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## College Algebra with Intermediate Algebra 1st Edition Solutions Manual By Beecher

Instant download College Algebra with Intermediate Algebra 1st Edition Solutions Manual By Beecher. You don’t have to wait as you’ll be able to download the files immediately after placing your order. All chapters are included with all the questions and correct answers. Get the College Algebra with Intermediate Algebra 1st Edition Solutions Manual By Beecher today and start getting better grades.

1. Consistent; independent; the pair (1, 1) is a solution of both equations in system C and is not a solution of any other system shown, so system C corresponds to this graph.
2. Consistent; independent; the system consists of a line with y-intercept (0, −1) and a horizontal line. Both lines pass through (4, −3). System E corresponds to this graph.
 −
1. Consistent; independent;  the  pair  (  1, 3) is  a  solution of both equations in system A and is  not  a  solution  of any other system shown, so system A corresponds to this graph.
 4               5
1. 3x + 2 = 2 x − 5
2. 20.3  + 2Σ =  20. 2        5Σ

RC2. True; we can use the substitution method to solve any system of two equations in two variables.

RC4. False; see Example 3 on page 161 in the text.

1. x = 8 − 4y, (1)

3x + 5y = 3  (2)

3(8 − 4y) + 5y = 3         Substituting for x in (2) 24 − 12y + 5y = 3

24 − 7y = 3

−7y = −21

y = 3

Substitute 3 for y in (1).

x = 8 − 4 · 3 = 8 − 12 = −4

The solution is (−4, 3).

1. 9x − 2y = 3, (1)

3x − 6 = y     (2)

9x − 2(3x − 6) = 3    Substituting for y in (1)

9x − 6x + 12 = 3

3x + 12 = 3

3x = −9

x = −3

Substitute −3 for x in (2).

3(−3) − 6 = y

−9 − 6 = y

−15 = y

4 x                 5 x

The solution is (−3, −15).

15x + 40 = 8x − 100

7x = −140

x = −20

### Exercise Set 3.2                                                                                                                                                                                              79

1. m − 2n = 3, (1)

4m + n = 1    (2)

We solve the ﬁrst equation for m. m − 2n = 3      (1)

m = 2n + 3   (3)

We substitute 2n + 3 for m in the second equation and solve for n.

4(2n + 3) + n = 1

8n + 12 + n = 1

9n + 12 = 1

9n = −11

11

n = − 9

 −         n

Now we substitute    11 for    in Equation (3).

9

.        Σ

Substitute 2p − 8 for q in Equation (2) and solve for p. 5p + 7(2p − 8) = 1

5p + 14p − 56 = 1

19p − 56 = 1

19p = 57

p = 3

Substitute 3 for p in Equation (3).

q = 2 · 3 − 8 = 6 − 8 = −2

The solution is (3, −2).

1. 3x + y = 4, (1)

12 − 3y = 9x  (2)

Solve the ﬁrst equation for y.

3x + y = 4

y = −3x + 4

m = 2 − 11 + 3 = − 22 + 27 = 5

Substitute −3x + 4 for y in the second equation and solve

9                     9        9       9

The solution is . 5 , 11 Σ.

for x.

12 − 3y = 9x

9      9

1. t = 4 − 2s, (1)

t + 2s = 6    (2)

(4 − 2s) + 2s = 6 Substituting for t in (2) 4 = 6

We get a false equation. The system has no solution.

1. 5x + 6y = 14, (1)

−3y + x = 7     (2)

We solve the second equation for x.

−3y + x = 7            (2)

x = 3y + 7   (3)

Substitute 3y + 7 for x in the ﬁrst equation and solve for

y.

5(3y + 7) + 6y = 14

15y + 35 + 6y = 14

21y + 35 = 14

21y = −21

y = −1

Substitute −1 for y in Equation (3).

12 − 3(−3x + 4) = 9x

12 + 9x − 12 = 9x

12 − 12 = 0

0 = 0

We have a true equation. Any value of x will make this equation true. Thus the system of equations has inﬁnitely many solutions.

1. 5x + 3y = 4, (1)

x − 4y = 3      (2)

We solve the second equation for x. x − 4y = 3    (2)

x = 4y + 3   (3)

Substitute 4y + 3 for x in the ﬁrst equation and solve for

y.

5x + 3y = 4         (1)

5(4y + 3) + 3y = 4

20y + 15 + 3y = 4

23y + 15 = 4

23y = −11

11

y = −

x = 3(−1) + 7 = −3 + 7 = 4                                                                                       11       23

The solution is (4, −1).

1. 4p − 2q = 16, (1)

Substitute − 23 for y in Equation (3).

x = 4. − 11 Σ + 3 = − 44  +  69  =  25

5p + 7q = 1      (2)

Solve the ﬁrst equation for q. 4p − 2q = 16          (1)

−2q = −4p + 16

q = 2p − 8        (3)

23                  23     23     23

 .    , −   Σ

The solution is      25     11 .

23     23

1. 4x + 13y = 5, (1)

−6x + y = 13  (2)

Solve the second equation for y.

