 times the number
=
of nickels. He has $1.60 in all. How many coins of each type does he
have?
let n = # of nickels
Number Labels. let q = # of quarters
Value Labels.
let .25q = value of quarters let .05n = value of nickels
q = 3n
3n
q  32
6
.25q + .05n = 1.60
25q  5n  160
25   5n  160
75n  5n  160
80n  160
n2
Clear the decimals.
Multiply both sides by 100.
Tom has 6 quarters and 2 nickels.
Example 5 The sum of two numbers is 100. Five times the smaller
number is 8 more than the larger number. What are the two numbers?
Assign Labels.
Equations.
Let s = smaller #
s + l = 100
Let l = larger #
5s = l + 8
 l 8
5
 5l  500  l  8
500  6l  8
492  6l
82  l
The larger number is 82 and the smaller number is 18.
s  l
 l  100
Example 6 One number is 12 more than half another number. The two
numbers have a sum of 60. Find the numbers.
Assign Labels.
Let x = first #
Equations.
1
x  y  12
2
Let y = second #
x  y  60
 1 y  12   y  60


2

1
1 y  12  60
2  12  12
 2  3 y  48  2 
 
 
3 2
3
y  32
One number is 28 and the other number is 32.
Example 7 If you buy six pens and one mechanical pencil, you’ll get $1
change from your $10 bill. But if you buy four pens and two mechanical
pencils, you’ll get $2 change. How much does each pen and pencil cost?
Assign
Labels. Let p = cost of a pen Let m = cost of a mechanical pencil
Equations.
6p + m = 10 - 1
m  6p  9
4p + 2m = 10 - 2
4p  2 
8
4p  12p  18  8
61.25  m  9
7.50  m  9
 7.50
 7.50
m  1.50
Pens cost $1.25 each and mechanical
pencils cost $1.50 each.
 8p  18  8
 18  18
 8p  10
8
8
5
p
4
p  1.25
5-A8 Handout A8.
WHEN THE GOING GETS TOUGH –
THE TOUGH GET GOING!
ask questions
pay attention
think, think, think
work with a buddy
pledge to do your
homework every day
do it and
CORRECT IT