Download Algebra 1 Notes Equations and Inequalities A.REI.A.1 Understand

Algebra 1 Notes
Equations and Inequalities
A.REI.A.1 Understand solving equations as a process of reasoning and explain the
reasoning.
Explain each step in solving a simple equation as following from the
equality of numbers asserted at the previous from the equality of numbers asserted at the
previous step, starting from the assumption that the original equation has a solution.
Construct a viable argument to justify a solution method.
A.REI.A.3 Understand solving equations as a process of reasoning and explain the
reasoning. Solve linear equations and inequalities in one variable, including equations
with coefficients represented by letters.
BIG IDEA: Algebraic properties enable students to arrive at a solution of an equation in a
logical step-by-step manner. Knowing the properties of equality can not only help you reduce
complex equations to simpler ones, but also give you the tools to justify your solution.
Solving linear equations and inequalities in one variable is a basic algebraic skill. Mastering this
skill will enable the students to solve higher-order equations, such as quadratic equations and
more advanced equations in future courses.
These standards will help students work the literal equations and inequalities, which are related
to formulas. After practicing rewriting formulas they will be able to focus on a given quantity.
Note: It is helpful to have the students always check their answers for accuracy and
reasonableness.
PREREQUISITE SKILLS:
 students must have a clear understanding of sign numbers and their operations
 students must understand meaning of operations and how they relate to one another
 students must be able to make reasonable estimates
 students must know how to write expressions and equations and simplify like terms
 students must have a clear understanding of the definition of absolute value
VOCABULARY:
 equivalent equation: two equations that have the same solution set
 equation: a mathematical statement that two expressions are equivalent
 variable: a symbol used to represent a quantity that can change
 linear equation in one variable- an equation that can be written in the form ax = b where
a and b
are constants and a  0
 term of an expression: the parts of the expression that are added or subtracted
 solution of an equation in one variable: a value or values that make the equation true
 evaluate: to find the value of an algebraic expression by substituting a number for each
variable and simplifying by using the order of operations
 formula: a literal equation that states a rule for a relationship among quantities
 literal equation: an equation that contains two or more variables
 solution set: the set of elements from the replacement set that makes an open sentence
true
Equations and Inequalities Notes
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Algebra 1 Notes




Equations and Inequalities
inequality: a sentence formed when an inequality symbol is placed between two
expressions
absolute value: the distance between the origin and the point representing the real
number
compound inequality: two inequalities connected by and or or
properties of inequality:
Addition Property of Inequality:
if a  b, then a  c  b  c
if a  b, then a  c  b  c
Subtraction Property of Inequality:
if a  b, then a  c  b  c
if a  b, then a  c  b  c
Multiplication Property of Inequality:
if a  b and c > 0, then ac > bc
if a  b and c > 0, then ac < bc
if a  b and c < 0, then ac < bc
if a  b and c < 0, then ac > bc
a
b
Division Property of Inequality:
if a  b and c > 0, then >
c
c
a
b
if a  b and c > 0, then <
c
c
a
b
if a  b and c < 0, then <
c
c
a
b
if a  b and c < 0, then >
c
c
SKILLS:
 write and solve equations and inequalities including the units in order to correctly label
real-world problems
 apply order of operation and properties of real numbers and equations to simplify
algebraic expressions and solve equations
 apply order of operation and properties of real numbers and inequalities to solve
algebraic inequalities and compound inequalities
 write and solve absolute value equations and inequalities in one variable
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
REVIEW AND EXAMPLES:
Ex 1: Justify the steps in solving
3x
 7 x  7  3(2 x  1) by using the properties of equality
2
and other properties.
Distribute the 3 on the right side of the equation: distributive property
3x
3x
 7 x  7  3(2 x  1) 5x  14 
 7x  7  6x  3
2
2
Multiply both sides of the equation by 2, then simplify: clearing the fractions
3x
2 (
 7 x  7)  2 (6 x  3)  3x +14x  14 = 12x +6
2
Combine like terms.
3x +14x 14 = 12x +6  17 x 14  12 x  6
Subtract 12x from both sides: subtraction property of equality
17 x  14  12 x  12 x  12 x  6  5x  14  6
Add 14 to both sides and simplify: addition property of equality
5x  14  14  6  14  5x  20
Divide both sides by 5 and simplify: division property of equality
5 x 20
 x=4

