The University of Texas at El Paso
Tutoring and Learning Center
ACCUPLACER
MATH 0311
OR
MATH 0120
http://www.academics.utep.edu/tlc
MATH 0311 OR MATH 0120
Page
Factoring
3
Factoring – Exercises
8
Factoring – Answer to Exercises
9
Rational Expressions
11
Rational Expressions – Exercises
15
Rational Expressions – Answers to Exercises
16
Linear Functions
18
Linear Functions – Exercises
24
Linear Functions – Answer to Exercises
25
Graphing Linear Equations
26
Systems of Linear Equations
29
Systems of Linear Equations – Exercises
33
Systems of Linear Equations – Answers to Exercises
34
Linear Inequalities
35
Equations Involving Absolute Value
37
Inequalities Involving Absolute Value
38
Linear Inequalities and Absolute Value – Exercises
40
Linear Inequalities and Absolute Value – Answers to Exercises
41
Graphing Linear Inequalities in Two Variables
42
Graphing Linear Inequalities in Two Variables – Exercises
45
Graphing Linear Inequalities in Two Variables – Answers to Exercises
46
Radicals
48
Radicals – Exercises
51
Radicals – Answers to Exercises
52
Quadratic Equations
53
Quadratic Equations – Exercises
60
Quadratic Equations – Answers to Exercises
61
FACTORING
Factoring is similar to breaking up a number into its multiples. For example, 10 = 5 · 2. The multiples
are 5 and 2. In a polynomial the procedure is more complicated because variables are involved in
addition to numbers. There are different ways of factoring an equation depending on the complexity of
the polynomial.
Factoring Out the Greatest Common Factor
The first thing to do when factoring is to look at all terms and break up each term into its multiples:
Examples:
Factor: 8 x 3 + 4 x 2 + 10 x
(
)
x 8 x 2 + 4 x + 10 →
In this polynomial the only variable common to all
is x.
2 x(4 x 2 + 2 x + 5) → The two is also common to all terms. Therefore,
this is as far as the polynomial can be factored.
Factor: 6 x 3 + 8 x 2 + 16 x
6 x 3 + 8 x 2 + 16 x
(
2 x 3x 2 + 4 x + 8
→
)
The common variable is x, and the smallest
exponent is 1. The common multiple is 2.
Therefore, the greatest common factor is 2x.
You can also look at it this way:
(2 x )(3x 2 ) = 6 x 3
(2 x )(4 x ) = 8 x 2
(2 x )(8) = 16 x
Factoring by Grouping
There will be occasions when there are no common factors for all terms. Some terms will have a
variable in common, while other terms will have a different variable in common. In these cases, we can
factor by grouping together the common terms.
Examples:
Factor: ax + ay + bx + by
ax + ay + bx + by
ax + bx + ay + by
x(a + b ) + y (a + b )
→
Notice how there are two variables, x and y. Group
together the x’s and the y’s and then
factor out the common factor in each group.
3
Factor: 4 − 2 x 2 + x 2 y 3 − 2 y 3
4 − 2x 2 + x 2 y 3 − 2 y 3 →
(
)
(
)
→
(
)
(
)
→
2 2 − x2 + y3 x2 − 2
2 2 − x2 − y3 2 − x2
(2 − x )(2 − y )
2
There are two variables present, x and y.
Start by grouping two terms at a time.
Notice how the two polynomials inside the
parentheses are similar. To make them the
same change the sign in the second
polynomial.
Now that they are the same, factor out the
common term.
Now it is completely factored out.
→
3
Factoring a Difference of Squares
For a difference of squares, a 2 − b 2 , the factors will be (a − b )(a + b ) .
x 2 − y 2 = (x − y )(x + y )
Examples:
( ) =(1 − x )(1 + x ) =(1 − x )(1 + x ) (1 + x )
1 − x 4 =(1) − x 2
2
2
2
2
2
Factoring the Sum and Difference of Two Cubes
For the addition or subtraction of two cubes, the following formulas apply:
(
= (a − b) (a
a 3 + b3 = ( a + b ) a 2 − ab + b 2
a 3 − b3
Examples:
2
+ a +bb 2
)
)
x3 + 8
3
3
= ( x ) + ( 2)
(
= ( x + 2) ( x
→ Manipulate to be in “ a 3 + b3 ” form.
=+
( x 2 ) x 2 − ( x )( 2 ) + 22
2
− 2x + 4
)
→
( 2 y ) − ( 3)
2
=
( 2 y − 3) ( ( 2 y ) + ( 2 y )( 3) + 32 )
3
=
4
→ Follow formula.
→ Manipulate to be in “ a 3 − b3 ” form.
8 y 3 − 27
=
)
3
( 2 y − 3) ( 4 y 2 + 6 y + 9 )
Follow formula.
Factoring Trinomials
Factoring trinomials is based on finding the two integers whose sum and product meet the given
requirements.
Example:
Find two integers whose sum is –11 and whose product is 30.
Step 1 – The product is positive. The sum is negative. Therefore, both integers must be
negative.
( − )( − ) =+ and ( − ) + ( − ) =−
Step 2 – Possible pairs of factors that will give a positive answer:
30
1. ( −1)( −30 ) =
30
3. ( −3)( −10 ) =
2. ( −2 )( −15 ) =
30
4. ( −5 )( −6 ) =
30
Step 3 – The pair whose sum is –11 is –5 and –6. Therefore, the critical integers are –5
and –6, ( −5 − 6 =−11) .
Factoring Trinomials of the Form x 2 + bx + c
To factor trinomials of the form x 2 + bx + c follow the same principle described above. Find two
integers whose sum equals the middle term and whose product equals the last term.
Examples:
Factor: x 2 − x − 6
Step 1 – Find two integers whose sum is equal to –1 and whose product is equal to –6.
The product is negative; therefore, we need integers that have different signs.
The sum is negative, so the bigger integer is negative.
Step 2 – Possible pairs of factors that will give us –6:
1. (− 3)(2 ) = −6
3. (− 1)(6 ) = −6
2. (3)(− 2 ) = −6
4. (1)(− 6 ) = −6
Step 3 – The pair whose sum is –1 is –3 and 2, (− 3 + 2 = −1) .
The critical integers are –3 and 2.
Step 4 – The factors for x 2 − x − 6 are ( x − 3)( x + 2) .
5
Factor: x 2 + 16 x + 60
Step 1 – Find two integers whose sum is 16 and whose product is 60. Both integers must
be positive since their sum and product are positive.
Step 2 – Possible pairs of factors:
1. (1)( 60 ) = 60
3. ( 3)( 20 ) = 60
5. ( 5 )(12 ) = 60
2. ( 2 )( 30 ) = 60
4. ( 4 )(15 ) = 60
6. ( 6 )(10 ) = 60
Step 3 – The pair whose sum is 16 is (6, 10). Therefore, the critical integers are 6, 10.
Step 4 – The factors for x 2 + 16 x + 60 are ( x + 6)( x + 10) .
Factoring Trinomials of the Form ax2 + bx + c
In this type of trinomial the coefficient of x 2 is not 1, therefore we must consider the product of a ⋅ c .
Examples:
Factor: 2 x 2 − 7 x − 4
−8 . The
Step 1 – Find two integers whose sum is –7 and whose product is ( 2 )( −4 ) =
integers must be of different signs since their product is a negative number.
The larger integer is negative since the sum is negative.
Step 2 – Possible pairs of factors:
−8
1. ( −1)( 8 ) =
−8
3. ( −2 )( 4 ) =
2. (1)( −8 ) =
−8
4. ( 2 )( −4 ) =
−8
Step 3 – The pair whose sum is –7 is 1 and –8. The critical integers are 1 and –8.
Step 4 – Use the critical integers to break the first degree term into two parts:
2 x2 − 7 x − 4 = 2 x2 − 8x + x − 4
Step 5 – Factor by grouping the first two and last two terms:
2 x 2 − 8 x + x − 4 = 2 x ( x − 4 ) + 1( x − 4 )
There are now two terms in the expression. The binomial factor in each of these
two terms should be the same; otherwise, there is an error. In this case, the
common binomial factor is ( x − 4 ) .
Step 6 – Factor out the common binomial factor: 2 x ( x − 4) + 1( x − 4) = ( x − 4)( 2 x + 1)
6
Factor: 6 x 2 − 23 x + 20
Step 1 – Find two integers whose sum is –23 and whose product is 6(20) = 120. Since
the product is positive and the sum is negative, the two integers are negative.
Step 2 – Possible pairs of factors:
120
1. ( −1)( −120 ) =
120
4. ( −4 )( −30 ) =
120
7. ( −8 )( −15 ) =
2. ( −2 )( −60 ) =
120
5. ( −5 )( −24 ) =
120
8. ( −10 )( −12 ) =
120
3. ( −3)( −40 ) =
120
6. ( −6 )( −20 ) =
60
Step 3 – The sum of –8 and –15 is –23. Therefore, the critical integers are –8 and –15.
Step 4 – Use the critical integers to break the first degree term into two parts:
6 x 2 − 23 x + 20 = 6 x 2 − 15 x − 8 x + 20
Step 5 – Factor, separately, the first two and last two terms:
6 x 2 − 15 x − 8 x + 20 = 3 x ( 2 x − 5) − 4( 2 x − 5)
Step 6 – Factor out the common binomial factor.
