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College Algebra
Unit 1: linear equations and inequalities, unit 2: graphs and forms of linear equations, unit 3: functions, unit 4: quadratics: multiplying and factoring, unit 5: quadratic functions and equations, unit 6: complex numbers, unit 7: exponents and radicals, unit 8: rational expressions and equations, unit 9: relating algebra and geometry, unit 10: polynomial arithmetic, unit 11: advanced function types, unit 12: transformations of functions, unit 13: rational exponents and radicals, unit 14: logarithms.
2.1 The Rectangular Coordinate Systems and Graphs
x -intercept is ( 4 , 0 ) ; ( 4 , 0 ) ; y- intercept is ( 0 , 3 ) . ( 0 , 3 ) .
125 = 5 5 125 = 5 5
( − 5 , 5 2 ) ( − 5 , 5 2 )
2.2 Linear Equations in One Variable
x = −5 x = −5
x = −3 x = −3
x = 10 3 x = 10 3
x = 1 x = 1
x = − 7 17 . x = − 7 17 . Excluded values are x = − 1 2 x = − 1 2 and x = − 1 3 . x = − 1 3 .
x = 1 3 x = 1 3
m = − 2 3 m = − 2 3
y = 4 x −3 y = 4 x −3
x + 3 y = 2 x + 3 y = 2
Horizontal line: y = 2 y = 2
Parallel lines: equations are written in slope-intercept form.
y = 5 x + 3 y = 5 x + 3
2.3 Models and Applications
C = 2.5 x + 3 , 650 C = 2.5 x + 3 , 650
L = 37 L = 37 cm, W = 18 W = 18 cm
2.4 Complex Numbers
−24 = 0 + 2 i 6 −24 = 0 + 2 i 6
( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i ( 3 −4 i ) − ( 2 + 5 i ) = 1 −9 i
5 2 − i 5 2 − i
18 + i 18 + i
−3 −4 i −3 −4 i
2.5 Quadratic Equations
( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1 ( x − 6 ) ( x + 1 ) = 0 ; x = 6 , x = − 1
( x −7 ) ( x + 3 ) = 0 , ( x −7 ) ( x + 3 ) = 0 , x = 7 , x = 7 , x = −3. x = −3.
( x + 5 ) ( x −5 ) = 0 , ( x + 5 ) ( x −5 ) = 0 , x = −5 , x = −5 , x = 5. x = 5.
( 3 x + 2 ) ( 4 x + 1 ) = 0 , ( 3 x + 2 ) ( 4 x + 1 ) = 0 , x = − 2 3 , x = − 2 3 , x = − 1 4 x = − 1 4
x = 0 , x = −10 , x = −1 x = 0 , x = −10 , x = −1
x = 4 ± 5 x = 4 ± 5
x = 3 ± 22 x = 3 ± 22
x = − 2 3 , x = − 2 3 , x = 1 3 x = 1 3
2.6 Other Types of Equations
{ −1 } { −1 }
0 , 0 , 1 2 , 1 2 , − 1 2 − 1 2
1 ; 1 ; extraneous solution − 2 9 − 2 9
−2 ; −2 ; extraneous solution −1 −1
−1 , −1 , 3 2 3 2
−3 , 3 , − i , i −3 , 3 , − i , i
2 , 12 2 , 12
−1 , −1 , 0 0 is not a solution.
2.7 Linear Inequalities and Absolute Value Inequalities
[ −3 , 5 ] [ −3 , 5 ]
( − ∞ , −2 ) ∪ [ 3 , ∞ ) ( − ∞ , −2 ) ∪ [ 3 , ∞ )
x < 1 x < 1
x ≥ −5 x ≥ −5
( 2 , ∞ ) ( 2 , ∞ )
[ − 3 14 , ∞ ) [ − 3 14 , ∞ )
6 < x ≤ 9 or ( 6 , 9 ] 6 < x ≤ 9 or ( 6 , 9 ]
( − 1 8 , 1 2 ) ( − 1 8 , 1 2 )
| x −2 | ≤ 3 | x −2 | ≤ 3
k ≤ 1 k ≤ 1 or k ≥ 7 ; k ≥ 7 ; in interval notation, this would be ( − ∞ , 1 ] ∪ [ 7 , ∞ ) . ( − ∞ , 1 ] ∪ [ 7 , ∞ ) .