5
5
The solution set is 4 .
Check your solution:
3(4)
12
 7(4)  7  3(2(4)  1) 
 28  7  3(8  1)  6  28  7  3(9)  27 = 27
2
2
Ex 2: Write an equation to represent the real-world problem and then use algebra to solve the
equation and find the answer.
Each household in the United States receives about 676 pieces of junk mail per year.
About how many pieces does a household expect to receive per week?
 2 pieces
(C)  676 pieces
 13 pieces
(D)  35,152 pieces
We start by listing the given information and writing the mathematical model that
represents our situation:
Let m represent the number of pieces of junk mail each household receives.
Given : -we know there are 52 weeks in a year, we know there are approximately 676
pieces of junk mail sent to each household every year.
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
The equation 52m  676 can be used to solve this problem.
52m 676
; m = 13 Therefore, the answer is 13 pieces of junk mail per week.

52
52
Ex 3: Find three consecutive integers whose sum is 84.
(A) 28,28,28
(C) 27,28,29
(B) 21,42,84
(D) 3.5,4.5,5.5
To solve we need to know the definition of consecutive integers. They are integers that
follow each other in order. If you assign the first integer to n, the second will be (n + 1),
and the third will be (n + 2).
n  (n  1)  (n  2)  84  3n  3  84  3n  3  3  84  3  3n  81 
3n 81

3
3
n  27 ; n + 1 =2 8 ; n + 2 = 29
Ex 4: Solve the equation.
3(x +5) = 3x + 15
(A) x = 0
(C) no solution
(B) x = 1
(D) all real solutions
To solve this expression we need to know how to simplify equations. First, we have to
use the distributive property.
 3x  15  3x  15
3(x +5) = 3x + 15
Subtract 3x from each side. 3x  3x  15  3x  3x  15  15 = 15
All values of x are solutions, because 15 = 15 is always true. The original equation is an
identity.
Ex 5: Solve the equation.
y2 y4
(A) 4
(C) no solution
(B) 2
(D) all real solutions
To solve this we will need to isolate the variable. The inverse of adding is subtracting
and so
y y2 y y4
2  4
The original answer has no solution, because 2  4 for any value of y.
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
Ex 6: Solve the linear inequality. Write your answer in set notation. Graph the solution
on the number line.
8x  1  15  8x  16

x2
set notation: {x x 
where x  2}
set notation: {x x 
where x  2}
Ex 7: Solve and graph the solution.
2 > n  4   2  4 > n  4  4  2  n or n < 2
Ex 8: Solve and graph the solution.
4.2m  6.3