3 x ( 2 x − 5) − 4 ( 2 x − 5) = ( 2 x − 5)( 3 x − 4)
7
FACTORING – EXERCISES
1. 3 x 2 − 10 x − 8
2. 2 x 2 (x − 3) − 5 x( x − 3) − 3( x − 3)
3. x 3 + 2 x 2 − 9 x − 18
4. x 2 − 10 x + 25 − y 2
5. x 6 − y 6
6. a 3 −
1
8
7. ( x + 5) 2 + 4( x + 5) + 4
8. 1 − 2t 3 + t 6
9. 3 x 4 − 18 x 3 + 27 x 2
10. 100 x 2 − 100 x − 600
11. 8 x 2 − 2 x − 15
12. x 2 + 3 xy + 2 y 2
13. 2 x 2 ( x + 2) + 13x( x + 2) + 15( x + 2)
14. x 2 + x − 12
15. x 2 y + 3 y + 2 x 2 + 6
16. 18a 2 − 50
17. x 2 + 8 x + 15
18. 16 x 3 − 20 x 2 − 50 x
19. 9 x 2 + 33 x + 30
20. 12 x 2 + 2 x − 2
8
FACTORING – ANSWER TO EXERCISES
2. 2 x 2 ( x − 3) − 5 x ( x − 3) − 3 ( x − 3)
1. 3 x 2 − 10 x − 8
3 x 2 − 12 x + 2 x − 8
3x ( x − 4 ) + 2 ( x − 4 )
( x − 3)(2 x 2 − 5 x − 3)
( x − 3)(2 x 2 − 6 x + x − 3)
( 3x + 2 )( x − 4 )
( x − 3)(2 x( x − 3) + 1( x − 3))
( x − 3)( 2 x + 1)( x − 3) = ( x − 3) ( 2 x + 1)
2
3. x 3 + 2 x 2 − 9 x − 18
4. x 2 − 10 x + 25 − y 2
( x 2 − 10 x + 25) − y 2
( x − 5) 2 − y 2
Difference of squares: (a 2 − b 2 ) = (a − b)(a + b)
x 2 ( x + 2) − 9( x + 2)
( x 2 − 9)( x + 2)
x 2 − 9 = ( x + 3)( x − 3)
where a =
y
( x − 5) and b =
( x − 5 − y )( x − 5 + y )
( x − 3)( x + 3)( x + 2)
6. a 3 −
5. x 6 − y 6
x 6 − y 6=
(x ) −( y )
3
2
3
2
Difference of cubes:
Difference of squares: (a 2 − b 2 ) = (a − b)(a + b)
(
x −y = x −y
6
3
3
)( x
3
+y
3
)
(
= (a − b) (a
a 3 + b3 = ( a + b ) a 2 − ab + b 2
a −b
3
2
+ a +bb
)
1
2
2
1  2

1 1 
=
 a −   a + ( a )   +   
2 

2 2 
1 
a 1

=  a −  a 2 + + 
2 
2 4

Sum & Difference of Cubes:
3
(
a 3 − b3 = ( a − b ) a 2 + a +bb 2
a = a and b =
3
where
=
a x=
and b y 3
6
1
8
2
)
)
= ( x + y )( x 2 − xy + y 2 )( x − y )( x 2 + xy + y 2 )
= ( x + y )( x − y )( x 2 + xy + y 2 )( x 2 − xy + y 2 )
7. ( x + 5 ) + 4 ( x + 5 ) + 4 =
2
( x + 7 )( x + 7 ) = ( x + 7 )
2
9
8. 1 − 2t 3 + t 6 =
(t
3
)(
) ( t − 1) ( t
2
−1 t3 −1 =
(
2
)
+ t +1
)
2
2
9. 3 x 4 − 18 x 3 + 27 x=
3x 2 x 2 − 6 x + 9= 3x 2 ( x − 3)
2
10. 100 x 2 − 100 x − 600= 100 ( x − 3)( x + 2 )
11. 8 x 2 − 2 x − 15 =
( 4 x + 5)( 2 x − 3)
12. x 2 + 3 xy + 2 y 2 =( x + 2 y )( x + y )
13. 2 x 2 ( x + 2 ) + 13 x ( x + 2 ) + 15 ( x + 2 ) = ( x + 2 )( 2 x + 3)( x + 5 )
14. x 2 + x − 12 =
( x + 4 )( x − 3)
15. x 2 y + 3 y + 2 x 2 + 6=
(x
2
)
+ 3 ( y + 2)
16. 18a 2 − 50= 2 ( 3a + 5 )( 3a − 5 )
17. x 2 + 8 x + 15 = ( x + 5 )( x + 3)
18. 16 x 3 − 20 x 2 − 50 x= 2 x ( 2 x − 5 )( 4 x + 5 )
19. 9 x 2 + 33 x + 30 = 3 ( x + 2 )( 3 x + 5 )
20. 12 x 2 + 2 x −=
2 2 ( 3 x − 1)( 2 x + 1)
10
RATIONAL EXPRESSIONS
Rational expressions can be described as a polynomial fraction, or as the ratio of two polynomials.
Examples:
x 2 − 2x + 4
x 4 + 6x 2 + 8
x
x+2
Working with rational expressions is similar to working with fractions. Simplify the fraction as much
as possible to work with simpler terms. In the case of rational expressions we simplify by factoring
each polynomial and canceling terms.
Reducing Rational Expressions to the Lowest Term
To reduce a rational expression to its lowest term, factor each polynomial and cancel like terms that are
both in the numerator and the denominator.
Examples:
Simplify
x−2
x2 − 4
x−2
(x + 2)(x − 2)
1
=
(x + 2)
→
Factor each polynomial, if possible.
→
Cancel like terms.
x2 − 9
to the lowest terms
x 2 + 6x + 9
(x + 3)(x − 3)
x2 − 9
=
2
x + 6 x + 9 (x + 3)(x + 3)
x−3
=
x+3
Reduce
Multiplication of Rational Expressions
Multiplying two rational expressions is done in the same way as fraction multiplication. Multiply the
numerators together and the denominators together. However, it is easier if you factor each polynomial
before multiplying them together. If you multiply two polynomials you’ll end up with a really big
polynomial, therefore it is easier to first factor and then combine.
11
Example:
Multiply
x+2
x2 − 9
x−3
x2 − 4
and
( x + 2 )( x − 3)
x + 2 x −3
⋅ 2
=
2
x −9 x −4
x2 − 9 x2 − 4
(
)(
( x + 2 )( x − 3)
=
( x − 3)( x + 3)( x − 2 )( x + 2 )
=
→
)
Factor each polynomial completely.
1
( x + 3)( x − 2 )
Division of Polynomials
As with multiplication, division of polynomials is similar to fractions. The same rules apply. Multiply
the first rational expression by the reciprocal of the second expression. Remember to factor each
polynomial before multiplying to simplify the polynomials.
x 2 − 5 x − 6 2 x 2 − 11x − 6
÷
x2 − 2x − 3
x2 − 9
Example:
→
Take the reciprocal of the second fraction
and multiply the expressions.
x2 − 5x − 6
x2 − 9
⋅
x 2 − 2 x − 3 2 x 2 − 11x − 6
=
→
( x − 6 )( x + 1) ⋅ ( x − 3)( x + 3)
( x − 3)( x + 1) ( x − 6 )( 2 x + 1)
=
=
Factor each polynomial completely.
→
Cancel common factors.
x+3
2x +1
Addition and Subtraction of Rational Expressions
To add or subtract rational expressions it is necessary to first find the LCD, or least common
denominator, as with fractions. The LCD for a set of rational expressions is the smallest quantity
divisible by each of the denominators.
Examples:
Add:
x −1
x+2
+ 2
2
x − 4 x − 5x + 6
Step 1 – Factor both denominators.
x 2 − 4 = ( x − 2 )( x + 2 )
x 2 − 5 x + 6 = ( x − 2 )( x − 3)
Step 2 – The LCD will consist of each different term obtained from the factors
of both denominators.
12
The LCD is: ( x − 2 )( x + 2 )( x − 3)
Step 3 – Change each rational expression to an equivalent expression having the
LCD as its denominator. Multiply the missing term from the LCD to
the numerator.
(x − 1)(x − 3)
(x − 2)(x + 2)(x − 3)
→
(x + 2)(x + 2)
(x − 2)(x + 2)(x − 3)
→
The LCD is ( x − 2 )( x + 2 )( x − 3) ;
the missing factor here is ( x − 3) .
The LCD is ( x − 2 )( x + 2 )( x − 3) ;
the missing factor here is ( x + 2 )
Step 4 – Now that we have a common denominator, an LCD, we can add the two
rational expressions together.
(x − 1)(x − 3) + (x + 2)(x + 2)
(x − 2)(x + 2)(x − 3) (x − 2)(x + 2)(x − 3)
=
(x − 1)(x − 3) + (x + 2)(x + 2)
(x − 2)(x + 2)(x − 3)
=
x 2 − 3x − x + 3 + x 2 + 2 x + 2 x + 4
→ Add common terms
(x − 2)(x + 2)(x − 3)
2x 2 + 7
=
(x − 2)(x + 2)(x − 3)
Subtract:
→ Eliminate parentheses by distribution.
→ Final answer.
2x −1
7
− 2
x + 5 x + 6 x − x − 12
2
Step 1 – Factor both denominators.
x2 + 5x + 6 =
x 2 − x − 12 =
( x + 3)( x + 2 )
( x − 4 )( x + 3)
Step 2 – The LCD will consist of each different term obtained from the factors
of both denominators.
The LCD is ( x + 3)( x + 2 )( x − 4 )
13
Step 3 – Change each rational expression to an equivalent expression having the
LCD as its denominator. Multiply the missing term from the LCD to
the numerator.
2x −1
=
x + 5x + 6
( x − 4)
2x −1
⋅ =
( x + 3)( x + 2 ) ( x − 4 )
7
=
x − x −1
+ 2)
7 ( x + 2)
( x=
( x2− 4 )( x + 3) ( x + 2 ) ( x + 3)( x + 2 )( x − 4 )
2
2
7
( 2 x − 1)( x − 4 )
( x + 3)( x + 2 )( x − 4 )
⋅
Step 4 – Now that we have a common denominator, an LCD, we can add the two
rational expressions together.
