Transcript 1-5

Solving
Equations
with with
Solving
Equations
1-5
1-5 Variables
on Both
Sides
Variables
on Both
Sides
Warm Up
Lesson Presentation
Lesson Quiz
Holt
HoltAlgebra
McDougal
1 Algebra
Algebra11
McDougal
Solving Equations with
1-5 Variables on Both Sides
Warm Up
Simplify.
1. 4x – 10x
–6x
2. –7(x – 3) –7x + 21
3.
2x + 3
4. 15 – (x – 2) 17 – x
Solve.
5. 3x + 2 = 8 2
6.
28
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Objective
Solve equations in one variable that
contain variable terms on both sides.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Vocabulary
identity
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
To solve an equation with variables on both sides,
use inverse operations to "collect" variable terms
on one side of the equation.
Helpful Hint
Equations are often easier to solve when the
variable has a positive coefficient. Keep this in
mind when deciding on which side to "collect"
variable terms.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Example 1: Solving Equations with Variables
on Both Sides
Solve 7n – 2 = 5n + 6.
7n – 2 = 5n + 6
–5n
–5n
2n – 2 =
+2
2n
=
To collect the variable terms on one
side, subtract 5n from both sides.
6
+2
8
Since n is multiplied by 2, divide both
sides by 2 to undo the multiplication.
n=4
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 1a
Solve 4b + 2 = 3b.
4b + 2 = 3b
–3b
–3b
b+2= 0
–2 –2
b = –2
Holt McDougal Algebra 1
To collect the variable terms on one
side, subtract 3b from both sides.
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 1b
Solve 0.5 + 0.3y = 0.7y – 0.3.
0.5 + 0.3y = 0.7y – 0.3
–0.3y –0.3y
0.5
= 0.4y – 0.3
+0.3
0.8
+ 0.3
= 0.4y
2=y
Holt McDougal Algebra 1
To collect the variable terms
on one side, subtract 0.3y
from both sides.
Since 0.3 is subtracted from
0.4y, add 0.3 to both sides
to undo the subtraction.
Since y is multiplied by 0.4,
divide both sides by 0.4 to
undo the multiplication.
Solving Equations with
1-5 Variables on Both Sides
To solve more complicated
equations, you may need to first
simplify by using the Distributive
Property or combining like terms.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Example 2: Simplifying Each Side Before
Solving Equations
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
4 – 6a + 4a = –1 –5(7 – 2a) Distribute –5 to the
expression in parentheses.
4 – 6a + 4a = –1 –5(7) –5(–2a)
4 – 6a + 4a = –1 – 35 + 10a
4 – 2a = –36 + 10a
+36
+36
40 – 2a =
+ 2a
40
=
Holt McDougal Algebra 1
10a
+2a
12a
Combine like terms.
Since –36 is added to 10a,
add 36 to both sides.
To collect the variable
terms on one side, add
2a to both sides.
Solving Equations with
1-5 Variables on Both Sides
Example 2 Continued
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
40 = 12a
Since a is multiplied by 12,
divide both sides by 12.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 2a
Solve
.
1
Distribute to the expression in
2
parentheses.
3=b–1
+1
+1
4=b
Holt McDougal Algebra 1
To collect the variable terms on
one side, subtract 1 b from
2
both sides.
Since 1 is subtracted from
b, add 1 to both sides.
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 2b
Solve 3x + 15 – 9 = 2(x + 2).
3x + 15 – 9 = 2(x + 2) Distribute 2 to the expression
in parentheses.
3x + 15 – 9 = 2(x) + 2(2)
3x + 15 – 9 = 2x + 4
3x + 6 = 2x + 4
–2x
–2x
x+6=
–6
x = –2
Holt McDougal Algebra 1
4
–6
Combine like terms.
To collect the variable terms
on one side, subtract 2x
from both sides.
Since 6 is added to x, subtract
6 from both sides to undo
the addition.
Solving Equations with
1-5 Variables on Both Sides
An identity is an equation that is true
for all values of the variable. An equation
that is an identity has infinitely many
solutions.
Some equations are always false. These
equations have no solutions.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Identities and False Equations
WORDS
Identity
When solving an equation, if you get an
equation that is always true, the original
equation is an identity, and it has
infinitely many solutions.
NUMBERS
2+1=2+1
3=3
ALGEBRA
2+x=2+x
–x
–x
2=2
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Identities and False Equations
WORDS
False Equations
When solving an equation, if you get a
false equation, the original equation has
no solutions.
NUMBERS
1=1+2
1=3
ALGEBRA
x= x+3
–x –x
0=3
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Example 3A: Infinitely Many Solutions or No
Solutions
Solve 10 – 5x + 1 = 7x + 11 – 12x.
10 – 5x + 1 = 7x + 11 – 12x
10 – 5x + 1 = 7x + 11 – 12x Identify like terms.