2.1 Section Exercises
Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.
The y -intercept is the point where the graph crosses the y -axis.
The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , 6 ) . ( 0 , 6 ) .
The x- intercept is ( 2 , 0 ) ( 2 , 0 ) and the y -intercept is ( 0 , −3 ) . ( 0 , −3 ) .
The x- intercept is ( 3 , 0 ) ( 3 , 0 ) and the y -intercept is ( 0 , 9 8 ) . ( 0 , 9 8 ) .
y = 4 − 2 x y = 4 − 2 x
y = 5 − 2 x 3 y = 5 − 2 x 3
y = 2 x − 4 5 y = 2 x − 4 5
d = 74 d = 74
d = 36 = 6 d = 36 = 6
d ≈ 62.97 d ≈ 62.97
( 3 , − 3 2 ) ( 3 , − 3 2 )
( 2 , −1 ) ( 2 , −1 )
( 0 , 0 ) ( 0 , 0 )
y = 0 y = 0
not collinear
A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 ) A: ( −3 , 2 ) , B: ( 1 , 3 ) , C: ( 4 , 0 )
d = 8.246 d = 8.246
d = 5 d = 5
( −3 , 4 ) ( −3 , 4 )
x = 0 y = −2 x = 0 y = −2
x = 0.75 y = 0 x = 0.75 y = 0
x = − 1.667 y = 0 x = − 1.667 y = 0
15 − 11.2 = 3.8 mi 15 − 11.2 = 3.8 mi shorter
6 .0 42 6 .0 42
Midpoint of each diagonal is the same point ( 2 , –2 ) ( 2 , –2 ) . Note this is a characteristic of rectangles, but not other quadrilaterals.
2.2 Section Exercises
It means they have the same slope.
The exponent of the x x variable is 1. It is called a first-degree equation.
If we insert either value into the equation, they make an expression in the equation undefined (zero in the denominator).
x = 2 x = 2
x = 2 7 x = 2 7
x = 6 x = 6
x = 3 x = 3
x = −14 x = −14
x ≠ −4 ; x ≠ −4 ; x = −3 x = −3
x ≠ 1 ; x ≠ 1 ; when we solve this we get x = 1 , x = 1 , which is excluded, therefore NO solution
x ≠ 0 ; x ≠ 0 ; x = − 5 2 x = − 5 2
y = − 4 5 x + 14 5 y = − 4 5 x + 14 5
y = − 3 4 x + 2 y = − 3 4 x + 2
y = 1 2 x + 5 2 y = 1 2 x + 5 2
y = −3 x − 5 y = −3 x − 5
y = 7 y = 7
y = −4 y = −4
8 x + 5 y = 7 8 x + 5 y = 7
Perpendicular
m = − 9 7 m = − 9 7
m = 3 2 m = 3 2
m 1 = − 1 3 , m 2 = 3 ; Perpendicular . m 1 = − 1 3 , m 2 = 3 ; Perpendicular .
y = 0.245 x − 45.662. y = 0.245 x − 45.662. Answers may vary. y min = −50 , y max = −40 y min = −50 , y max = −40
y = − 2.333 x + 6.667. y = − 2.333 x + 6.667. Answers may vary. y min = −10 , y max = 10 y min = −10 , y max = 10
y = − A B x + C B y = − A B x + C B
The slope for ( −1 , 1 ) to ( 0 , 4 ) is 3. The slope for ( −1 , 1 ) to ( 2 , 0 ) is − 1 3 . The slope for ( 2 , 0 ) to ( 3 , 3 ) is 3. The slope for ( 0 , 4 ) to ( 3 , 3 ) is − 1 3 . The slope for ( −1 , 1 ) to ( 0 , 4 ) is 3. The slope for ( −1 , 1 ) to ( 2 , 0 ) is − 1 3 . The slope for ( 2 , 0 ) to ( 3 , 3 ) is 3. The slope for ( 0 , 4 ) to ( 3 , 3 ) is − 1 3 .
Yes they are perpendicular.
2.3 Section Exercises
Answers may vary. Possible answers: We should define in words what our variable is representing. We should declare the variable. A heading.