7 ( x + 2)
( 2 x − 1)( x − 4 ) −
( 2 x − 1)( x − 4 ) − 7 ( x + 2 )
=
( x + 3)( x + 2 )( x − 4 ) ( x + 3)( x + 2 )( x − 4 )
( x + 3)( x + 2 )( x − 4 )
14
=
2 x 2 − 8 x − x + 4 − 7 x − 14
(x + 3)(x + 2)(x − 4)
=
2 x 2 − 16 x − 10
(x + 3)(x + 2)(x − 4)
=
2 x 2 − 8x − 5
(x + 3)(x + 2)(x − 4)
(
)
RATIONAL EXPRESSIONS – EXERCISES
Reduce to lowest terms:
1.
x−2
x2 − 4
2.
5 x + 25
x 2 − 25
3.
x2 − 2 x + 1
x −1
4.
x−3
2
x − 6x + 9
5.
x2 − 4
x2 − 4 x + 4
6.
2 x 2 + 5x − 3
x2 − 9
Perform the indicated operations:
7.
x−3
x+2
⋅ 2
2
x − 4 x − 6x + 9
9.
3x 2 − 2 x − 8 x 2 − 4
÷
2 x 2 + 3x − 2 3x + 4
8.
x + y x2 − 2 x + 1
⋅
x − 1 x2 − y2
10.
x 2 + 7 x + 12 x 2 + 9 x + 18
÷ 2
x −5
x − 7 x + 10
11.
x+3
x −1
+
2x −1 2x −1
12.
1
5
+ 3
2
3x
2x
13.
1
3
+ 3
x − 3 x − 27
14.
x
3
+ 2
x − 6 x + 9 2 x − 5x − 3
15.
2
4
− 2
x − x − 12 x + 6 x + 9
16.
x
1
− 2
2 x − 3 x − 20 2 x + 7 x + 5
2
2
2
15
RATIONAL EXPRESSIONS – ANSWERS TO EXERCISES
1.
x−2
x−2
1
=
=
2
x − 4 ( x − 2 )( x + 2 ) x + 2
2.
5 x + 25
5( x + 5)
5
=
=
2
x − 25 ( x − 5)( x + 5) x − 5
3.
x 2 − 2 x + 1 ( x − 1)( x − 1)
=
= x −1
x −1
x −1
x −3
4. =
2
x − 6x + 9
x −3
1
=
( x − 3)( x − 3) x − 3
5.
x2 − 4
( x − 2 )( x + 2 ) x + 2
=
=
2
x − 4 x + 4 ( x − 2 )( x − 2 ) x − 2
6.
2 x 2 + 5 x − 3 ( x + 3)( 2 x − 1) 2 x − 1
=
=
x2 − 9
( x − 3)( x + 3)
x−3
7.
x−3
x+2
1
( x − 3)( x + 2 )
⋅ 2
=
=
2
x − 4 x − 6 x + 9 ( x − 2 )( x + 2 )( x − 3)( x − 3) ( x − 2)( x − 3)
8.
x + y x 2 − 2 x + 1 ( x + y )( x − 1)( x − 1) x − 1
⋅
=
=
x − 1 x2 − y2
( x − 1)( x + y )( x − y ) x − y
9.
3x 2 − 2 x − 8 x 2 − 4 3x 2 − 2 x − 8 3x + 4
÷
=
⋅
2 x 2 + 3x − 2 3x + 4 2 x 2 + 3x − 2 x 2 − 4
=
10.
x 2 + 7 x + 12 x 2 + 9 x + 18 x 2 + 7 x + 12 x 2 − 7 x + 10
=
⋅ 2
÷ 2
x −5
x − 7 x + 10
x −5
x + 9 x + 18
=
16
( 3 x + 4 )( x − 2 )( 3 x + 4 )
( 3x + 4) 2
=
( 2 x − 1)( x + 2 )( x − 2 )( x + 2 ) ( 2 x − 1)( x + 2 ) 2
( x + 4 )( x + 3)( x − 5)( x − 2 ) ( x + 4 )( x − 2 )
=
( x − 5)( x + 6)( x + 3)
x+6
11.
x+3
x − 1 ( x + 3) + ( x − 1) 2 x + 2 2 ( x + 1)
+
=
=
=
2x −1 2x −1
2x −1
2x −1
2x −1
12.
1
5
2 x 15 2 x + 15
+ 3= 3+ 3=
2
3x
2x
6x 6x
6x3
13.
1
3
1
3
+ 3
=
+
2
x − 3 x − 27 x − 3 ( x − 3)( x + 3x + 9)
=
14.
1( x 2 + 3x + 9)
3
x 2 + 3x + 12
+
=
( x − 3)( x 2 + 3x + 9) ( x − 3)( x 2 + 3x + 9) ( x − 3)( x 2 + 3x + 9)
x
x
3
3
+ 2
=
+
x − 6 x + 9 2 x − 5 x − 3 ( x − 3)( x − 3) ( 2 x + 1)( x − 3)
2
x ( 2 x + 1) + 3 ( x − 3)
=
( x − 3)( x − 3)( 2 x + 1)
15.
2 x2 + 4 x − 9
( x − 3) ( 2 x + 1)
2
2
4
2
4
− 2
=
−
x − x − 12 x + 6 x + 9 ( x − 4 )( x + 3) ( x + 3)( x + 3)
2
2 ( x + 3) − 4 ( x − 4 )
=
( x − 4 )( x + 3)( x + 3)
16.
2 x2 + 4 x − 9
=
( x − 3)( x − 3)( 2 x + 1)
−2 x + 22
=
( x − 4 )( x + 3)( x + 3)
−2 ( x − 11)
( x − 4 )( x + 3)
2
x
x
1
1
− 2
=
−
2 x − 3 x − 20 2 x + 7 x + 5 ( 2 x + 5)( x − 4 ) ( 2 x + 5)( x + 1)
2
=
x ( x + 1) − 1( x − 4)
x2 + 4
=
( 2 x + 5)( x − 4 )( x + 1) ( 2 x + 5)( x − 4 )( x + 1)
17
LINEAR FUNCTIONS
As previously described, a linear equation can be defined as an equation in which the highest exponent
of the equation variable is one. A linear function is a function of the form f ( x=
) ax + b . The graph of
a linear equation is a graphic view of the set of all points that make the equation true. The graph of any
linear function is a straight line.
A linear function can be represented in two ways, standard form and slope-intercept form. Standard
form is a formal way of writing a linear equation, while slope-intercept form makes the equation easier
to graph.
Form
Equation
Note
Standard
Ax + By =
C
A and B are not 0. A > 0
Slope-intercept
=
y mx + b
m is the slope of the line
and b is the y-intercept.
To graph a linear function we must first identify the x and y intercepts. The x-intercept is the point
where the graph crosses the x-axis and the y-intercept is the point where the graph crosses the y-axis.
To find the x-intercept:
1. Set y = 0 in the equation.
2. Solve for x. The value obtained is the x-coordinate of the x-intercept.
3. The x-intercept is the point (x, 0), with x the value found in step 2.
18
To find the y-intercept:
1. Set x = 0 in the equation.
2. Solve for y. The value obtained is the y-coordinate of the y-intercept.
3. The y-intercept is the point (0, y), with y the value found in step 2.
Vertical Lines
Equations of the form x = a are vertical lines. The x-coordinate of every point on the vertical line x = a
has the value "a," or any given number.
Horizontal Lines
Equations of the form y = a are horizontal lines. The y-coordinate of every point on the horizontal line
y = a has the value "a," or any given number.
19
Slope
The slope of a line refers to the slant or inclination of the line. The slope is the ratio of the vertical
change to the horizontal change between two points on the line. The slope can also be called the rise
over run ratio because it tells how many spaces to move up or down and how many spaces to move to
the right. A positive sign will move the line up and a negative sign will move the line down. One
important thing to remember is that the run will always be to the right, regardless of the sign.
=
m
rise change in y ∆x
=
=
run change in x ∆y
The formula to find the slope of a line passing through the points ( x1 , y1 ) and
m=
( x2 , y2 ) is:
y2 − y1
x2 − x1
Note: A horizontal line has slope of 0, while a vertical line has an undefined slope.
Example:
Find the slope of the line:
Use any two points on a line to calculate its slope. Your answer will be the same no
matter which points you choose.
y
5
Choosing the points (1, 2) and (2, 0):
4
3
m=
y 2 − y1 0 − 2 − 2
=
=
= −2
x 2 − x1 2 − 1
1
2
1
x
−3
−2
1
−1
−1
−2
−3
4
20
2
3
4
5
6
Parallel Lines
In the y-intercept form equation, y = mx + b , m is the slope and b is the y-intercept. Two lines are
parallel if their slopes are the same ( m1 = m2 ) and their y-intercepts are different ( b1 ≠ b2 ).
Example:
1) y = 2 x + 9
m1 = 2
b1 = 9
2) y = 2 x − 1
m2 = 2
b2 = −1
In the previous example the slope of both equations is the same and their y-intercepts are different.
Therefore, the lines are parallel.
y
8
6
4
2
x
−8
−6
−4
−2
2
4
6
8
10
−2
−4
−6
−8
10
Perpendicular Lines
Two lines are perpendicular if the slopes are the negative reciprocal of each other: m1 = −
Example:
m1 = 4
1) y = 4 x + 7
1
2) y = − x + 2
4
1
.
m2
b1 = 7
1
4
m2 = −
b2 = 2
The slope of equation 2 is the negative reciprocal of the slope of equation 1. Therefore, the lines are
perpendicular.
y
8
6
4
2
x
−8
−6
−4
−2
2
4
6
8
10
−2
−4
−6
−8
10
21
Finding the Equation of a Line
To find the equation of a line when only a set of points or a slope is given, use the point-slope form of
a linear equation formula:
y − y1 = m(x − x1 )
where m is the slope of the line and ( x1 , y1 ) is a point on the line.
