### Introductory Algebra - Math TV Schools Media Server

```Introductory Algebra
Name _________________________
Problem Set 7.2
Solutions to Every
Odd-Numbered Problem
Date _________________________
7.2 The Elimination Method
1.
5.
9.
11.
3.
2x = 4
x=2
Substituting into the first equation:
2+y= 3
y =1
The solution is ( 2,1) .
7.
!2 y = 10
y = !5
Substituting into the first equation:
x ! (!5) = 7
x+5=7
x=2
The solution is ( 2, !5 ) .
0=0
The lines coincide (the system is dependent).
Multiplying the first equation by 2:
13.
6x ! 2y = 8
2 x + 2 y = 24
8 x = 32
x=4
Substituting into the first equation:
3(4) ! y = 4
12 ! y = 4
! y = !8
y=8
The solution is ( 4, 8 ) .
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2 y = 14
y=7
Substituting into the first equation:
x + 7 = 10
x=3
The solution is ( 3, 7 ) .
4 x = !4
x = !1
Substituting into the first equation:
!1 + y = !1
y=0
The solution is ( !1, 0 ) .
Multiplying the second equation by –3:
5 x ! 3y = !2
!30 x + 3y = !3
!25 x = !5
1
x=
5
Substituting into the first equation:
! 1\$
5 # & ' 3y = '2
" 5%
1 ' 3y = '2
'3y = '3
y =1
!1 \$
The solution is # ,1& .
"5 %
This worksheet may be printed and used for educational purposes
only. It cannot be used for commercial purposes without the written
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15.
19.
23.
Introductory Algebra
Name _________________________
Problem Set 7.2
Solutions to Every
Odd-Numbered Problem
Date _________________________
Multiplying the second equation by 4:
11x ! 4 y = 11
20 x + 4 y = 20
31x = 31
x =1
Substituting into the second equation:
5(1) + y = 5
5+y=5
y=0
The solution is (1, 0 ) .
Multiplying the first equation by –2:
2 x + 16 y = 2
!2 x + 4 y = 13
20 y = 15
3
y=
4
Substituting into the first equation:
" 3%
! x ! 8 \$ ' = !1
# 4&
! x ! 6 = !1
!x = 5
x = !5
3%
"
The solution is \$ !5, ' .
#
4&
8 x = !24
x = !3
Substituting into the second equation:
2(!3) + y = !16
!6 + y = !16
y = !10
The solution is ( !3, !10 ) .
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17.
21.
Multiplying the second equation by 3:
3x ! 5 y = 7
!3x + 3y = !3
!2 y = 4
y = !2
Substituting into the second equation:
! x ! 2 = !1
!x = 1
x = !1
The solution is ( !1, !2 ) .
Multiplying the first equation by 2:
!6 x ! 2 y = 14
6 x + 7 y = 11
5 y = 25
y=5
Substituting into the first equation:
!3x ! 5 = 7
!3x = 12
x = !4
The solution is ( !4, 5 ) .
This worksheet may be printed and used for educational purposes
only. It cannot be used for commercial purposes without the written
consent of MathTV.com, Inc. Created by Ross Rueger.
25.
Introductory Algebra
Name _________________________
Problem Set 7.2
Solutions to Every
Odd-Numbered Problem
Date _________________________
Multiplying the second equation by 3:
x + 3y = 9
6 x ! 3y = 12
7 x = 21
x=3
Substituting into the first equation:
3 + 3y = 9
3y = 6
y=2
The solution is ( 3, 2 ) .
29.
27.
Multiplying the second equation by 2:
x ! 6y = 3
8 x + 6 y = 42
9 x = 45
x=5
Substituting into the second equation:
4(5) + 3y = 21
20 + 3y = 21
3y = 1
1
y=
3
1
!
\$
The solution is # 5, & .
" 3%
Multiplying the second equation by –3:
2x + 9y = 2
!15 x ! 9 y = 24
!13x = 26
x = !2
Substituting into the first equation:
2(!2) + 9 y = 2
!4 + 9 y = 2
9y = 6
2
y=
3
2%
"
The solution is \$ !2, ' .
