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Holt Algebra 1 2-4 Solving Equations with Variables on Both Sides 2-4 Solving Equations with Variables on Both Sides Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides 2-4 Solving Equations with
Variables on Both Sides
Holt Algebra 1
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Warm Up
Simplify.
1. 4x – 10x
2. –7(x – 3)
3.
4. 15 – (x – 2)
Solve.
5. 3x + 2 = 8
6.
–6x
–7x + 21
17 – x
2x + 3
2
28
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve equations in one variable that contain variable terms on both sides.
Objective
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Vocabulary
identity
contradiction
Holt Algebra 1
2-4 Solving Equations with
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 Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 7n – 2 = 5n + 6.
Example 1: Solving Equations with Variables
on Both Sides
To collect the variable terms on one
side, subtract 5n from both sides.
7n – 2 = 5n + 6
–5n –5n
2n – 2 = 6
Since n is multiplied by 2, divide both
sides by 2 to undo the multiplication.
2n = 8
+ 2 + 2
n = 4
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 4b + 2 = 3b.
Check It Out! Example 1a
To collect the variable terms on one
side, subtract 3b from both sides.
4b + 2 = 3b
–3b –3b
b + 2 = 0
b = –2
– 2 – 2
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 0.5 + 0.3y = 0.7y – 0.3.
Check It Out! Example 1b
To collect the variable terms
on one side, subtract 0.3y
from both sides.
0.5 + 0.3y = 0.7y – 0.3
–0.3y –0.3y
0.5 = 0.4y – 0.3
0.8 = 0.4y
+0.3 + 0.3
2 = y
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.
Holt Algebra 1
2-4 Solving Equations with
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 Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
Example 2: Simplifying Each Side Before
Solving Equations
Combine like terms.
Distribute –5 to the
expression in parentheses. 4 – 6a + 4a = –1 –5(7 – 2a)
4 – 6a + 4a = –1 –5(7) –5(–2a)
4 – 6a + 4a = –1 – 35 + 10a
4 – 2a = –36 + 10a
+36 +36
40 – 2a = 10a
+ 2a +2a
40 = 12a
Since –36 is added to 10a,
add 36 to both sides.
To collect the variable
terms on one side, add
2a to both sides.
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
Example 2 Continued
40 = 12a
Since a is multiplied by 12,
divide both sides by 12.
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve .
Check It Out! Example 2A
Since 1 is subtracted from
b, add 1 to both sides.
Distribute to the expression in
parentheses.
1 2
+ 1 + 1
3 = b – 1
To collect the variable terms on
one side, subtract b from
both sides.
1 2
4 = b
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 3x + 15 – 9 = 2(x + 2).
Check It Out! Example 2B
Combine like terms.
Distribute 2 to the expression
in parentheses. 3x + 15 – 9 = 2(x + 2)
3x + 15 – 9 = 2(x) + 2(2)
3x + 15 – 9 = 2x + 4
3x + 6 = 2x + 4
–2x –2x
x + 6 = 4
– 6 – 6
x = –2
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.
Holt Algebra 1
2-4 Solving Equations with
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.
A contradiction is an equation that is not true for any value of the variable. It has no solutions.
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
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
Identities and Contradictions
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Contradiction
When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions.
WORDS
x = x + 3
–x –x
0 = 3
1 = 1 + 2
1 = 3
ALGEBRA
NUMBERS
Identities and Contradictions
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 10 – 5x + 1 = 7x + 11 – 12x.
Example 3A: Infinitely Many Solutions or No
Solutions
Add 5x to both sides.
Identify like terms. 10 – 5x + 1 = 7x + 11 – 12x
11 – 5x = 11 – 5x
11 = 11
+ 5x + 5x
True statement.
Combine like terms on the left and the right.
10 – 5x + 1 = 7x + 11 – 12x
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 Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 12x – 3 + x = 5x – 4 + 8x.
Example 3B: Infinitely Many Solutions or No
Solutions
Subtract 13x from both sides.
Identify like terms. 12x – 3 + x = 5x – 4 + 8x
13x – 3 = 13x – 4
–3 = –4
–13x –13x
False statement.
Combine like terms on the left and the right.
12x – 3 + x = 5x – 4 + 8x
The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 4y + 7 – y = 10 + 3y.
Check It Out! Example 3a
Subtract 3y from both sides.
Identify like terms. 4y + 7 – y = 10 + 3y
3y + 7 = 3y + 10
7 = 10
–3y –3y
False statement.
Combine like terms on the left and the right.
4y + 7 – y = 10 + 3y
The equation 4y + 7 – y = 10 + 3y is a contradiction. There is no value of y that will make the equation true. There are no solutions.
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Solve 2c + 7 + c = –14 + 3c + 21.
Check It Out! Example 3b
Subtract 3c both sides.
Identify like terms. 2c + 7 + c = –14 + 3c + 21
3c + 7 = 3c + 7
7 = 7
–3c –3c
True statement.
Combine like terms on the left and the right.
2c + 7 + c = –14 + 3c + 21
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 Algebra 1
2-4 Solving Equations with
Variables on Both Sides
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?
Example 4: Application
Person Bulbs
Jon 60 bulbs plus 44 bulbs per hour
Sara 96 bulbs plus 32 bulbs per hour
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Example 4: Application Continued
Let b represent bulbs, and write expressions for the number of bulbs planted.
60 bulbs
plus
44 bulbs each hour
the same
as
96 bulbs plus
32 bulbs each hour
When is ?
60 + 44b = 96 + 32b
60 + 44b = 96 + 32b
– 32b – 32b
To collect the variable terms
on one side, subtract 32b
from both sides.
60 + 12b = 96
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Example 4: Application Continued
Since 60 is added to 12b,
subtract 60 from both
sides.
60 + 12b = 96
–60 – 60
12b = 36 Since b is multiplied by 12,
divide both sides by 12 to
undo the multiplication.
b = 3
Holt Algebra 1
2-4 Solving Equations with
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 Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg?
Check It Out! Example 4
Let g represent Greg's age, and write expressions for his age.
four times Greg's age
decreased by
3
is equal
to
three times Greg's age
increased by
7 .
4g – 3 = 3g + 7
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Check It Out! Example 4 Continued
4g – 3 = 3g + 7 To collect the variable terms
on one side, subtract 3g
from both sides.
g – 3 = 7
–3g –3g
Since 3 is subtracted from g,
add 3 to both sides. + 3 + 3
g = 10
Greg is 10 years old.
Holt Algebra 1
2-4 Solving Equations with
Variables on Both Sides
Lesson Quiz
Solve each equation.
1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4
3. 6 – 7(a + 1) = –3(2 – a)
4. 4(3x + 1) – 7x = 6 + 5x – 2
5.
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?
3 8
all real numbers
1
20 hours