Examples:
Find the equation of the line which passes through the point (–4, 2) and whose slope is 5.
Using the point-slope form equation we obtain:
y − y1 = m(x − x1 )
y − 2 = 5( x − (− 4 ))
y − 2 = 5 x + 20
y = 5 x + 22
← Resultant Equation
Find the equation of the line through the points (–2, 5) and (–6, 4).
First find the slope between these two points using the slope equation:
m=
y 2 − y1
−1 1
4−5
=
=
=
x 2 − x1 − 6 − (− 2 ) − 4 4
With the slope obtained and one of the two points given, use the point-slope form equation to
find the equation of the line.
y − y1 = m( x − x1 )
1
(x − (− 2))
4
1
1
y −5 = x+
4
2
1
11
y = x+
4
2
y −5 =
← Resultant Equation
Eventually, you may be asked to find the equation of a line that is parallel or perpendicular to a given
line. Remember that the slopes of two parallel lines are exactly the same and that the slopes of two
perpendicular lines are the negative reciprocals of each other.
22
Examples:
Find the equation of a line that is parallel to =
y 4 x + 2 and that passes through the point (2, 7).
=
y 4x + 2
m1 = 4
← Since the equation is in slope-intercept form, the coefficient of x is
the slope of the line. Remember y=mx+b (m is the slope!).
Since the lines must be parallel, use the same slope and the given point to find the equation of
the parallel line:
m
=
m=
4
1
2
y − y1= m ( x − x1 )
y − 7= 4 ( x − 2 )
y − 7 = 4x − 8
← Parallel line to =
=
y 4x −1
y 4x + 2 .
Find the equation of a line perpendicular to 4 x + 2 y =
9 and that passes through the point
(–1, –1).
4x + 2 y = 9
2 y = −4 x + 9
− 4x + 9
y=
2
9
y=
−2 x +
2
← This equation is in standard form. We need to convert it
to slope-intercept form: y=mx+b.
← The slope is m1 = −2 .
Since the second equation must be perpendicular to the first equation, find the negative
reciprocal.
1 1
1
→
m2 =
−
=
m2 = −
−2 2
m1
1
and the point (–1, –1):
2
y − y1= m ( x − x1 )
Using the slope m2 =
1
( x − ( −1) )
2
1
1
y + 1=
x+
2
2
1
1
← Perpendicular line to 4 x + 2 x =
=
y
x−
9.
2
2
y − ( −1=
)
23
LINEAR FUNCTIONS – EXERCISES
1. Identify the false statement below
a) The standard form of a linear equation is =
y ax + b
b) The point (4, 0) is on the graph of 3 x − 4 y =
12
c) The graph of 3 x − 4 y =
12 passes the vertical line test for functions
d) The y-intercept of 3 x − 4 y =
12 is (0, –3)
2. Identify the false statement below
a) To find the x-intercept of a graph, set y equal to zero and solve the equation for x
b) The graph of x = 3 is a horizontal line three units above the x-axis
c) Generally speaking, it is a good idea to plot three points when constructing the graph of a linear
function
d) To find the y-intercept of a graph, set x equal to zero and solve the equation for y
3. The standard form of the equation 5 y − 6 x =
30 is
6
a) −6 x + 5 y =
b) =
y
x+6
30
5
c) 5 y − 6 x =
d) 6 x − 5 y =
30
−30
4. Identify the true statement below
a) All linear graphs are functions
b) The graph of y = 4 is a function
c) To find the y-intercept of a function, let y equal to zero and solve for x
d) The graph of x = 4 is a function
5. Determine the slope and y-intercept of each equation:
a) 6 x + 7 y =
b) 2 x + 5 =
0
14
c) 3 y + 9 =
0
6. Use the point slope form to find the equation of the line which passes through (2, –1) and whose
3
slope is .
7
7. Write in standard form the equation of the line passing through the point (–3, –4) with slope equal to
3
m= .
2
8. Which of the following is the equation of a line in standard form passing through the point (3, 0) and
perpendicular to the line y − 2 x =
−1 ?
a) 2 x + y =
b) x + 2 y =
6
3
c) 2 x − y =
d) x − 2 y =
6
3
9. Find the equation of a line in standard form that passes through the points (–1, 4) and (3, 2).
10. Determine whether the two lines are parallel, perpendicular or either: 3 x=
− 5 y 6 and 6 x =
− 10 y 7
24
LINEAR FUNCTIONS – ANSWERS TO EXERCISES
1. A is the false statement. The standard form of a linear equation is Ax + By =
C.
2. B is the false statement. The graph of x = 3 is a vertical line 3 units to the right of the y-axis.
3. D: 6 x − 5 y =
−30 , A > 0
4. B: y = 4 is an example of a linear function.
6
5. a) m =
− and b =
2
7
b) m = undefinded and b does not exist
c) m = 0 and b = −3
6. =
y
3
13
x−
7
7
7. 3 x − 2 y =
−1
8. B: x + 2 y =
−1
3 is a perpendicular line to y − 2 x =
9. x + 2 y =
7
10. The lines are parallel.
25
GRAPHING LINEAR EQUATIONS
A linear equation can be defined as an equation in which the highest exponent of the equation variable
is one. When graphed, the equation is shown as a single line. A linear equation has only one solution.
The solution of a linear equation is equal to the value of the unknown variable that makes the linear
equation true.
A linear equation in the form of ax + by = c is said to be in standard form. A linear equation in the form
of =
y mx + b is said to be in slope intercept form. The solution set for linear equations in either form
is not just a single number but a set of ordered pairs of numbers ( x, y ) . The solution set can be
represented graphically on a Cartesian coordinate system, and its graph is a straight line.
Graphing a Linear Equation in Standard Form
The easiest way to graph a linear equation in standard form is by plotting its points. Different set of
points can be found by giving one of the variables a value and solving for the other variable. At the end
all the points can be plotted in the graph and joined together by a straight line.
Example:
Draw a graph of 3 x + 2 y =
12
Step 1 – Assign values to one unknown and calculate the corresponding values
for the other unknown.
x
0
2
4
y
6
3
0
Step 2 – Plot these points on a graph. Connect the points with a straight line.
If the points do not lie in a straight line there is probably an error in one
of the points used.
y
7
6
5
4
3
2
1
x
−3
−2
−1
−1
−2
−3
4
26
1
2
3
4
5
6
Graphing Linear Equations in Slope Intercept Form
Linear equations in slope intercept form can also be graphed using the points system. However, the
form in which the equation is written clearly tells us the slope and the y-intercept and thus makes it
easier to plot with just two points. Remember that in the form y = mx + b , m is the slope and b is the yintercept. This is the slope intercept form.
Example:
Graph 3 x + 2 y = 12
Step 1 – Since the equation is in standard form, solve for y to get the equation in
slope-intercept form =
y mx + b
y=
− 3 x + 12 − 3 x 12
3
=
− x+6
=
+
2
2
2
2
Step 2 – Remember that in the slope-intercept form equation,=
y mx + b , m is
the slope and b is the y-intercept. By looking at the equation we can
know our slope and y-intercept:
m= −
3
2
→ Slope
b=6
→ y-intercept
Step 3 – Using the slope and the y-intercept, graph the equation of the line. Start
by finding the y-intercept on the graph. From there find the slope.
Remember that the slope is rise over run, in this case the slope is
m = − 3 2 , which means that from the y-intercept point we need to
move three spaces down since the slope is negative and then two places
to the right. Remember that you always move to the right. The negative
sign does not indicate moving to the left; it is only an indicator for
going up or down. Always move to the right.
y
9
8
7
slope – starting from your y-intercept, move three
6
y-intercept
(0, 6)
units down (because of negative sign), and then two
units to the right m = − 3 2
5
4
3
2
1
x
−4 −3 −2 −1
−1
1
2
3
4
5
6
7
8
9 10
−2
−3
−4
5
Step 4 – After you have found the two points, draw a straight line connecting the
two points together. This is the graph of your equation.
27
y
9
8
7
draw a connecting line between the two
points obtained
6
5
4
3
2
1
x
−4 −3 −2 −1
−1
−2
−3
−4
5
28
1
2
3
4
5
6
7
8
9 10
SYSTEMS OF LINEAR EQUATIONS
Systems of linear equations refer to a set of two or more linear equations used to find the value of the
unknown variables. If the set of linear equations consist of two equations there will be two unknown
variables. If the set consists of three equations there will be three variables and so on. If there are more
variables present than linear equations, the system cannot be solved.
There are three general methods for solving systems of linear equations: graphically, by method of
elimination and by substitution. Remember that in order to be able to solve the system, it is necessary
that you have the same number of equations and unknown variables.
Solving a System of Linear Equations Graphically
If you are given a system of two linear equations with two unknowns the system can be solved and will
have two answers, one for each of the variables. The graph of each linear equation will be a straight
line, and the point of intersection of the two straight lines represents the solution to the system of
equations. Thus, the solution to a system of two linear equations with two unknowns is an ordered pair
of numbers ( x, y ) . It is called a consistent system.
Example:
2x + 3y =
5
5x + 2 y =
7
Thus, the solution is x = 1 and y = 1, or (1, 1).
29
Method of Elimination
This method consists on eliminating one variable by addition or subtraction of the linear equations. To
be able to cancel a variable, the variable needs to have the same coefficient in both equations, however,
most of the time this will not be the case. If the variables have a different coefficient, multiply the
coefficients of each equation with the opposite equation.
Example:
3x − y = 4
− 4x + 2 y = 2
Step 1 – Start by canceling one of the variables. You can choose whichever
variable you want to start with. In this case we are going to start by
cancelling the x variable.