4.2m 6.3

dividing by  4.2 will reverse inequality symbol  m  1.5
4.2
4.2
set notation: {x x 
where x  1.5}
Ex 9: Solve 2  3x  8  10. Graph the solution.
In order to solve this inequality you will have to isolate the variable x between the two
inequality symbols.
2  3x  8  10
Write the original inequality
6  3x  18
Add 8 to each expression
2 x6
Divide each expression by 3
The solution is all real numbers that are greater than or equal to 2 and less than or equal
to 6.
set notation: {x x  where 2  x  6}
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
Ex 10: Solve 3x  1 < 4 or 2 x  5 > 7. Graph the solution.
A solution of this inequality is a solution of either of its simple parts. You can solve each
part separately.
3x  1 < 4
or
2x  5 > 7
3x < 3
or
2 x > 12
x <1
or
x  6
The solution is all real numbers that are less than 1 or greater than 6.
set notation: {x x  where x  1 or x  6}
Ex 11: Find the speed and the velocity of the object: a helicopter is descending for a landing at a
rate of 6 feet per second.
To solve we will need to know the definitions of velocity and speed. Velocity indicates
both speed and direction (up is positive and down is negative). The speed of an object is
the absolute value of its velocity. Velocity will then be 6 ft./sec. and speed will be 6
ft./sec because it is always positive.
Ex 12: Ice cream should be stored at 12° F with an allowance for 3°F. Write and solve an
equation to find the maximum and minimum temperatures at which the ice cream should
be stored.
To solve we will need to know this is an absolute value equation because of the
allowance of a 3°F difference. Therefore, t  12 will be the first part of the equation.
The allowance of temperature is what it will be equal to, which is 3 degrees.
t  12  3
Students should be able to identify that there could be two possible answers and so we
write the original equation (case 1) where the expression inside the absolute value symbol
is positive or zero and we write the opposite (case 2) where the expression inside the
absolute value symbol is negative.
Solve each equation.
Case 1: t  12  3
Case 2: (t  12)  3
 t  15
 t  12  3  t  9  t  9
Therefore, the minimum temperature the ice cream could be stored at is 9° F and the
maximum temperature it should be stored at is 15°F.
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
A second way to solve absolute value equations is to make each case the same
distance away from the given point, therefore, making one solution positive and one
solution negative.
Ex 13: Solve the equation, if possible; 2 x  7  9 .
This equation states that 2x – 7 is 9 spaces from zero. So, that means that 2x – 7 is
either equal to 9 or -9 because both values are 9 spaces from zero. To find the solution
set, we solve both equations.
Case 1: 2 x  7  9
Case 2:
2 x  7  9
Solve each equation.
2 x  7  9  2 x  16  x  8
2x  7  9  2x  2  x  1
The equation has two solutions: -1 and 8.
set notation: x = {-1, 8}
A third way to solve absolute value equations is to directly translate what the statement
tells us and graph it. Then write the solution set based on the graph.
Ex 14: x  4  2 This statement tells us that the difference between a number (x) and 4 is less
than 2. So, the distance between x and 4 is less than 2. Let’s graph that:
x  2 and x  6
set notation: {x x 
All of the solutions to this
inequality will be values that fall
between 2 and 6, because all of
these values are within 2 spaces
from 4.
where 2  x  6}
Ex 15: x  1  3 This statement tells us about the sum of a number (x) and 1, so let’s rewrite
the statement so that it tells us about the difference between x and 1.
x  ( 1)  3 Now the statement tells us that the difference between a number (x) and -1
is greater than or equal to 3. So, the distance between x and -1 is greater than or equal to
3. Let’s graph that:
All of the solutions to this
inequality will be values that fall
outside of and including -4 and 2,
because all of these values are
greater than or equal to 3 spaces
from -1.
Equations and Inequalities Notes
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Algebra 1 Notes
x  4 or x  2
Equations and Inequalities
set notation: {x x 
where x  4 or x  2}
Ex 16: 3x  1  8  3 This statement tells us about the sum of a product (3x) and 1, so let’s
rewrite the statement so that it tells us about the difference between 3x and 1.
3x  (1)  8  3 Next, isolate the absolute value statement…
3x  ( 1)  8  3
+8
8
3x  ( 1)  5
Let’s clear the product so that x has a coefficient of 1.
3x  ( 1)  5
3
3
3
1
5
x  ( ) 
3
3
Now, the statement tells us that the difference between x and 
5
. So, Let’s graph that:
3

All of the solutions to this
inequality will be values that fall
4
between and including -2 and ,
3
because all of these values are less
5
1
than or equal to spaces from  .
3
3
1
3
x  2 and x 
4
3
1
is less then or equal to
3
set notation: {x x 
4
where  2  x  }
3
Ex 17: Solve this equation, if possible 2w  3  7 .
Think about the definition of absolute value; the distance between the origin and point(s)
on the horizontal number line. Distance must be a positive value. This statement tells us
that 2w + 3 is -7 spaces from the origin. Since there is no way to be -7 spaces from
anywhere, there is no solution to this equation.
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
Ex 18: Solve for the indicated variable.
C  2 r; r
To solve we will have to isolate the variable r.
C 2 (r )