11 – 5x = 11 – 5x Combine like terms on the left and the right.
+ 5x
+ 5x Add 5x to both sides.
11
= 11 
True statement.
The equation 10 – 5x + 1 = 7x + 11 – 12x is an
identity. All values of x will make the equation
true. All real numbers are solutions.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Example 3B: Infinitely Many Solutions or No
Solutions
Solve 12x – 3 + x = 5x – 4 + 8x.
12x – 3 + x = 5x – 4 + 8x
12x – 3 + x = 5x – 4 + 8x Identify like terms.
13x – 3 = 13x – 4 Combine like terms on the left and the right.
Subtract 13x from both sides.
–13x
–13x
–3 =
–4 False statement.
The equation 12x – 3 + x = 5x – 4 + 8x is a false
equation. There is no value of x that will make the
equation true. There are no solutions.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 3a
Solve 4y + 7 – y = 10 + 3y.
4y + 7 – y = 10 + 3y
4y + 7 – y = 10 + 3y
3y + 7 = 3y + 10
–3y
–3y
7 =
10
Identify like terms.
Combine like terms on the left and the right.
Subtract 3y from both sides.
False statement.
The equation 4y + 7 – y = 10 + 3y is a false
equation. There is no value of y that will make the
equation true. There are no solutions.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 3b
Solve 2c + 7 + c = –14 + 3c + 21.
2c + 7 + c = –14 + 3c + 21
2c + 7 + c = –14 + 3c + 21 Identify like terms.
3c + 7 = 3c + 7
Combine like terms on the left and the right.
Subtract 3c both sides.
–3c
–3c
7=
7  True statement.
The equation 2c + 7 + c = –14 + 3c + 21 is an
identity. All values of c will make the equation
true. All real numbers are solutions.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Example 4: Application
Jon and Sara are planting tulip bulbs. Jon has
planted 60 bulbs and is planting at a rate of 44
bulbs per hour. Sara has planted 96 bulbs and
is planting at a rate of 32 bulbs per hour. In
how many hours will Jon and Sara have planted
the same number of bulbs? How many bulbs
will that be?
Person
Bulbs
Jon
60 bulbs plus 44 bulbs per hour
Sara
Holt McDougal Algebra 1
96 bulbs plus 32 bulbs per hour
Solving Equations with
1-5 Variables on Both Sides
Example 4: Application Continued
Let b represent bulbs, and write expressions for
the number of bulbs planted.
When is
60
plus
bulbs
60
+
44
bulbs
each
hour
the
same
as
96
bulbs
44b
=
96
plus
+
32
bulbs
?
each
hour
32b
60 + 44b = 96 + 32b To collect the variable terms
on one side, subtract 32b
– 32b
– 32b
from both sides.
60 + 12b = 96
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Example 4: Application Continued
60 + 12b = 96
–60
– 60
12b = 36
b=3
Holt McDougal Algebra 1
Since 60 is added to 12b,
subtract 60 from both
sides.
Since b is multiplied by 12,
divide both sides by 12 to
undo the multiplication.
Solving Equations with
1-5 Variables on Both Sides
Example 4: Application Continued
After 3 hours, Jon and Sara will have planted the
same number of bulbs. To find how many bulbs they
will have planted in 3 hours, evaluate either
expression for b = 3:
60 + 44b = 60 + 44(3) = 60 + 132 = 192
96 + 32b = 96 + 32(3) = 96 + 96 = 192
After 3 hours, Jon and Sara will each have planted
192 bulbs.
Holt McDougal Algebra 1
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 4
Four times Greg's age, decreased by 3 is equal
to 3 times Greg's age increased by 7. How old is
Greg?
Let g represent Greg's age, and write expressions
for his age.
four
times
Greg's
age
4g
decreased
by
3
–
3
Holt McDougal Algebra 1
is
equal
to
=
three
times
Greg's
age
3g
increased
by
+
7 .
7
Solving Equations with
1-5 Variables on Both Sides
Check It Out! Example 4 Continued
4g – 3 = 3g + 7
–3g
–3g
g–3=
+3
g=
7
+3
10
Greg is 10 years old.
Holt McDougal Algebra 1
To collect the variable terms
on one side, subtract 3g
from both sides.
Since 3 is subtracted from g,
add 3 to both sides.
Solving Equations with
1-5 Variables on Both Sides
Lesson Quiz
Solve each equation.
1. 7x + 2 = 5x + 8 3
2. 4(2x – 5) = 5x + 4 8
3. 6 – 7(a + 1) = –3(2 – a)
4. 4(3x + 1) – 7x = 6 + 5x – 2 all real numbers
5.
1
6. A painting company charges $250 base plus $16
per hour. Another painting company charges $210
base plus $18 per hour. How long is a job for which
the two companies costs are the same? 20 hours
Holt McDougal Algebra 1