#
3&
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This worksheet may be printed and used for educational purposes
only. It cannot be used for commercial purposes without the written
consent of MathTV.com, Inc. Created by Ross Rueger.
31.
33.
35.
Introductory Algebra
Name _________________________
Problem Set 7.2
Solutions to Every
Odd-Numbered Problem
Date _________________________
To clear each equation of fractions, multiply the first equation by 12 and the second equation by 6:
1 \$
1 %
!1
! 7\$
"3
" 7%
12 # x + y& = 12 # &
6 \$ x ! y' = 6 \$ '
"3
%
"
%
#
&
# 3&
4
6
2
3
4 x + 3y = 14
9 x ! 2 y = 14
The system of equations is:
4 x + 3y = 14
9 x ! 2 y = 14
Multiplying the first equation by 2 and the second equation by 3:
8 x + 6 y = 28
27 x ! 6 y = 42
35 x = 70
x=2
Substituting into 4 x + 3y = 14 :
4(2) + 3y = 14
8 + 3y = 14
3y = 6
y=2
The solution is ( 2, 2 ) .
Multiplying the first equation by –2:
!6 x ! 4 y = 2
6x + 4y = 0
0=2
Since this statement is false, the two lines are parallel, so the system has no solution.
Multiplying the first equation by 2 and the second equation by 3:
22 x + 12 y = 34
15 x ! 12 y = 3
37 x = 37
x =1
Substituting into the second equation:
5(1) ! 4 y = 1
5 ! 4y =1
!4 y = !4
y =1
The solution is (1,1) .
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This worksheet may be printed and used for educational purposes
only. It cannot be used for commercial purposes without the written
consent of MathTV.com, Inc. Created by Ross Rueger.
37.
39.
41.
45.
Introductory Algebra
Name _________________________
Problem Set 7.2
Solutions to Every
Odd-Numbered Problem
Date _________________________
To clear each equation of fractions, multiply the first equation by 6 and the second equation by 6:
1 \$
1 %
!1
! 1\$
"
" 1%
6 # x + y& = 6 # &
6 \$ ! x ! y' = 6 \$ ! '
"2
%
"
%
#
&
# 6&
6
3
3
3x + y = 2
!6 x ! 2 y = !1
The system of equations is:
3x + y = 2
!6 x ! 2 y = !1
Multiplying the first equation by 2:
6x + 2y = 4
!6 x ! 2 y = !1
0=3
Since this statement is false, the two lines are parallel, so the system has no solution.
Multiplying the second equation by 100 (to eliminate decimals):
x + y = 22
5 x + 10 y = 170
Multiplying the first equation by –5:
!5 x ! 5 y = !110
5 x + 10 y = 170
5 y = 60
y = 12
Substituting into the first equation:
x + 12 = 22
x = 10
The solution is (10,12 ) .
Solving the equation:
43.
Solving the equation:
x + ( 2 x ! 1) = 2
2 ( 3y ! 1) ! 3y = 4
x + 2x !1 = 2
6 y ! 2 ! 3y = 4
3x ! 1 = 2
3y ! 2 = 4
3x = 3
3y = 6
x =1
y=2
Solving the equation:
4 x + 2 ( !2 x + 4 ) = 8
4x ! 4x + 8 = 8
8=8
Since this statement is true, the solution is all real numbers.
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This worksheet may be printed and used for educational purposes
only. It cannot be used for commercial purposes without the written
consent of MathTV.com, Inc. Created by Ross Rueger.
47.
49.
51.
53.
55.
Introductory Algebra
Name _________________________
Problem Set 7.2
Solutions to Every
Odd-Numbered Problem
Date _________________________
Solving for x:
x ! 3y = !1
x = 3y ! 1
Solving for y: y = 2 (1) ! 1 = 2 ! 1 = 1
Solving for x: x = 3( 2 ) ! 1 = 6 ! 1 = 5
Substituting x = 13: y = 1.5 (13) + 15 = 19.5 + 15 = 34.5
Substituting x = 12: y = 0.75 (12 ) + 24.95 = 9 + 24.95 = 33.95