To cancel the x multiply the coefficient of x in the first equation with
the entire second equation, and vice versa. For this system we do not
need to worry about changing signs since they are different already.
4(3 x − y = 4 )
3(− 4 x + 2 y = 2 )
→
12 x − 4 y = 16
− 12 x + 6 y = 6
Step 2 – Add both equations together to cancel the x and solve for y.
12 x − 4 y = 16
− 12 x + 6 y = 6
2 y = 22
22
y=
2
y = 11
Step 3 – Use the variable that we just found, y, and plug it into any of the
equations to obtain x.
3x − y = 4
3 x − 11 = 4
3 x = 4 + 11
15
3
x=5
x=
The solution is x = 5 and y = 11, or (5, 11).
30
Method of Substitution
The method of substitution consists of solving one of the equations for any variable and then
substituting the resultant equation into the other equation, thus leaving a one-equation, one-unknown
system.
Example:
3x − y = 4
− 4x + 2 y = 2
Step 1 – Pick one equation and solve for one variable. For this system we are
going to use the first equation and solve for y.
3x − y = 4
3x − 3x − y = 4 − 3x
− y = 4 − 3x
y = 3x − 4
Step 2 – Use the new equation to substitute the value of y into the second
equation.
− 4x + 2 y = 2
− 4 x + 2(3 x − 4 ) = 2
− 4x + 6x − 8 = 2
2x − 8 + 8 = 2 + 8
2 x = 10
10
x=
2
x=5
Step 3 – Plug in the value of the x variable into the equation obtained in step 1.
y = 3x − 4
y = 3(5) − 4
y = 15 − 4
y = 11
The solution is x = 5 and y = 11, or (5, 11).
31
Parallel Lines
If two lines are parallel they have the same slope and thus no point of intersection since they run in the
same direction and they never meet. When a system of linear equations consist of two parallel lines the
system is said to have no solution since the lines have no point of intersection. The system is said to be
inconsistent.
Example:
2 x + 3 y =1 → − 2 ( 2 x + 3 y =1) → − 4 x − 6 y =−2
4x + 6 y = 7 →
4x + 6 y = 7 →
← Parallel lines
4x + 6 y = 7
← No solution
0 + 0 =5
Coinciding Lines
When a system of linear equations consist of two lines that have the same straight line when graphed,
then the two equations are equivalent and the lines are said to be coinciding lines. Since they are
basically the same line, any point that satisfies one equation will satisfy the second equation. Therefore
the system has an infinite number of solutions. The system is said to be dependent.
Example:
2x − 4 y = 8
→
2x − 4 y = 8
→
x − 2y =
4 → − 2( x − 2y =
4) →
2x − 4 y =
8
−2 x + 4 y =
−8
0=0
32
← Coinciding lines
← Infinite number of
solutions
SYSTEM OF LINEAR EQUATIONS – EXERCISES
1.
3x + 4 y =
10
−6 x + 3 y =
−9
2.
7x + 4 y =
−5
−2 x + 5 y =
26
3.
2x + 4 y =
−12
3x + 5 y =
−16
4.
x + 2y =
5
2x + 4 y =
1
6.
6x + 5 y =
28
7x + 2 y =
2
3x + 2 y =
1
4x − 3y =
2
7x − 2 y =
2
5. 1
1
1
x− y =
2
7
7
7.
3x − 2 y =
1
9x − 6 y =
3
8.
9.
3x − 2 y =
4
12 x − 8 y =
6
10.
2x + 4 y =
11
6x + 2 y =
3
33
SYSTEM OF LINEAR EQUATIONS – ANSWERS TO EXERCISES
1.
3 x + 4 y = 10 → 6 x + 8 y = 20
−6 x + 3 y =
−9 → −6 x + 3 y =
−9
2.
7x + 4 y =
−5 → 14 x + 8 y =
−10
y 26 → −14 x + 35=
y 182
−2 x + 5=
11 y = 11
y =1
43 y = 172
y=4
3 x + 4 (1) = 10 → x = 2
Solution (2, 1)
3.
2x + 4 y =
−12
3 x + 5 y =−16
7 x + 4 ( 4) =
−5 → x =
−3
Solution (–3, 4)
→
6 x + 12 y =
−36
→ −6 x − 10 y =32
4.
x + 2y =
5
2x + 4 y = 2
→ − 2x − 4 y =
−10
6→ 2 x + 4 y = 1
2y = −4
y = −2
0=9
2 x + 4 ( −2 ) =
−12 → x =
−2
Solution (–2, –2)
5.
7x – 2y = 2
7x − 2 y = 2 →
1
1
1
x− y =
→ –7x + 2y = –2
2
7
7
0+0=0
No Solution
6.
7x + 2y = 2
→
→
–9x + 6y = –3
9x – 6y = 3
0+0=0
Lines Coincide
12x + 10y = 56
→ –35x – 10y = –10
–23x = 46
x = –2
6(–2) + 5y = 28 → y =
8
Solution (–2, 8)
Lines Coincide
Infinitely Many Solutions
7. 3x – 2y = 1
9x – 6y = 3
6x + 5y = 28 →
8.
3x + 2y = 1
4x – 3y = 2
→
→
1
( 17 ) + 2 y =→
3 7
Infinitely Many Solutions
9x + 6y = 3
8x – 6y = 4
17x = 7
x=7
−2
y=
(
Solution 7
9.
3x – 2y = 4 →
12x – 8y = 6 →
–12x + 8y = – 16
12x – 8y = 6
0 = –10
No Solution
10.
17
, −2
17
)
2x + 4y = 11 → 2x + 4y = 11
6x + 2y = 3 → –12x – 4y = –6
–10x = 5
x= −1
2
2(− 1 2 ) + 4 y = 11 → y = 3
Solution − 1 , 3
2
(
34
17
17
)
LINEAR INEQUALITIES
When we use the equal sign in an equation we are stating that both sides of the equation are equal to
each other. In an inequality, we are stating that both sides of the equation are not equal to each other.
It can also be seen as an order relation; that is, it tells us which one of the two expressions is smaller, or
larger, than the other one. A linear inequality is an equation in which the highest variable exponent is
one.
Examples:
→
→
→
→
x<3
2x + 5 ≤ x − 3
x > −2
3x − 2 ≥ x − 6
less than, <
less than or equal to, ≤
greater than, >
greater than or equal to, ≥
The solution to an inequality is the value of the variable which makes the statement, or the inequality,
true.
Examples:
x<3
→
This inequality is telling us that “x is less than 3”. Therefore, any
number less than three is a possible solution. Remember that on
the number line, any number to the left is less than a given
number, and any number to the right of that given number is
greater.
Notice how 3 is marked with an
empty dot, this means that the answer
does not include the number 3.
–3 – 2 –1 0
2x + 5 ≤ x − 3
→
2x + 5 ≤ x − 3
2x + 5 − 5 ≤ x − 3 − 5
2x ≤ x − 8
2x − x ≤ x − x − 8
x ≤ −8
1
2
3
This inequality is telling us that the equation 2 x + 5 is
less than or equal to the equation x − 3 . Solve for x to find
which number, or numbers, make this equation true.
→
Solve for x by leaving the variables on one side
and the numbers on the other side, just as you
would solve a linear equation.
Notice how –8 is marked with a black
dot, this means that the answer does
include the number –8.
–9 –8 –7 –6 –5 –4 –3 –2 –1
0
35
x > −2
→
This inequality is telling us that “x is greater than –2”. Therefore,
any number greater than negative two is a possible solution.
–3 –2 –1
0
→
3x − 2 ≥ x − 6
3x − 2 + 2 ≥ x − 6 + 2
3x ≥ x − 4
3x − x ≥ x − x − 4
2 x ≥ −4
2x − 4
≥
2
2
x ≥ −2
1
–3 –2 –1
2
3
Solve for x by leaving the variables on one side
and the numbers on the other side.
0
1
2
3
As you can see, solving linear inequalities is very similar to solving linear equations. There is only one
thing you need to keep in mind when multiplying or dividing a negative number, the direction or sense
of the inequality is reversed. For example, if the sign is >, after multiplying or dividing by a negative
number the sign changes to <.
Example:
1
x+4≥ x−2
2
1
x+4−4≥ x−2−4
2
1
x ≥ x−6
2
1
x− x ≥ x− x−6
2
1
− x ≥ −6
2
2 1
2
− ⋅ − x ≥ −6 ⋅ −
1 2
1
x ≤ 12
36
→
Solve for x as in the previous examples.
→
Multiply times the negative reciprocal to
leave the x alone and get an answer.
However, since we are multiplying by a
negative the inequality sign must change
from ≥ to ≤ .
EQUATIONS INVOLVING ABSOLUTE VALUE
The absolute value of a number is always the positive value of that number. Since the absolute value is
the distance of a number from the origin, and since distances are always positive, the absolute value is
always a positive value. The same definition applies to equations involving absolute value. Any
equation inside the absolute value bars must always be equal to a positive number.
Example:
x − 4 = −8
→
This absolute value equation does not have a real answer.
Any equation inside the absolute value bars must always be equal to a positive
number. In this case, for x − 4 to be equal to − 8 , x must be − 4 : − 4 − 4 = −8
However, since the equation x − 4 is inside the absolute value bars it will
always yield a positive answer − 8 = 8 and never a negative answer.
When the equation involving absolute value is equal to a positive number, the value of the variable
inside the absolute value bars can be negative or positive; therefore, the equation must be made equal
to a positive and a negative answer. The answer will have two different possible values of x, and both
will make the statement valid.
Examples:
− 3 x + 6 = 12
− 3 x + 6 = 12
− 3 x + 6 − 6 = 12 − 6
− 3x = 6
6
− 3x
=
−3 −3
x = −2
→
Make the equation equal to 12 and − 12 and solve
for x in each case.
and
− 3 x + 6 = −12
− 3 x + 6 − 6 = −12 − 6
− 3 x = −18
− 3 x − 18
=
−3
−3
x=6
The answer to − 3 x + 6 = 12 is x = −2 and x = 6 .