−6x + y = 13             (2)

y = 6x + 13  (3)

### 80                                                                                                                                                           Chapter 3: Systems of Equations

Substitute 6x + 13 for y in Equation (1) and solve for x. 4x + 13(6x + 13) = 5

4x + 78x + 169 = 5

82x + 169 = 5

82x = −164

x = −2

5x + 2y = a

5x + 5y = 5b    Multiplying by −5 7y = a − 5b

y = a 5b

7

We obtain . a + 2b , a − 5b Σ. This checks, so it is the so-

Substitute −2 for x in Equation (3).

y = 6(−2) + 13 = −12 + 13 = 1

The solution is (−2, 1).

1. Let l = the length and w = the width. Solve: 2l + 2w = 340,

l = w + 50

The solution is (110, 60), so the length is 110 m and the width is 60 m.

1. Let x and y represent the angles. Solve:

x + y = 90,

y = 5x + 6

The solution is (14, 76), so the measures of the angles are 14and 76.

1. Let x = the number of coach-class seats and y = the num- ber of ﬁrst-class seats. Solve:

x + y = 152,

x = 5 + 6y

The solution is (131, 21), so there are 131 coach-class seats and 21 ﬁrst-class seats.

1. −9(y + 7) − 6(y − 4) = −9y − 63 − 6y + 24 = −15y − 39

lution.                7           7

1. Let b = the number of ounces of baking soda and v = the number of ounces of vinegar to be used. Solve:

b = 4v, b + v = 16

The solution is (12.8, 3.2), so 12.8 oz of baking soda and

3.2 oz of vinegar should be used.

# Exercise Set 3.3

RC2. If a system of equations has no solution, then it is inconsistent.

RC4. If the graphs of the equations in a system of two equa- tions in two variables are the same line, then the equa- tions are dependent.

RC6. If the graph of the equations in a system of two equa- tions in two variables intersect at one point, then the equations are independent.

1. x + y = 9,    (1)

2x y = −3    (2)

1. m = 3 (4) = 7

−2 − (−5)     3

1. −12(2x − 3) = 16(4x − 5)

−24x + 36 = 64x − 80

116 = 88x

29  = x

22

1. 5x + 2y = a,

x y = b

We multiply by 2 on both sides of the second equation and then add.

5x + 2y = a

2x 2y = 2b      Multiplying by 2 7x +    0 = a + 2b                   Adding

7x = a + 2b

a + 2b

x = 2

Substitute 2 for x in (1).

2 + y = 9

y = 7

The solution is (2, 7).

1. 2x − 3y = 18, (1)

2x + 3y = −6   (2)

x = 3

Substitute 3 for x in (2). 2 · 3 + 3y = −6

6 + 3y = −6

3y = −12

y = −4

x =      7

 −

Next we multiply by 5 on both sides of the second equa- tion and then add.

The solution is (3, −4).

### Exercise Set 3.3                                                                                                                                                                                              81

1. 2x + 3y = −9, (1)

5x − 6y = −9    (2)

4x + 6y = −18 Multiplying (1) by 2

5 · 2 + 4y = 2

 3

10  + 4y = 2

3

4

5x 6y = 9

9x        = −27

x = −3

Substitute −3 for x in (1).

4y = − 3

1

y = − 3

The solution is . 2 , 1 Σ.

2(−3) + 3y = −9

−6 + 3y = −9

3y = −3

y = −1

The solution is (−3, −1).

1. 2a + 3b = 11, (1)

4a − 5b = −11  (2)

−4a − 6b = −22 Multiplying (1) by −2

4a −  5b = 11

− 11b = −33

b = 3

Substitute 3 for b in (1).

2a + 3 · 3 = 11

2a + 9 = 11

2a = 2

a = 1

The solution is (1, 3).

1. 3x − 2y = 1, (1)

−6x + 4y = −2   (2)

6x − 4y = 2 Multiplying (1) by 2

6x + 4y = 2 (2)

0 = 0 True for all (x, y) There are inﬁnitely many solutions.

1. 5x + 4y = 2, (1)

2x − 8y = 4    (2)

10x + 8y = 4     Multiplying (1) by 2

2x 8y = 4

12x        = 8

3     3

1. 5x + 3y = 25,  (1)

3x + 4y = 26    (2)

20x + 12y = 100     Multiplying (1) by 4

−9x − 12y = −78 Multiplying (2) by −3

11x         = 22

x = 2

Substitute 2 for x in (1).

5 · 2 + 3y = 25

10 + 3y = 25

3y = 15

y = 5

The solution is (2, 5).

1. 3x − 5y = −2

−3x + 5y = 7 Rearranging

0 = 5 False equation There is no solution.

1. 10x + y = 306,   (1)

10y + x = 90        (2)

y = −10x + 306   (3)   Solving (1) for y

10(−10x + 306) + x = 90 Substituting for y

in (2)

−100x + 3060 + x = 90

−99x = −2970

x = 30

Substitute 30 for x in (3).

y = −10 · 30 + 306 = 6

The solution is (30, 6).

1. 2x + 1 y = −11,

2                                                                                               3         7

x = 3

1         1

x y = −10

Substitute  2  for  x in (1).                                                                            7        3

3                                                                                                  14x + 3y = −231,      (1)    Multiplying by 21

3x − 7y = −210        (2)    to clear fractions

98x + 21y = −1617   Multiplying (1) by 7

9x − 21y =   −630     Multiplying  (2) by 3

107x         = −2247

x =     −21

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