2 , 000 − x 2 , 000 − x
v + 10 v + 10
Ann: 23 ; 23 ; Beth: 46 46
20 + 0.05 m 20 + 0.05 m
90 + 40 P 90 + 40 P
50 , 000 − x 50 , 000 − x
She traveled for 2 h at 20 mi/h, or 40 miles.
$5,000 at 8% and $15,000 at 12%
B = 100 + .05 x B = 100 + .05 x
R = 9 R = 9
r = 4 5 r = 4 5 or 0.8
W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14 W = P − 2 L 2 = 58 − 2 ( 15 ) 2 = 14
f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21 f = p q p + q = 8 ( 13 ) 8 + 13 = 104 21
m = − 5 4 m = − 5 4
h = 2 A b 1 + b 2 h = 2 A b 1 + b 2
length = 360 ft; width = 160 ft
A = 88 in . 2 A = 88 in . 2
h = V π r 2 h = V π r 2
r = V π h r = V π h
C = 12 π C = 12 π
2.4 Section Exercises
Add the real parts together and the imaginary parts together.
Possible answer: i i times i i equals -1, which is not imaginary.
−8 + 2 i −8 + 2 i
14 + 7 i 14 + 7 i
− 23 29 + 15 29 i − 23 29 + 15 29 i
8 − i 8 − i
−11 + 4 i −11 + 4 i
2 −5 i 2 −5 i
6 + 15 i 6 + 15 i
−16 + 32 i −16 + 32 i
−4 −7 i −4 −7 i
2 − 2 3 i 2 − 2 3 i
4 − 6 i 4 − 6 i
2 5 + 11 5 i 2 5 + 11 5 i
1 + i 3 1 + i 3
( 3 2 + 1 2 i ) 6 = −1 ( 3 2 + 1 2 i ) 6 = −1
5 −5 i 5 −5 i
9 2 − 9 2 i 9 2 − 9 2 i
2.5 Section Exercises
It is a second-degree equation (the highest variable exponent is 2).
We want to take advantage of the zero property of multiplication in the fact that if a ⋅ b = 0 a ⋅ b = 0 then it must follow that each factor separately offers a solution to the product being zero: a = 0 o r b = 0. a = 0 o r b = 0.
One, when no linear term is present (no x term), such as x 2 = 16. x 2 = 16. Two, when the equation is already in the form ( a x + b ) 2 = d . ( a x + b ) 2 = d .
x = 6 , x = 6 , x = 3 x = 3
x = − 5 2 , x = − 5 2 , x = − 1 3 x = − 1 3
x = 5 , x = 5 , x = −5 x = −5
x = − 3 2 , x = − 3 2 , x = 3 2 x = 3 2
x = −2 , 3 x = −2 , 3
x = 0 , x = 0 , x = − 3 7 x = − 3 7
x = −6 , x = −6 , x = 6 x = 6
x = 6 , x = 6 , x = −4 x = −4
x = 1 , x = 1 , x = −2 x = −2
x = −2 , x = −2 , x = 11 x = 11
z = 2 3 , z = 2 3 , z = − 1 2 z = − 1 2
x = 3 ± 17 4 x = 3 ± 17 4
One rational
Two real; rational
x = − 1 ± 17 2 x = − 1 ± 17 2
x = 5 ± 13 6 x = 5 ± 13 6
x = − 1 ± 17 8 x = − 1 ± 17 8
x ≈ 0.131 x ≈ 0.131 and x ≈ 2.535 x ≈ 2.535
x ≈ − 6.7 x ≈ − 6.7 and x ≈ 1.7 x ≈ 1.7
a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a a x 2 + b x + c = 0 x 2 + b a x = − c a x 2 + b a x + b 2 4 a 2 = − c a + b 4 a 2 ( x + b 2 a ) 2 = b 2 − 4 a c 4 a 2 x + b 2 a = ± b 2 − 4 a c 4 a 2 x = − b ± b 2 − 4 a c 2 a
x ( x + 10 ) = 119 ; x ( x + 10 ) = 119 ; 7 ft. and 17 ft.
maximum at x = 70 x = 70
The quadratic equation would be ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. ( 100 x −0.5 x 2 ) − ( 60 x + 300 ) = 300. The two values of x x are 20 and 60.