6 + 5 x − 6 = 10
→
5x − 6 = 4
5x − 6 = 4
5x − 6 + 6 = 4 + 6
5 x = 10
and
x=2
Leave the absolute value expression by itself
on the left side of the equation. Then solve
for x as in the previous example.
5 x − 6 = −4
5 x − 6 + 6 = −4 + 6
5x = 2
x=2
5
The answer is x = 2 and x = 2 .
5
37
INEQUALITIES INVOLVING ABSOLUTE VALUE
Inequalities involving absolute value are solved in a similar manner to equations involving absolute
value. Since the variable inside the absolute value sign can be either positive or negative, both positive
and negative values are used to solve for the variable. However, in the case of linear inequalities
involving absolute value, the inequality is placed between the two values rather than in two different
equations.
Examples:
x <2
→
To solve this linear inequality use both positive and
negative values of 2 and place the inequality in the middle
without the absolute value sign. The inequality sign will
be the same on both sides.
−2< x< 2
→
This notation is read: x is greater than − 2 but less than 2
– 3 –2 –1
0
1
2
3
− 4 x + 10 ≥ 2
→ Write both positive and negative values of 2 on each
side of the inequality and leave it in the middle.
− 2 ≥ −4 x + 10 ≥ 2
→ Solve for x by moving each term to both sides of
the inequality.
− 2 − 10 ≥ −4 x + 10 − 10 ≥ 2 − 10
− 12 ≥ −4 x ≥ −8
− 12 − 4 x − 8
→ Remember that when dividing by a negative number
≥
≥
−4
−4 −4
the direction of the inequality sign changes.
3≤ x≤2
→
This notation is read: x is greater than 3 but less than 2.
– 3 –2 –1
38
0
1
2
3
4
5
We explained previously that equations that contain an absolute value cannot be equal to a negative
number since an absolute value will always give you a positive number. In the case of linear
inequalities, if an inequality is greater than or equal to a negative number, then there are infinite
solutions. If the inequality is less than or equal to a negative number, then there is no real solution to
the inequality.
Examples:
x + 2 ≥ −2
→
The absolute value will give you a positive number which
will always be greater than any negative number.
The inequality is true and has infinite solutions.
x + 2 ≤ −2
→
The absolute value will give you a positive number and
no positive number will ever be less than or equal to a
negative number.
The inequality is false and has no real solution.
39
LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS – EXERCISES
Solve for x:
1.
3 x ≤ 27
2.
−5 x ≤ 40
3.
−6 x + 3 ≥ 21
4.
2
− x ≥ 10
3
5.
3 − 2 x ≥ −8 + 3 x
6.
3( 2 x + 1) < −8
7.
− ( x + 1) − 4 x > 4 x − 3
8.
−3 ≤ 4 x + 2 ≤ 10
9.
1
−3 ≤ 2 x + x + 4 ≤ 6
3
10.
3 x − 7 − 5 = 12
11.
2−
12.
x ≥4
13.
2x − 3 ≤ 6
40
x
=4
3
LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS – ANSWER TO
EXERCISES
1. 3 x ≤ 27
x≤9
2. −5 x ≤ 40
x ≥ −8
2
4. − x ≥ 10
3
5.
3. −6 x + 3 ≥ 21
−6 x ≥ 21 − 3
−6 x ≥ 18
x ≤ −3
6. 3( 2 x + 1) < −8
3 − 2 x ≥ −8 + 3 x
−2 x − 3 x ≥ −8 − 3
−5x ≥ −11
11
x≤
5
−2 x ≥ 30
x ≤ −15
7. − ( x + 1) − 4 x > 4 x − 3
8. −3 ≤ 4 x + 2 ≤ 10
−x −1 − 4x > 4x − 3
−9 x > −2
−3 − 2 ≤ 4 x ≤ 10 − 2
−5 ≤ 4 x ≤ 8
2
x<
9
5
− ≤x≤2
4
6 x + 3 < −8
6 x < −11
11
x<−
6
1
9. −3 ≤ 2 x + x + 4 ≤ 6
3
−9 ≤ 6 x + x + 12 ≤ 18
−9 − 12 ≤ 7 x ≤ 18 − 12
−21 ≤ 7 x ≤ 6
−3 ≤ x ≤
11. 2 −
10. 3x − 7 − 5 = 12
3 x − 7 = 12 + 5
3x − 7 =
17
3x − 7 =
17
3 x = 24
x =8
12.
x ≥4
−4≥ x≥ 4
or
x
=4
3
x
2− =
4
3
x
− =
2
3
or
x = −6
6
7
3 x − 7 =−17
3 x = −10
10
x= −
3
2−
x
=
−4
3
x
=
−6
3
x = 18
−
13. 2 x − 3 ≤ 6
−6 ≤ 2 x − 3 ≤ 6
−6 + 3 ≤ 2 x ≤ 6 + 3
− 3 ≤ 2x ≤ 9
3
9
− ≤x≤
2
2
41
GRAPHING LINEAR EQUATIONS IN TWO VARIABLES
To graph a linear inequality with one variable we use the number line. To graph a linear inequality
with two variables we use a two-dimensional graph. As with linear equations, linear inequalities can be
written in standard form or slope intercept form.
Standard Form:
Ax + By < C or Ax + By > C
Ax + By ≤ C or Ax + By ≥ C
y < mx + b or y > mx + b
y ≤ mx + b or y ≥ mx + b
Slope-Intercept Form:
To graph a linear inequality with two variables it is probably easier and recommended to change it to
slope intercept form. There are two important rules to remember when graphing a linear inequality. If
the inequality sign is > or <, the line is dashed. If the inequality is ≥ or ≤ , then the line is solid. Also,
for inequalities that are greater than > or greater or equal to ≥ the graph needs to be shaded above the
line. For inequalities that are less than < or less or equal to ≤ the graph needs to be shaded below the
line.
Examples:
Graph the solutions of 3 x + 2 y ≥ 6 :
Step 1 – Since the equation is in standard form, change it to slope intercept form.
3x + 2 y ≥ 6
→
2 y ≥ −3 x + 6
−3 x + 6
y≥
2
3
y ≥ − x+3
2
← Slope-Intercept Form
Step 2 – Graph the slope-intercept form equation. Since the inequality has the
sign ≥ the line is solid.
42
Step 3 – Since the equation has the sign ≥ , the graph is shaded above the line.
← The shaded area in the
graph is the graphic
answer to the inequality.
It means that any point
above the line, including
the line, satisfies the
inequality and can be a
possible answer.
Graph the solutions of x − 3 y < 4
Step 1 – Since the equation is in standard form, change it to slope intercept form.
x − 3y < 4
→
− 3y < −x + 4
−x + 4
y>
−3
Remember to switch the
inequality sign when
dividing by a negative.
y>
1
4
x−
3
3
← Slope-Intercept Form
Step 2 – Graph the slope-intercept form equation. Since the inequality has the
sign > the line is dashed.
Step 3 – Since the equation has the sign >, the graph is shaded above the line.
43
Solving a System of Linear Inequalities
To solve a system of linear inequalities we need to do it graphically since the solution to a system of
linear inequalities is the set of points whose coordinates satisfy all the inequalities in the system, or
where both shaded areas overlap. Graph the lines as previously shown.
1
y <− x+2
Example:
Determine the solution to the following system of inequalities
2
x− y ≤4
Step 1 – Change the second inequality to slope-intercept form.
x− y≤4
x− x− y ≤ 4− x
− y ≤ −x + 4
y ≥ x−4
← Remember to switch the inequality sign since
we are dividing by a negative number.
Step 2 – Graph both inequalities in slope-intercept form and shade the solution
for each inequality. Since the first inequality has the sign < the line is
dashed. The second inequality has the sign ≥ so the line is solid.
y ≥ x−4
Solution to the
system of
inequalities
y<−
44
1
x+2
2
GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES – EXERCISES
Graph each inequality:
1
1. y ≥ − x
2
2. y < 2 x + 1
3. 10 ≥ 5 x − 2 y
Determine the solution to each system of inequalities:
2x − y < 4
y ≥ −x + 2
5.
y < 3x − 4
6x ≥ 2 y + 8
x≥0
1
1
7. x + y ≥ 2
2
2
2 x − 3 y ≤ −6
x≥0
y≥0
8.
2x + 3y ≤ 6
4x + y ≤ 4
x≥0
y≥0
9.
5 x + 4 y ≤ 16
x + 6 y ≤ 18
4.
6.
− 3 x + 2 y ≥ −5
y ≤ −4 x + 7
x≥0
y≥0
10. x ≤ 15
30 x + 25 y ≤ 750
10 x + 40 y ≤ 800
45
GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES – ANSWERS TO EXERCISES
1.
2.
3.
4.
5.
6.
7.
46
8.
9.
10.
47
RADICALS
For every operation there is an opposite that can be used to cancel the operation or make it equal to
zero.
Example:
+
×
inverse
inverse
x2
–
÷
x
3
inverse
inverse
3
x
x
The opposite of exponents are called radicals. If the exponent is a square, or 2, the opposite would be a
square root; if the exponent is 3, the opposite would be a cubic root, and so on.
Examples:
22 = 4
23 = 8
35 = 243
inverse
inverse
inverse
3
5
4=2
8=2
243 = 3
Radicals can also be written in exponent notation; however, in this case the exponent would be a
fraction.
4 = 41 2
Examples:
3
8 = 81 3
5
243 = 2431 5
The expression 3 8 is called a radical expression, where 3 is called the index,
is the radical sign,
and 8 is called the radicand. The index of a radical expression must always be a positive integer
greater than 1. When no index is written it is assumed to be 2, or a square root.
