Chapter 1
Linear Equations and Mathematical Concepts

Exercises 1.1

1. 1.
2. 3.
3. 5.
4. 7.
5. 9.
6. 11.
7. 13.
8. 15.
9. 17.
10. 19.
11. 21.
12. 23.
13. 25. multiply by to eliminate the fraction
    1. (Check the result. Multiplication by a factor such as can introduce an extraneous solution.)
14. 27.
15. 29.
16. 31.
17. 33.
18. 35.
19. 37. Let monthly installment ($). Since Sally paid $300 down, she owes . Therefore, or is the monthly installment.
20. 39. The consumption function is . The slope is the "marginal propensity to consume." Therefore, . The disposable income, when consumption is yields and . The consumption function is .
21. 41.
    1. a.
    2. b. and
22. 43. The tax is 6.2% or 0.062 in decimal form, so , where is .
23. 45.
    1. a.
    2. b.

Exercises 1.2

1. 1. Setting determines the -intercept and setting determines the -intercept.
   1. a. -intercept 3, -intercept
   2. b. -intercept 5/4, -intercept
   3. c. -intercept 12, -intercept 8
   4. d. -intercept 2, -intercept
   5. e. -intercept 4, no -intercept(vertical line)
   6. f. no -intercept (horizontal line), -intercept
2. 3. The slope is
   1. a. (3, 6) and
   2. b. (1, 6) and (2, 11)
   3. c. (6, 3) and (12, 7)
   4. d. (2, 3) and (2, 7)
   5. e. (2, 6) and (5, 6)
   6. f. (5/3, 2/3) and (10/3, 1)
3. 5. a) -intercept 5/2 and -intercept
   b) -intercept 4 and no -intercept
   c) x-intercept 5 and y-intercept 3
   d) x-intercept 7 and y-intercept 2
   
4. 7.
   1. a. and ; the slope of the first line is 5/3. Solving for in the second equation yields . This slope is also 5/3. The slopes are both (5/3) so the lines are parallel (with different intercepts).
   2. b. and . The slope of the second line is easily determined (line in slope intercept form) as 1/3. Again, solve for in the first equation to determine . The slope is . The slopes are negative reciprocals; the lines are perpendicular.
   3. c. and . Solving for y in each equation, one determines that and . These lines have the same slope (and different intercepts) making them parallel.
   4. d. and . The slope of the first line is 5 and solving for in the second equation, , the slope is 3. These slopes are neither the same nor negative reciprocals. They are neither parallel nor perpendicular.
   5. e. is a horizontal line while is a vertical line. The two lines are perpendicular.
5. 9. A linear equation has a single -intercept except for (the -axis) with an infinite number of -intercepts. Any horizontal line except has no -intercepts. Generally, lines do not have more than one -intercept. The exception is (the -axis) with an infinite number of -intercepts. Any vertical line with the exception of has no -intercepts.
6. 11. The ordered pairs of "time" and "machine value" are (0, 75,000) and (9, 21,000), respectively. The slope is

   . The -intercept is the purchase price, $75,000. The equation to model the straight-line depreciation is , where is the machine value ($) at time .

7. 13. The ordered pairs (gallons, miles) are (7, 245) and (12, 420). The slope is with gallons and miles. Use either pair with the point slope-formula.

   Therefore, or .

8. 15. Total cost reflects both fixed and variable costs. The fixed cost is monthly rent ($1100). The variable cost is , where is monthly production. Therefore, total cost is .
9. 17.
   1. a. Here, the fixed cost is $50/day and variable cost $0.30/mile. To rent the car for a single day costs $50 to which the mileage cost must be added. The cost is .
   2. b. If a person has $110 for rental, the equation to solve for the travel distance is .

      Solving yields,

      The person can rent the car and travel 200 miles with $110.

10. 19. Since is to be a function of , the ordered pairs are . The two ordered pairs are (70, 84) and (40, 48). The slope is . Using either pair with the slope to yield or .

Exercises 1.3

1. 1. The ordered pair must satisfy each equation to be a solution to the system.
   1. a. is true but is not. Therefore, (3, 1) is not a solution to the system.
   2. b. is true and so is . Therefore, (2, 3) is a solution to the system.
   3. c. is true but is not. Therefore, is not a solution to the system.
2. 3.
   1. a. and , and . Since the slopes differ, this is a consistent system.
   2. b. and . Since both the slopes and intercepts are the same, the two equations are the same line. It is a dependent system.
   3. c. and . Here, the slopes are the same and the intercepts differ. The lines are parallel and the system is inconsistent.
3. 5. The graphs and solution to each system are: a) b) c)
4. 7.
   1. a. Given and substituting in the second equation yields

      , , and . The ordered pair solution is (3, 2).

   2. b. Using, , . Substituting into the second equation yields , , and . The ordered pair solution is (3, 2).
   3. c. The second equation, already solved for , substituted in the first equation yields or . This is false, so the system is inconsistent and has no solution.
5. 9.
   1. a. Here, can be eliminated by simply adding the two equations as written.

      Next, use to determine . The ordered pair solution is (1, 3).

   2. b. Here, can be eliminated by multiplying the second equation by 3. The system is rewritten as

      Adding the two equations yields or . Using this value for yields that is also 5. (Check the solution in the original equation to see that (5, 5) is correct.)

   3. c. The second equation must first be rewritten in standard form , so the system to solve is

      Multiplying the first equation by will allow to be eliminated from the system.

      Using , it is determined that . (Check the solution in the original system to verify the solution (1, 3)).

   4. d. Rewriting the first equation in standard form and multiplying by 2 to eliminate x yields

      If , then and checking in the original system indicates that (0, 4) is the correct solution.

6. 11. Let represent the number of boxes of cookies and the boxes of candy. The system to be solved is

   Solving yields sales of 1500 boxes of cookies and 900 boxes of candy.

7. 13. Let of regular and premium. The system of equations to be solved (by either substitution or elimination) is

   Using elimination,

8. 15. The Market equilibrium occurs when is about 30 and about 260.
   Using substitution
   so,
9. 17.

   The price is about $75 and quantity is 50 at market equilibrium.

Exercises 1.4

1. 1. The area of a circle is , so,

   The radius is 1 unit, so the diameter is two units.

2. 3.
   1. a. Pi day is March 14 (3/14)
   2. b. Professor Yasumasa Kanada of the University of Tokyo.
   3. c. Albert Einstein was born on March 14.
   4. d. It is an illness thought to be attached to trying to square a circle.
   5. e. Pi is known as die Ludolphschezahl in Germany.
3. 5. a)

   b)

   c)

Exercises 1.5

1. 1.
   1. a. The expression must involve powers of 2 or 3. In this case, 8 is a power of 2, so one has .
   2. b. Rewrite as .
   3. c. Rewrite as .
2. 3.
   1. a.
   2. b.
   3. c.
3. 5.
   1. a.
   2. b.
   3. c.
4. 7.
5. 9.
6. 11.
7. 13. Since the bases are the same, exponents are equal. Therefore, yields .
8. 15. Equate bases as to yield and .
9. 17. Factoring yields

   Since cannot be zero, and .

10. 19. Factoring yields . Simplifying yields . Therefore, . The solutions are or .
11. 21.
12. 23.
13. 25. This is a difference of cubes
14. 27. Rewrite as to determine ordered pairs and graph as
15. 29. so , which yields .
16. 31. so , which yields .
17. 33. so , which yields
18. 35....