INTERMEDIATE ALGEBRA
ANSWERS
TEST CHAPTER 6
WEB SITE
1.
Look at the denominator only. Factor the denominator. x³ - 9x = x(x² - 9 ) = x(x – 3)(x+3) Set
each of these equation to equal zero. x = 0; x – 3 = 0; x + 3 = 0. Therefore x = 0, 3, (-3). The
domain can equal all rational number except 0, 3 or (-3).
2.
All rational numbers.
3.
x  5 , solve
Look at the denominator only. You do not want the denominator to equal zero.
x + 5 > 0. x > (-5) Therefore the domain can equal all reationl numbers greater than (-5).
In problems 4 – 9, factor all possible statements first then do the operation indicated.
4.
5(2 y  1)
y 5 5y
6y

 =
= simplify
4
4
1 4
6(2 y  1)
5.
x
( x  9)( x  5)
x( x  9)

= simplify
x 8
( x  9)( x  9)
( x  8)( x  5)
6.
5(2  x)
 5( x  2)
(2 x  1)( x  1)
2(3x  4)


= invert and multiply
=
( x  7)( x  5)
( x  7)( x  5) (2 x  1)( x  1)
2(3x  4)
 10( x  2)(3x  4)
( x  7)( x  5)( 2 x  1)( x  1)
7.
note x
note: y
 -1/2
note: x
 -9, -5, 8
 -4/3, 7, 5, ½, -1
8x  1
 4  7x
+
= since the denominators are the same, combine the
( x  7)( x  5) ( x  7)( x  5)
Numerators
8.
1
x5
=
( x  7)( x  5) x  7
Rewrite –16 + x² to x² - 16;
note: x
7
x
7
= Find the common
( x  4)( x  4) ( x  4)( x  1)
denominator;
The common denominator will be (x + 4)(x – 4)( x + 1)
x( x  1)
7( x  4)
= Remember the subtraction sign between the
( x  4)( x  4)( x  1) ( x  4)( x  4)( x  1)
groups
x² + x –7x + 28
(x - 4)(x + 4)(x + 1)
=
x² - 6x + 28
(x – 4)(x + 4)(x + 1)
note: x
 4,-4,-1
PAGE 2
9.
y7
6
9
6
2( y  7)


=
= Reduce 6 and 9 by 3.
9
y
y7 y
3y
10.
1
1
1
mz mz


=
=
mz
mz mz
1
mz
note: m
note: y
 0, -7
 z, -z
In problems 11 – 15, factor the denominators first, then find the least common denominator.
11.
1 3
7
=
The least common denominator is 4x, or you may use (2x)(4) as your common
1 2x 4
denominator.
(1)( 2 x )( 4) (3)( 2 x)( 4) (7)( 2 x)( 4)
=
In each unit cancel the information that is the same
1
2x
4
in the numerator and the denominator.
(2x)(4) – (3)(4) = (7)(2x)
8x - 12 = 14x
8x – 8x – 12 = 14x – 8x
-12 = 6x
 12 6x
=
6
6
-2 = x
12.
4
3
1 x
=
The least common denominator for this problem is
( x  1)( x  2) x  2 x  1
(1  x)( x  1)( x  2) (4)( x  1)( x  2) (3)( x  1)( x  2)
(x-1)(x-2).
=
x2
x 1
( x  1)( x  2)
1+x
13.
=
(4)(x - 1)
- (3)(x – 2)
1 + x = 4x – 4 –3x + 6
1+x=x+2
1–1+x=x+2–1
x=x+1
x–x=x–x+1
0x = 1
NO SOLUTION (you can not divide by zero)
4
y3
=
The least common denominator for this problem is y(y – 5)(y – 2)
y ( y  2) ( y  5)( y  2)
( y  3)( y )( y  5)( y  2) (4)( y )( y  5)( y  2)
=
y ( y  2)
( y  5)( y  2)
(y + 3)(y – 5)
= (-4)(y)
y² - 5y + 3y – 15 = -4y
y² - 2y – 15
= -4y
y² - 2y + 4y – 15 = -4y + 4y
note: y
 0, 2, 5
PAGE 3
13. Continued.
14.
y² + 2y – 15 = 0
(y + 5)(y – 3) = 0
y + 5 = 0; y – 3 = 0
y + 5 – 5 = 0 – 5; y – 3 + 3 = 0 + 3
y = -5 ; 3
6
1
1
+
=
The least common denominator is
( x  4)( x  1) ( x  4)( x  1) ( x  1)( x  1)
(x + 4)(x – 1)(x + 1)
1( x  4)( x  1)( x  1) 1( x  4)( x  1)( x  1) 6( x  4)( x  1)( x  1)
+
=
( x  4)( x  1)
( x  4)( x  1)
( x  1)( x  1)
1( x + 1)
+
1(x – 1)
=
6(x + 4)
x+1+x–1=
6x + 24
2x + 0 = 6x + 24
2x – 6x = 6x – 6x + 24
-4x = 24
 4x
24
=
4
4
x = -6
15.
x3 5
2
=
The least common denominator is (1)(x + 1)
x 1 1 x 1
( x  3)(1)( x  1) (5)(1)( x  1) ( 2)(1)( x  1)
=
x 1
1
x 1
(x + 3)(1) – (5)(x + 1)
= (2)(1)
x + 3 –5x –5 = 2
-4x - 2 = 2
-4x - 2 + 2 = 2 + 2
-4x = 4
 4x
4
=
4
4
x = -1
But x
NO SOLUTION
 - 1. Answer to this problem is