Negative Radicals – The only restriction that exists for negative signs and radicals is that there cannot
be a negative sign under an even root since there is no real solution to this problem. However, a
negative sign can exist in front of a radical or under odd roots and still be able to obtain a real number.
Examples:
3
− 16
← Negative sign under even root, no real answer.
(4)(4) = 16 or
(− 4)(− 4) = 16
−8
← Negative sign under odd root, real answer.
(− 2)(− 2)(− 2) = −8
− 81
48
← Negative sign is outside the radical, the sign does not affect the
Calculations. It is only carried out to the final answer.
− 81 = −(9 )(9 ) = −81
Radicals Containing Large Numbers – When working with large numbers under a radical sign, it is
always easier to break down the number and work in smaller parts rather than trying to find an exact
root.
Example:
← Break down the number into smaller numbers.
525
21 ⋅
← Find the root of each of the smaller numbers.
25
← Write final answer with single digits first and radicals
at the end.
21 ⋅ 5
5 21
Sometimes it is not easy to find two small numbers that when multiplied give you the large numbers.
In these cases, you may break down the large number little by little.
Example:
← Follow the same procedure as described above. It
takes a little longer, but at the end the same result is
obtained.
525
105 ⋅ 5
105 ⋅
5
21 ⋅ 5 ⋅
21 ⋅
5
25
21 ⋅ 5
5 21
Multiplication and Division of Radicals
Radical multiplication is similar to polynomial multiplication, the same distributive property applies.
For instance, if you had to multiply 3(x + 2), you would take the 3 through the parentheses to get 3x +
6. With radicals it works the same way.
(
)
3 2 3+ =
5
Examples:
(
)
3 2 3 + 3
( 5)
→ Distribute the outside radical to each
term inside the parentheses.
= 2 9 + 15 = 2 ⋅ 3 + 15
= 6 + 15
(
3+ 5
)(
)
3 − 6 = 9 + 15 − 18 − 30
49
= 3 + 15 − 9 ⋅ 2 − 30
=
3 + 15 − 3 2 − 30
Rationalizing the Denominator
When dealing with fractions, a final answer cannot contain radicals in the denominator; therefore, it is
necessary to eliminate any radical from the denominator. The process of removing the radical from the
denominator is called rationalizing the denominator.
Examples:
Simplify the following expression:
2
3
To simplify the expression we need to eliminate the radical in the denominator by
rationalizing the denominator. It is necessary to remove the radical sign from 3 and
leave the 3 by itself. To do this, multiply 3 by another 3 and you'll get 3.
Remember that x = x1 2 , and when you multiply the exponents add together.
Therefore, x ⋅ x = x or x1 2 ⋅ x1 2 = x . However, don’t forget to multiply the
numerator as well to compensate the change.
2
2
3 2 3
=
⋅
=
3
3
3 3
2
3+ 6
→
→ Multiply the denominator by 3 to eliminate the radical.
When the denominator is a little more complex, multiply the
denominator by its conjugate and the radicals will disappear. The
conjugate is the same denominator but with a different sign in
between: in this case 3 − 6 .
2
2
3− 6
=
⋅
3+ 6 3+ 6 3− 6
=
=
50
(
2 3− 6
)
9−3 6 +3 6 − 3 6
6−2 6 6−2 6
=
9−6
3
RADICALS - EXERCISES
Simplify:
1.
27
2.
72
3.
4
4.
5.
16
50
3
250
6.
1
5
7.
3
5
8.
( 2 + 3 )( 2 − 3 )
9.
5
2+ 3
10.
11.
6
5− 2
x− y
x+ y
12. 4 2 −
1
+ 32
8
13. 6 + 4 2 − 2 + 3
14.
15.
3 + 27
(3
6− 3
)
2
51
RADICALS - ANSWERS TO EXERCISES
1.
27 = 3 ⋅ 9 = 3 3
2.
72 = 36 ⋅ 2 = 6 2
3.
4
50 = 25 ⋅ 2 = 5 2
4.
5.
16 = 2
3
250 = 3 125 ⋅ 2 = 53 2
6.
5
5
7.
15
5
8. 1
9. 10 − 5 3
10. 2 5 + 2 2
11.
x2 − 2x y + y
x2 − y
12.
31 2
4
13. 9 + 3 2
14. 4 3
15. 57 − 18 2
52
QUADRATIC EQUATIONS
A quadratic equation is always written in the form of:
ax2 + bx + c = 0
where a ≠ 0
The form ax2 + bx + c = 0 is called the standard form of a quadratic equation.
Examples:
x 2 − 5x + 6 = 0
This is a quadratic equation written in standard form.
x 2 + 4 x = −4
This is a quadratic equation that is not written in standard form but
can be once we set the equation to: x 2 + 4 x + 4 = 0 .
x2 = x
This too can be a quadratic equation once it is set to 0.
x 2 − x = 0 (standard form with c=0).
Solving Quadratic Equations by Square Root Property
When x 2 = a , where a is a real number, then your x = ± a
Examples:
x2 − 9 = 0
x2 = 9
y2 + 3 =
28
y 2 + 3 − 3 = 28 − 3
y 2 = 25
x =± 9
y = ± 25
x = ±3
y = ±5
x2 − 9 = 0
Solving Quadratic Equations by Factoring
It can also be solved by factoring the equation. Remember to always check your solutions. You can use
direct substitution of the solutions in the equation to see if the solutions satisfy the equation.
Examples:
x 2 − 5x + 6 = 0
(x − 3)(x − 2) = 0
x−3= 0
x=3
x−2 = 0
x=2
← Factoring x
← Set it equal to 0 and solve for x
53
Now check if, x = 3 and x = 2 are the solutions of x 2 − 5 x + 6 = 0
Check:
32 − 5( 3) + 6 = 0
2 2 − 5( 2) + 6 = 0
9 − 15 + 6 = 0
4 − 10 + 6 = 0
2 x2 + 7 x − 4 = 0
( 2 x − 1)( x + 4 ) = 0
2x −1 = 0
2x = 1
1
x=
2
x+4 = 0
x = −4
Another method of checking the solutions is by using one of the following
statements:
c
b
or
The product of the solutions =
a
a
where a, b, and c are the coefficients in ax2 + bx + c = 0 .
The sum of the solutions = −
Now we check if x =
Check:
1
and x = − 4 are the solutions of 2 x 2 + 7 x − 4 = 0
2
Using the sum of the solutions =
1
7
+ ( −4 ) = −
2
2
Based on the original equation = −
b
7
=−
a
2
Now by using the product of the solutions =
Based on the original equation =
1
( −4 ) = −2
2
c −4
=
= −2
a
2
2 8
0 ← Rewrite in standard form by multiplying each side of the
− =
x x2
equation by x 2
x2 + 2 x − 8 = 0
( x + 4 )( x − 2 ) = 0
1+
x+4=0
54
x−2=
0
x = −4
Check:
x=2
2
8
 Solutions must be checked in the
−
=
0
−4 ( −4 )2
1+
original equation to avoid any errors.
1 1
1− − =
0
2 2
2 8
1+ − 2 =
0
2 2
1+1–2=0
Solution Using the Quadratic Formula
Factoring is useful only for those quadratic equations which have whole numbers. When you
encounter quadratic equations that can not be easily factored out, use the quadratic formula to find the
value of x:
x=
Examples:
x 2 − 8 = −2 x
x2 + 2 x − 8 = 0
x=
−b ± b 2 − 4 ac
2a
← Rewrite in standard form, where a = 1, b = 2, and c = −8
−2 ± 4 − 4 (1)( −8)
2 (1)
−2 ± 36
2 (1)
−2 ± 6
=
2
= 2 , −4
← Plug in numbers into the equation
=
← The two rational solutions
3 x 2 − 13 x + 4 = 0
− ( −13) ± ( −13) 2 − 4( 3)( 4)
x=
2( 3)
13 ± 121
=
6
13 ± 11
=
6
1
24 2
← The two rational solutions
=
, = 4,
3
6 6
In some cases you encounter repeated rational solutions. And to prove you have the right values you
use the discriminant which gives you information about the nature of the solutions to the equation.
55
Based on the expression b 2 − 4 ac , which is under the radical in the quadratic formula it can be found
in the equation ax2 + bx + c = 0 .
I. When the discriminant is equal to 0, the equation has repeated rational solutions.
Example:
x2 − 2 x + 1 = 0
By using the discriminant b 2 − 4 ac = ( −2 ) 2 − 4 (1)(1) = 0
− ( −2 ) ± ( −2 ) 2 − 4 (1)(1)
2 (1)
2± 0
=
2
x=
x = 1,1
← Repeated rational solutions
II. When the discriminant is positive and a perfect square, the equation has two distinct rational
solutions.
Example:
x2 − 4 x + 3 = 0
By discriminant b 2 − 4 ac = ( −4 ) 2 − 4 (1)( 3) = 4
− ( −4) ± ( −4 ) 2 − 4 (1)( 3)
2 (1)
4± 4
=
2
x=
x = 3,1
← Two distinct rational solutions
III. When the discriminant is positive but not a perfect square, the equation has two irrational solutions.
Example:
x2 + 4x − 6 = 0
The discriminant b 2 − 4 ac = ( 4 ) 2 − 4 (1)( −6) = 40
−4 ± ( 4 ) 2 − 4 (1)( −6)
x=
2 (1)
−4 ± 40
=
2
x =−2 ± 10
← Two irrational solutions
IV. When the discriminant is negative, the equation has two complex number solutions.
Example:
56
x2 + 4 x + 6 = 0
The discriminant b 2 − 4 ac = ( 4 ) 2 − 4 (1)( 6) = −8
x=
−4 ± ( 4 ) 2 − 4 (1)( 6)
2 (1)
x=
−4 ± −8
2
← Two complex number solutions
Solution by Completing the Square
One more method of solving quadratic equations is by completing the square.