2.6 Section Exercises
This is not a solution to the radical equation, it is a value obtained from squaring both sides and thus changing the signs of an equation which has caused it not to be a solution in the original equation.
He or she is probably trying to enter negative 9, but taking the square root of −9 −9 is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in −27. −27.
A rational exponent is a fraction: the denominator of the fraction is the root or index number and the numerator is the power to which it is raised.
x = 81 x = 81
x = 17 x = 17
x = 8 , x = 27 x = 8 , x = 27
x = −2 , 1 , −1 x = −2 , 1 , −1
y = 0 , 3 2 , − 3 2 y = 0 , 3 2 , − 3 2
m = 1 , −1 m = 1 , −1
x = 2 5 , ±3 i x = 2 5 , ±3 i
x = 32 x = 32
t = 44 3 t = 44 3
x = −2 x = −2
x = 4 , −4 3 x = 4 , −4 3
x = − 5 4 , 7 4 x = − 5 4 , 7 4
x = 3 , −2 x = 3 , −2
x = 1 , −1 , 3 , -3 x = 1 , −1 , 3 , -3
x = 2 , −2 x = 2 , −2
x = 1 , 5 x = 1 , 5
x ≥ 0 x ≥ 0
x = 4 , 6 , −6 , −8 x = 4 , 6 , −6 , −8
2.7 Section Exercises
When we divide both sides by a negative it changes the sign of both sides so the sense of the inequality sign changes.
( − ∞ , ∞ ) ( − ∞ , ∞ )
We start by finding the x -intercept, or where the function = 0. Once we have that point, which is ( 3 , 0 ) , ( 3 , 0 ) , we graph to the right the straight line graph y = x −3 , y = x −3 , and then when we draw it to the left we plot positive y values, taking the absolute value of them.
( − ∞ , 3 4 ] ( − ∞ , 3 4 ]
[ − 13 2 , ∞ ) [ − 13 2 , ∞ )
( − ∞ , 3 ) ( − ∞ , 3 )
( − ∞ , − 37 3 ] ( − ∞ , − 37 3 ]
All real numbers ( − ∞ , ∞ ) ( − ∞ , ∞ )
( − ∞ , − 10 3 ) ∪ ( 4 , ∞ ) ( − ∞ , − 10 3 ) ∪ ( 4 , ∞ )
( − ∞ , −4 ] ∪ [ 8 , + ∞ ) ( − ∞ , −4 ] ∪ [ 8 , + ∞ )
No solution
( −5 , 11 ) ( −5 , 11 )
[ 6 , 12 ] [ 6 , 12 ]
[ −10 , 12 ] [ −10 , 12 ]
x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 , ( − 2 , + ∞ ) x > − 6 and x > − 2 Take the intersection of two sets . x > − 2 , ( − 2 , + ∞ )
x < − 3 or x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ [ 1 , ∞ ) x < − 3 or x ≥ 1 Take the union of the two sets . ( − ∞ , − 3 ) ∪ [ 1 , ∞ )
( − ∞ , −1 ) ∪ ( 3 , ∞ ) ( − ∞ , −1 ) ∪ ( 3 , ∞ )
[ −11 , −3 ] [ −11 , −3 ]
It is never less than zero. No solution.
Where the blue line is above the orange line; point of intersection is x = − 3. x = − 3.
( − ∞ , −3 ) ( − ∞ , −3 )
Where the blue line is above the orange line; always. All real numbers.
( − ∞ , − ∞ ) ( − ∞ , − ∞ )
( −1 , 3 ) ( −1 , 3 )
( − ∞ , 4 ) ( − ∞ , 4 )
{ x | x < 6 } { x | x < 6 }
{ x | −3 ≤ x < 5 } { x | −3 ≤ x < 5 }
( −2 , 1 ] ( −2 , 1 ]
( − ∞ , 4 ] ( − ∞ , 4 ]
Where the blue is below the orange; always. All real numbers. ( − ∞ , + ∞ ) . ( − ∞ , + ∞ ) .