2
2
r
C
2
Ex 19: Rewrite the equation 3x  2 y  4  2  5  y  so that y is a function of x. Then, use the
result to find y when x = 2,0,1,and 5.
To solve this we will have to isolate the y variable. y is on both sides of the equation and
so each side will have to be simplified before you can solve for it.
3x  2 y  4  2  5  y 
Original Equation
3x  2 y  4  10  2 y
Distributive Property
3x  2 y  4  2 y  10  2 y  2 y
Subtraction Property of Equality
3x  4  10  4 y
Simplifying (combining like terms)
3x  4 10  10  4 y 10
Subtraction Property of Equality
3x  14  4 y
Simplifying (combining like terms)
3x 14 4 y


4 4 4
Division Property of Equality
3
7
x  y
4
2
Simplifying
Now that you have solved so that y is a function of x, you substitute in the given domain
values for the x and evaluate.
y
3
7
 2   5
4
2
y
3
7
7
 0  
4
2
2
y
3
7 11
1  
4
2
4
y
3
7
1
5  
4
2
4
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
ASSESSMENT ITEMS:
1. Solve the equation if possible.
5
 9  w  10
3
2
3
B. 15
C. no solution
D. infinite solutions
A. 6
ANS: B
2. How many solutions does the equation 2v  3(4  v)  12  5v have?
A. one
B. two
C. infinite solutions
D. no real solutions
ANS: C
3. You sell lemonade for $1.25 a glass. Which equation can you use to find how many glasses
of lemonade you sold if you collected a total of $60?
A. 1.25x  60
B. 60 x  1.25
x
C.
 60
1.25
D. (1.25)(60)  x
ANS:
A
4. Solve the formula for h, V  lwh
Vh
h
lw
A.
B. h  l  w  V
V
h
lw
C.
h