Example:
Solve x 2 + 6 x + 5 =
0 by completing the square.
1) If the leading coefficient is not 1, use the multiplication (or division) property of
equality to make it 1:
x2 + 6x + 5 =
0 ← In this case the leading coefficient is already 1
2) Rewrite the equation by sending the constant to the right side of the equation:
0
x2 + 6x + 5 =
x2 + 6x + 5 − 5 = 0 − 5
x2 + 6x =
−5
3) Divide the numerical coefficient the middle term by 2, then square it, and add it to
both sides of the equation, but leave the square form on the left side of the equation:
x2 + 6x =
−5
x 2 + 6 x + ( 3) =−5 + ( 3)
2
x 2 + 6 x + ( 3) =−5 + 9
2
2
Middle term coefficient = 6
6
2
= 3 → ( 3)
2
x 2 + 6 x + ( 3) =
4
2
4) Once you found the squared number rewrite the equation as follows:
x 2 + 6 x + ( 3) =
4
2
( x + 3)
2
=
4
← Bring down the variable x and put it inside the parentheses
← Use the sign of the middle term. In this case it is +.
← Write the squared number. In this case it is 3.
The resultant binomial is ( x + 3) =
4
2
57
5) Using the square root property clear the term.
( x + 3)
2
=
± 4
← The square root of a squared term is the term by itself.
x + 3 =±2
6) Solve for the variable x.
← The ± notation is used because the square root can have both
positive and negative answers.
x + 3 =±2
x+3=
2
x + 3 =−2
x= 2 − 3
x = −1
x =−2 − 3
x = −5
( −1 + 3=
) ( 2=
)
2
2
4 And ( −5 + 3) = ( −2 ) = 4
7) Check your solution.
x = −1
2
2
x = −5
0
x2 + 6x + 5 =
( −1)
2
0
x2 + 6x + 5 =
0
+ 6 ( −1) + 5 =
← Both solutions are true:
1− 6 + 5 =
0
0=0
( −5)
2
0
+ 6 ( −5 ) + 5 =
25 − 30 + 5 =
0
0=0
Let’s keep practicing with one more.
Example:
← One way to make the leading coefficient 1 is
4x2 − 2x − 5 =
0
1
1
4 x 2 − 2 x − 5 =( 0 )
4
4
1
5
x2 − x − =
0
2
4
1
5
x2 − x =
2
4
(
)
2
x2 −
1
5 1
1
x+  = + 
2
4 4
4
by multiplying both sides of the equation by
← Move the constant to the right side of the equation
2
← Divide the middle term coefficient by 2, square it,
and add it to both sides of the equation:
58
1
4
1 2 1 1 1
1
÷ = × =
→  
2 1 2 2 4
4
2
2
x2 −
1
5 1
1
x+  = +
2
4 16
4
2
1
21
1
x − x+  =
2
 4  16
2
2
1
21

x−  =
4  16

← Write the squared number in binomial form.
2
1
21

±
x−  =
4
16

← Find the square root of both sides and don’t forget the
± sign.
x−
1
21
=
±
4
16
← Send the other number to the right side of the
equation.
x−
1
21
=
±
4
4
← Try to solve it using the square root. If not possible
leave it in radical form.
21 1
x=
±
+
4
4
1 ± 21
x=
4
← Solve for x.
← Final answer.
59
QUADRATIC EQUATIONS – EXERCISES
Solve each of the following equations by the method of your choice and check your solutions.
1.
x2 − 2 x + 1 = 0
2.
x 2 + 9 x + 20 = 0
3.
3 x 2 − 5 x − 12 = 0
4.
6x2 + 9 x − 6 = 0
5.
x 2 + 3 x − 28 = 0
6.
3x 2 − 2 x = 2 x + 7
7.
4 x 2 − 12 x = 16
8.
x 2 + 3x = 0
9.
3+
10.
3y2 − y − 4 = 0
11.
y2 + 2 y + 1 = 0
12.
x2 − 2 x − 8 = 0
13.
x2 + 4 = 0
14.
x 2 + x = −1
15.
9 y2 + 6 y − 8 = 0
16.
y 2 − 25 = 0
17.
6 y 2 − 13 y + 6 = 0
1 10
=
x x2
4 −1
=
3x 3
4 21
19. x − =
x 5
5
1
20. 3 +
= 2
2x x
1 3
21. 4 − = 2
x x
18.
x−
8x = x2
23. ( x − 5)( x + 8) = −20
22.
24.
( x + 6)( x − 3) = 10
25.
2
x
−
=
1
x +5 x −5
60
QUADRATIC EQUATIONS – ANSWER TO EXERCISES
1. x 2 − 2 x + 1 = 0
( x − 1)( x − 1) = 0
2. x 2 + 9 x + 20 = 0
( x + 5)( x + 4 ) = 0
x −1 = 0
x −1 = 0
x +5= 0
x+4 = 0
x =1
x =1
x = −5
x = −4
3. 3 x 2 − 5 x − 12 = 0
( 3 x + 4 )( x − 3) = 0
3x + 4 = 0
4
x=−
3
x−3= 0
x=3
5. x 2 + 3 x − 28 = 0
( x + 7 )( x − 4 ) = 0
x+7 = 0
x = −7
x−4 = 0
x=4
4. 6 x 2 + 9 x − 6 = 0
3( 2 x 2 + 3 x − 2 ) = 0
3( 2 x − 1)( x + 2 ) = 0
2x −1 = 0
x+2 = 0
1
x=
x = −2
2
6. 3 x 2 − 2 x = 2 x + 7
3x 2 − 2 x − 2 x − 7 = 0
3x 2 − 4 x − 7 = 0
( 3 x − 7 )( x + 1) = 0
3x − 7 = 0
7
x=
3
7. 4 x 2 − 12 x = 16
4 x 2 − 12 x − 16 = 0
4( x 2 − 3x − 4) = 0
4 ( x − 4 )( x + 1) = 0
x−4 = 0
x=4
8. x 2 + 3 x = 0
x ( x + 3) = 0
x=0
1 10
=
x x2
3 x 2 + x = 10
3 x 2 + x − 10 = 0
x = −3
10. 3 y 2 − y − 4 = 0
( 3 y − 4 )( y + 1) = 0
3y − 4 = 0
4
y=
3
( 3 x − 5)( x + 2 ) = 0
11. y 2 + 2 y + 1 = 0
x+3= 0
x +1 = 0
x = −1
9. 3 +
3x − 5 = 0
5
x=
3
x +1 = 0
x = −1
y +1 = 0
y = −1
x+2 = 0
x = −2
12. x 2 − 2 x − 8 = 0
61
( y + 1) 2 = 0
( x − 4 )( x + 2 ) = 0
y +1= 0
x−4 = 0
x+2 = 0
y = −1, −1
x=4
x = −2
13. x 2 + 4 = 0
−0 ± 0 − 4 (1)( 4 )
x=
2
x= ±
−16
2
14. x 2 + x = −1
x2 + x + 1 = 0
x=
x=
15. 9 y 2 + 6 y − 8 = 0
−6 ± 36 − 4 ( 9 )( −8)
x=
2( 9)
6 + 18
18
2
x=
3
62
4 21
=
x 5
16. y 2 − 25 = 0
( y − 5)( y + 5) = 0
y +5= 0
y = −5
6 − 18
18
−4
x=
3
4
1
=−
3x
3
2
3x − 4 = − x
3x 2 + x − 4 = 0
18. x −
( 3 y − 2 )( 2 y − 3) = 0
19. x −
−1 ± −3
2
x=
17. 6 y 2 − 13 y + 6 = 0
3y − 2 = 0
2
y=
3
2 (1)
y −5= 0
y=5
−6 ± 324
18
−6 ± 18
=
18
=
x=
−1 ± 1 − 4 (1)(1)
2y − 3 = 0
3
y=
2
( 3 x + 4 )( x − 1) = 0
3x + 4 = 0
4
x=−
3
20. 3 +
5
1
= 2
2x x
x −1 = 0
x =1
5 x 2 − 20 = 21x
5 x 2 − 21x − 20 = 0
6 x 2 + 5x = 2
6 x 2 + 5x − 2 = 0
−5 ± 25 − 4 ( 6)( −2 )
x=
12
(5 x + 4 )( x − 5) = 0
5x + 4 = 0
4
x=−
5
x −5= 0
x=
x=5
1 3
=
x x2
4 x2 − x = 3
4 x2 − x − 3 = 0
( 4 x + 3)( x − 1) = 0
21. 4 −
4x + 3 = 0
3
x=−
4
22. 8 x = x 2
x2 − 8x = 0
x ( x − 8) = 0
x=0
x −1 = 0
25.
x −8 = 0
x=8
x =1
23. ( x − 5)( x + 8) = −20
x 2 + 3 x − 40 + 20 = 0
x 2 + 3 x − 20 = 0
−3 ± 9 − 4 (1)( −20)
x=
2
x=
−5 ± 73
12
24. ( x + 6)( x − 3) = 10
x 2 + 3 x − 18 − 10 = 0
x 2 + 3 x − 28 = 0
( x + 7 )( x − 4 ) = 0
x+7 = 0
x = −7
−3 ± 89
2
x−4 = 0
x=4
2
x
−
=
1
x +5 x −5
2 ( x − 5) − x ( x + 5) = x 2 − 25
2 x − 10 − x 2 − 5 x − x 2 + 25 = 0
−2 x 2 − 3 x + 15 = 0
2 x 2 + 3 x − 15 = 0
−3 ± 9 − 4 ( 2 )( −15)
x=
4
x=
−3 ± 129
4
63