Where the blue is below the orange; ( 1 , 7 ) . ( 1 , 7 ) .
x = 2 , − 4 5 x = 2 , − 4 5
( −7 , 5 ] ( −7 , 5 ]
80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400 80 ≤ T ≤ 120 1 , 600 ≤ 20 T ≤ 2 , 400
[ 1 , 600 , 2 , 400 ] [ 1 , 600 , 2 , 400 ]
Review Exercises
x -intercept: ( 3 , 0 ) ; ( 3 , 0 ) ; y -intercept: ( 0 , −4 ) ( 0 , −4 )
y = 5 3 x + 4 y = 5 3 x + 4
72 = 6 2 72 = 6 2
620.097 620.097
midpoint is ( 2 , 23 2 ) ( 2 , 23 2 )
x = 4 x = 4
x = 12 7 x = 12 7
y = 1 6 x + 4 3 y = 1 6 x + 4 3
y = 2 3 x + 6 y = 2 3 x + 6
females 17, males 56
x = − 3 4 ± i 47 4 x = − 3 4 ± i 47 4
horizontal component −2 ; −2 ; vertical component −1 −1
7 + 11 i 7 + 11 i
−16 − 30 i −16 − 30 i
−4 − i 10 −4 − i 10
x = 7 − 3 i x = 7 − 3 i
x = −1 , −5 x = −1 , −5
x = 0 , 9 7 x = 0 , 9 7
x = 10 , −2 x = 10 , −2
x = − 1 ± 5 4 x = − 1 ± 5 4
x = 2 5 , − 1 3 x = 2 5 , − 1 3
x = 5 ± 2 7 x = 5 ± 2 7
x = 0 , 256 x = 0 , 256
x = 0 , ± 2 x = 0 , ± 2
x = 11 2 , −17 2 x = 11 2 , −17 2
[ − 10 3 , 2 ] [ − 10 3 , 2 ]
( − 4 3 , 1 5 ) ( − 4 3 , 1 5 )
Where the blue is below the orange line; point of intersection is x = 3.5. x = 3.5.
( 3.5 , ∞ ) ( 3.5 , ∞ )
Practice Test
y = 3 2 x + 2 y = 3 2 x + 2
( 0 , −3 ) ( 0 , −3 ) ( 4 , 0 ) ( 4 , 0 )
( − ∞ , 9 ] ( − ∞ , 9 ]
x = −15 x = −15
x ≠ −4 , 2 ; x ≠ −4 , 2 ; x = − 5 2 , 1 x = − 5 2 , 1
x = 3 ± 3 2 x = 3 ± 3 2
( −4 , 1 ) ( −4 , 1 )
y = −5 9 x − 2 9 y = −5 9 x − 2 9
y = 5 2 x − 4 y = 5 2 x − 4
5 13 − 14 13 i 5 13 − 14 13 i
x = 2 , − 4 3 x = 2 , − 4 3
x = 1 2 ± 2 2 x = 1 2 ± 2 2
x = 1 2 , 2 , −2 x = 1 2 , 2 , −2
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- Authors: Jay Abramson
- Publisher/website: OpenStax
- Book title: College Algebra
- Publication date: Feb 13, 2015
- Location: Houston, Texas
- Book URL: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
- Section URL: https://openstax.org/books/college-algebra/pages/chapter-2
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Alexander E. Guterman
Russian version
Mailing address: Faculty of Algebra, Department of Mathematics and Mechanics, Moscow State University, 119991, GSP-1, Russia, Moscow
E-mail address: guterman at list dot ru
Tel./Fax: 007(495)9391611
Position: Professor at the Moscow State University, Faculty of Algebra, Department of Mathematics and Mechanics
Academic background
Major research interests
Courses presented
Tentative list of publications
Recent invited addresses
Conferences
Other professional services
Other interests: tourism/hiking, see Don't Worry;-) site
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Introduction to Systems of Equations and Inequalities; 7.1 Systems of Linear Equations: Two Variables; 7.2 Systems of Linear Equations: Three Variables; 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables; 7.4 Partial Fractions; 7.5 Matrices and Matrix Operations; 7.6 Solving Systems with Gaussian Elimination; 7.7 Solving Systems with Inverses; 7.8 Solving Systems with Cramer's Rule
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Mailing address: Faculty of Algebra, Department of Mathematics and Mechanics, Moscow State University, 119991, GSP-1, Russia, Moscow. E-mail address: guterman at list dot ru. Tel./Fax: 007 (495)9391611. Position: Professor at the Moscow State University, Faculty of Algebra, Department of Mathematics and Mechanics. Academic background.
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