V
l  w
D.
ANS: C
5. Which ordered pair is a solution to 3x  5 y  15
A. (0, 5)
B. (1, 2)
C. (8, 1)
D.  5, 0 
ANS: A
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
6. A local business has agreed to donate no more than half as much as the senior class
raises. Which inequality shows how much money b the business will contribute if the
seniors raised $870?
1
870   b
A. 2
1
870  b
2
B.
1
870   b
C. 2
1
D. 870  b
2
ANS: A
7. If x  5 is a solution of 2 x  tx  5  30 , what is the value of t?
A. 9
B. 7
C. 3
D. 9
ANS: A
8. The sum of three consective whole numbers is 18. What is the greatest of the three numbers?
A. 3
B. 6
C. 7
D. 18
ANS: C
9. The movie theater donates at least 10% of its sales to charity. From Cassie’s purchases, the
theater will donate at least $2.15. Which inequality below shows the amount of money m that
Cassie spent at the refreshment stand?
A. m  21.50
B. m  21.50
C. m  215
D. m  215
ANS: B
10. The elevation of Lone Mountain is 27 feet more than 16 times the lowest point in the state. If
the elevation of the lowest point in the state is 257 feet, what is the elevation of Lone
Mountain?
A. 4139 ft.
B. 4166 ft.
C. 11,024 ft.
D. 12,245 ft.
ANS: B
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Algebra 1 Notes
Equations and Inequalities
11. Barb has saved $91 from last year and would like to baby-sit ($5.50 per hour) to earn
enough to buy a mountain bike. She needs a good quality bike to bicycle through Red
Rock Canyon. A bike like this costs at least $300. What numbers of hours h can Barb babysit to reach her goal?
A. h ≥ 71
B. h ≥ 38
C. h ≥ 23
D. h ≥ 14
ANS: B
12. Find the value of x so that the figures have the same area.
A.
B.
C.
D.
10
12
13
15
25 ft
15 ft
(x – 6)ft
x ft
ANS: D
13. Lara’s car needed a particular part that costs $75. The mechanic charges $50 per hour to
install the part. If the total cost was $350, how many hours did it take to install part?
A. 2 hours
B. 2.5 hours
C. 5.5 hours
D. 7 hours
ANS: C
14. A panda has a lifespan of 14 to 20 years. The inequality x  17  c gives the number of years
a panda may live. What is the value of c?
A. 20
B. 14
C. 6
D. 3
ANS: D
15. The formula for the lateral area of a cylinder is A  2 rh where r is the radius and h is the
height. Solve the equation for h.
ANS: h 
A
2 r
16. Solve the equation. Then graph the solution set.
p4  6
ANS: 2,10
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Algebra 1 Notes
Equations and Inequalities
-2
10
17. Solve 5x – 3y = 9 for y.
ANS: y 
5
x3
3
18. Evaluate the expression 3t  2u  5v if t  2, u  5, and v  3 .
ANS: x = 1
19. Your cousin earns $25 per week babysitting and received one $5 bonus. You earn $15 per
week mowing lawns and $12 per week running errands. After working the same number of
weeks, you have $11 more than your cousin. How many weeks have the two of you worked?
ANS: 8 weeks
R
20.
pV
nT ; solve for p.
ANS: p 
21. Solve the equation, if possible.
RnT
V
17  5  3p
ANS: p  4
22. 24a  8 10a  2(4  7a)
ANS: all real numbers
23. Sam can spend no more than $100 on school supplies. He has to spend $12 on paper and
pencils. What is the maximum abount Sam can spend on the graphing calculator he wants?
ANS: let m represent the money left for his graphing calculator
one possible answer: m  12  100
24. It costs $5 to have a tote bag monogrammed with up to 12 letters and $0.50 for each
additional letter. Summer wants her backpack monogrammed, but has a budget of $8. What
is the maximum amount of letters that she can have embroidered into the backpack.
Let l equal the number of letters that she can have monogrammed.
We know the first 12 are free (with the $5 charge)
We know each additional letter is $0.50
We know the maximum amount she can spend is $8
5  0.50l < $8
l  12 will be the maximum number of letters that can be monogrammed onto her backpack
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Algebra 1 Notes
Equations and Inequalities
25. Solve the inequality and graph the solution.
x
3 < x + 6 or
>4
3
ANS:
x >  9 or x <  12
x x 
where x  12 or x  9
26. Solve the inequality and graph the solution.
10  4 x 18  30
ANS:
10  18  4 x  18  18  30  18
8
 4x
 48
8

4
-2 
4
x
4
x
48
4
 12
 12 
x
set notation: {x x 

 2
where  12  x  2}
27. Solve the absolute value equation and graph the solution.
x  3.6  4.6
x  3.6  4.6
x  3.6  4.6
ANS:
x  1.0
x  8.2
28. Solve the inequality. Then graph the solution.
9 x  7
ANS:
Case 1:
9 x7
Case 2 :
9  x  7
x  2
x  16
 16  x  2
x x 
where x  16 and x  2
Equations and Inequalities Notes
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Algebra 1 Notes
Equations and Inequalities
3x  9 > 5
29. Solve the inequality. Then graph the solution.
ANS:
Case 1:
3x  9 > 5
Case 2 :
3 x > 14
3x 14
>
3
3
14
x>
3
3x < 4
3x
4
<
3
3
4
x<
3
x>

x x 

where x <
3x  9 <  5
14
3
or x <
4
3
4
14 
or x >

3
3
30. Test scores in your class range from 60 to 100. Write an absolute value inequality describing
the range of the test scores.
ANS:
100  60
 80
 x  80
2
100  60
 20
 20
2
x  80  20 would be the equation for the range of test scores.
Equations and Inequalities Notes
page 15 of 15
9/4/2014