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

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  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

3x       = 6        Adding

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)

4x        = 12     Adding

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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