1. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
(For help, go to Lessons 2-4 and 6-2.)
Solve each equation.
1. 2n + 3 = 5n – – 4z = 2z – q – 12 = 3q + 23
Graph each pair of equations on the same coordinate plane.
4. y = 3x – y = 6x + 1
y = –x y = 6x – 4
6. y = 2x – y = x + 5
6x – 3y = y = –3x + 5
7-1

2. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
n + 3 = 5n – – 4z = 2z – 13
2n – 2n + 3 = 5n – 2n – – 4z + 4z = 2z + 4z – 13
3 = 3n – = 6z – 13
5 = 3n 21 = 6z
1 = n 3 = z
q – 12 = 3q + 23
8q – 3q – 12 = 3q – 3q + 23
5q – 12 = 23
5q = 35
q = 7
Solutions 2 3 1 2
7-1

3. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
4. y = 3x – y = 6x + 1
y = –x y = 6x – 4
y = 2x – y = x + 5
6x – 3y = y = –3x + 5
–3y = –6x – 15
y =
y = 2x – 5
Solutions (continued)
–6x  15 –3
7-1

4. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solve by graphing. Check your solutions.
y = 2x + 1
y = 3x – 1
y = 2x + 1 The slope is 2. The y-intercept is 1.
y = 3x – 1 The slope is 3. The y-intercept is –1.
Graph both equations on the same coordinate plane.
7-1

5. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
(continued)
Find the point of intersection.
The lines intersect at (2, 5), so (2, 5) is the solution of the system.
y = 2x y = 3x – 1
5 2(2) + 1 Substitute (2, 5) for (x, y) (2) – 1
– 1
5 = = 5
Check: See if (2, 5) makes both equations true.
7-1

6. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Suppose you plan to start taking an aerobics class. Non-members pay $4 per class while members pay $10 a month plus an additional $2 per class. After how many classes will the cost be the same? What is that cost?
Define: Let = number of classes.
Let = total cost of the classes. c
T(c)
Relate: cost is membership plus cost of classes
fee attended
Write: member =
non-member =
T(c) c
7-1

7. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
(continued)
Method 1: Using paper and pencil.
T(c) = 2c + 10 The slope is 2. The intercept on the vertical axis is 10.
T(c) = 4c The slope is 4. The intercept on the vertical axis is 0.
Graph the equations.
T(c) = 2c + 10
T(c) = 4c
The lines intersect at (5, 20). After 5 classes, both will cost $20.
7-1

8. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
(continued)
Method 2: Using a graphing calculator.
First rewrite the equations using x and y.
T(c) = 2c y = 2x + 10
T(c) = 4c y = 4x
Then graph the equations using a graphing calculator.
Set an appropriate range.
Then graph the equations.
Use the key to find the coordinates of the intersection point.
The lines intersect at (5, 20). After 5 classes, both will cost $20.
7-1

9. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solve by graphing. y = 3x + 2
y = 3x – 2
Graph both equations on the same coordinate plane.
y = 3x – 2 The slope is 3. The y-intercept is –2.
y = 3x + 2 The slope is 3. The y-intercept is 2.
The lines are parallel. There is no solution.
7-1

10. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solve by graphing. 3x + 4y = 12
y = – x + 3 3 4
Graph both equations on the same coordinate plane.
y = – x + 3 The slope is – . The y-intercept is 3. 3 4
3x + 4y = 12 The y-intercept is 3.
The x-intercept is 4.
The graphs are the same line. The solutions are an infinite number of ordered pairs (x, y), such that
y = – x + 3. 3 4
7-1

11. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
pages 343–345  Exercises
1. Yes, (–1, 5) makes
both equations true.
2. No, (–1, 5) makes only
one equation true.
3. Yes, (–1, 5) makes both
equations true.
4. Yes, (–1, 5) makes both
5. (0, 2);
6. (0, 0);
7. (1, 1); 
8. (1, 5);
9. (6, –1); 
7-1

12. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
10. (2, 2);
11. (4, 0);
12. (2, 3);
13. a. 3 weeks    b. $35
14. 7 weeks
15. no solution;
16. no solution;
17. infinitely many solutions;
7-1

13. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
25. (20, 60);
26. Answers may vary.
Sample: y = –1; x = 2
27. Answers may vary.
Sample: y = 2x – 1, y = 2x + 5
28. Answers may vary.
Sample: x + y = 3, 3x + 3y = 9
18. no solution;
19. no solution; same slope,
different y-int.
20. inf. many solutions;
equivalent equations
21. one solution; different slopes
22. inf. many solutions;
23. A
24. 5 min
7-1

14. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
29. (2, 20)
30. (15, 40)
31. (12, 30)
32. (–20, 0)
33. a. time on the horizontal and
distance on the vertical
b. Red represents the tortoise
because it shows distance
changing steadily over
time. Blue represents the
hare because it is steeper
than the other line at the
ends but shows no change
in distance while the hare
is napping.
c. The point of intersection
shows when the tortoise
passed the sleeping hare.
34. (–12, –16)
35. (–2, 10)
7-1

15. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
36. (–30, –2.5)
37. (–0.9, 1.6)
38. (2, 3)
39. a. c = t; c = t;
(2, 200);
b. The cost of renting either studio
for 2 h is the same, $200.
40. a. no values
b. w = v
c. w = v
41. a. sometimes
b. never
42. (–9, –2)
43. D
44. F
45. [2] a. Answers may vary.
Sample: x – 2y = 6
b. Since the lines do not intersect,
the lines are parallel. Parallel lines
have the same slope but
different intercepts.
[1] incorrect equation OR incorrect
explanation /
7-1

16. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
46. [4] a. y = x
y = x
b. $500
c. cellular phone sales
[3] appropriate methods but
one computational error
[2] incorrect system solved
correctly OR correct system
solved incorrectly
[1] no work shown
47. It is translated up 2 units.
48. It is translated 3 units left.
49. It is translated 5 units up
and 2 units right.
50. 25% increase
% decrease
52. 20% increase
% increase
% decrease
55. 25% increase
56. 10% increase
% decrease 1 3 1 3
7-1

17. Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solve by graphing.
1. y = –x – 2 2. y = –x y = 3x + 2
y = x y = 2x – 6 6x – 2y = –4
4. 2x – 3y = –2x + 4y = 12
y = x – 5 – x + y = –3 2 3
(3, 1)
(3, 0)
Infinitely many solutions 1 2
(6, 1)
no solution
7-1

18. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(For help, go to Lessons2-4 and 7-1.)
Solve each equation.
1. m – 6 = 4m n = 9 – 2n t + 5 = 10 1 3
For each system, is the ordered pair a solution of both equations?
4. (5, 1) y = –x (2, 2.4) 4x + 5y = 20
y = x – 6 2x + 6y = 10
7-2

19. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solutions
m – 6 = 4m n = 9 – 2n
m – m – 6 = 4m – m n + 2n = 9 – 2n + 2n
–6 = 3m n = 9
–14 = 3m n = 1
–4 = m
t + 5 = 10
t = 5
t = 15 1 2 2 3 1 3 1 3
7-2

20. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solutions (continued)
4. (5, 1) in y = –x + 4 (5, 1) in y = x – 6
– – 6
1 = – = –1
no no
No, (5, 1) is not a solution of both equations.
5. (2, 2.4) in 4x + 5y = 20 (2, 2.4) in 2x + 6y = 10
4(2) + 5(2.4) (2) + 6(2.4) 10
20 = = 10
yes no
No, (2, 2.4) is not a solution of both equations. / / /
7-2

21. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solve using substitution. y = 2x + 2
y = –3x + 4
Step 1: Write an equation containing only one variable and solve.
y = 2x + 2 Start with one equation.
–3x + 4 = 2x + 2 Substitute –3x + 4 for y in that equation.
4 = 5x + 2 Add 3x to each side.
2 = 5x Subtract 2 from each side.
0.4 = x Divide each side by 5.
Step 2: Solve for the other variable.
y = 2(0.4) + 2 Substitute 0.4 for x in either equation.
y = Simplify.
y = 2.8
7-2

22. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Since x = 0.4 and y = 2.8, the solution is (0.4, 2.8).
Check: See if (0.4, 2.8) satisfies y = –3x + 4 since y = 2x + 2 was used in Step 2.
–3(0.4) + 4 Substitute (0.4, 2.8) for (x, y) in the equation. –
2.8 = 2.8
7-2

23. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solve using substitution.
–2x + y = –1
4x + 2y = 12
Step 1: Solve the first equation for y because it has a coefficient of 1.
–2x + y = –1
y = 2x –1 Add 2x to each side.
Step 2: Write an equation containing only one variable and solve.
4x + 2y = 12 Start with the other equation.
4x + 2(2x –1) = 12 Substitute 2x –1 for y in that equation.
4x + 4x –2 = 12 Use the Distributive Property.
8x = 14 Combine like terms and add 2 to each side.
x = 1.75 Divide each side by 8.
7-2

24. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Step 3: Solve for y in the other equation.
–2(1.75) + y = 1 Substitute 1.75 for x.
–3.5 + y = –1 Simplify.
y = 2.5 Add 3.5 to each side.
Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5).
7-2

25. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
A youth group with 26 members is going to the beach. There will also be five chaperones that will each drive a van or a car. Each van seats 7 persons, including the driver. Each car seats 5 persons, including the driver. How many vans and cars will be needed?
Let v = number of vans and c = number of cars.
Drivers v c = 5
Persons v c = 31
Solve using substitution.
Step 1: Write an equation containing only one variable.
v + c = 5 Solve the first equation for c.
c = –v + 5
7-2

26. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Step 2: Write and solve an equation containing the variable v.
7v + 5c = 31
7v + 5(–v + 5) = 31 Substitute –v + 5 for c in the second equation.
7v – 5v + 25 = 31 Solve for v.
2v + 25 = 31
2v = 6
v = 3
Step 3: Solve for c in either equation.
3 + c = 5 Substitute 3 for v in the first equation.
c = 2
7-2

27. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Three vans and two cars are needed to transport 31 persons.
Check: Is the answer reasonable? Three vans each transporting 7 persons is 3(7), of 21 persons. Two cars each transporting 5 persons is 2(5), or 10 persons. The total number of persons transported by vans and cars is , or 31. The answer is correct.
7-2

28. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
pages 349–352  Exercises
1. D
2. C
3. B
4. A
5–16. Coordinates given in
alphabetical order.
5. (9, 28)
6. (– , –4 )
7. (6 , – )
8. (2, 4 )
9. (4, 20)
10. ( , 9 )
11. (2, 0)
12. (7 , )
13. (6, –2)
14. (3, –2)
15. (8, –7)
16. (–3, 9.4)
17. 4 cm by 13 cm
18. 4 wk
19. (15, 15)
20. (9, 126)
21. (–4, 4) 3 4 3 8
video rentals
acres flax,
160 acres sunflowers
24. 9 yr
25. estimate: ( , 1); 
; ( , 1) 7 17 8 17 1 2 1 2 1 2 1 2 1 3 1 3 1 2
7-2

29. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
28. estimate: (–3.5, –3.5);
(– , – )
29. estimate: (– , 4 );
(– , )
26. estimate: (–2, 3); 
; (–2, 3)
27. estimate: (–1, 1);
(–1, 1) 10 3 11 3 3 4 3 4 2 3 14 3
7-2

30. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
30. estimate: (–4, 0);
(– , – )
31. a. Let n = number of nickels,
let d = number of dimes.
        n + d = 28
0.05n d = 2.05
b. Solve the first eq. for either var.
Sub. the expression into the second
eq. Solve this eq. for the other var.,
and then sub. its value into the first
eq. and solve for the first var.
c. (15, 13)
32. Answers may vary.
Sample: y = x and y = –3x + 2;
( , )
33. a. (x, y) such that y = 0.5x + 4 b.
c. Graphing shows only one line.
Substitution results in a true
equation with no variables. 1 2 1 2 50 11 2 11
7-2

31. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
38. (2, )
39. (– , 0)
40. (4, –2)
41. inf. many solutions
42. no solution
43. 1 solution
44. a. g = b
b. (b, g) = (572, 598)
c. 26
45. a. (t, d) = (9, 79.2)
b. yes
46. (r, s, t) = (7, 9, 4) 1 2
50. [2] 7(–7) – 4(–2) 29
      –
           –
No, (–2, –7) must
satisfy both
equations to be a
solution of the
system.
[1] no explanation
given
34. a. no solution b.
c. Graphing shows 2 parallel lines. Substitution results in a false equation with no variables.
35. (2, 4)
36. (– , – )
37. (2, –4) 1 2 = / 23 22 1 2 1 2
7-2

32. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
51. (12, 10);
52. (2, 1);
53. (4, 2);
54.
55.
56.
57.
58.
59.
7-2

33. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solve each system using substitution.
1. 5x + 4y = x + y = m – 2n = 7
y = 5x 2x – y = 6 3m + n = 4
(0.2, 1)
(2, 2)
(1.25, 0.25)
7-2

34. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(For help, go to Lesson 7-2.)
Solve each system using substitution.
1. y = 4x – 3 2. y + 5x = y = –2x + 2
y = 2x y = 7x – x – 17 = 2y
7-3

35. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solutions
1. y = 4x – 3
y = 2x + 13
Substitute 4x – 3 for y in the second equation.
4x – 3 = 2x + 13
4x – 2x – 3 = 2x – 2x + 13
2x – 3 = 13
2x = 16
x = 8
y = 4x – 3 = 4(8) – 3 = 32 – 3 = 29
Since x = 8 and y = 29, the solution is (8, 29).
7-3

36. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solutions (continued)
2. y + 5x = 4
y = 7x – 20
Substitute 7x – 20 for y in the first equation.
y + 5x = 4
7x – x = 4
12x – 20 = 4
12x = 24
x = 2
y = 7x – 20 = 7(2) – 20 = 14 – 20 = –6
Since x = 2 and y = –6, the solution is (2, –6).
7-3

37. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solutions (continued)
3. y = –2x + 2
3x – 17 = 2y
Substitute –2x + 2 for y in the second equation.
3x – 17 = 2(–2x + 2)
3x – 17 = –4x + 4
7x – 17 = 4
7x = 21
x = 3
y = –2x + 2 = –2(3) + 2 = –6 + 2  –4
Since x = 3 and y = –4, the solution is (3, –4).
7-3

38. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve by elimination. 2x + 3y = 11
–2x + 9y = 1
Step 1: Eliminate x because the sum of the coefficients is 0.
2x + 3y = 11
–2x + 9y =1
0 + 12y = 12 Addition Property of Equality
y = 1 Solve for y.
Step 2: Solve for the eliminated variable x using either original equation.
2x + 3y = 11 Choose the first equation.
2x + 3(1) = 11 Substitute 1 for y.
2x + 3 = 11 Solve for x.
2x = 8
x = 4
7-3

39. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Since x = 4 and y = 1, the solution is (4, 1).
Check: See if (4, 1) makes true the equation not used in Step 2.
–2(4) + 9(1) Substitute 4 for x and 1 for y into the second equation. –
1 = 1
7-3

40. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
On a special day, tickets for a minor league baseball game were $5 for adults and $1 for students. The attendance that day was 1139, and $3067 was collected. Write and solve a system of equations to find the number of adults and the number of students that attended the game.
Define: Let a = number of adults
Let s = number of students
Relate: total number at the game total amount collected
Write: a s = a s = 3067
Solve by elimination.
7-3

41. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 1: Eliminate one variable.
a + s = 1139
5a + s = 3067
–4a = –1928 Subtraction Property of Equality
a = 482 Solve for a.
Step 2: Solve for the eliminated variable using either of the original equations.
a + s = 1139 Choose the first equation.
s = 1139 Substitute 482 for a.
s = 657 Solve for s.
There were 482 adults and 657 students at the game.
Check: Is the solution reasonable? The total number at the game was , or The money collected was $5(482), or $2410, plus $1(657), or $657, which is $3067. The solution is correct.
7-3

42. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve by elimination. 3x + 6y = –6
–5x – 2y = –14
Step 1: Eliminate one variable.
Start with the given
system.
3x + 6y = –6
–5x – 2y = –14
To prepare to eliminate y, multiply the second equation by 3.
3x + 6y = –6
3(–5x – 2y = –14)
Add the equations to eliminate y.
3x + 6y = –6
–15x – 6y = –42
–12x – 0 = –48
Step 2: Solve for x.
–12x = 48
x = 4
7-3

43. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 3: Solve for the eliminated variable using either of the original equations.
3x + 6y = –6 Choose the first equation.
3(4) + 6y = –6 Substitute 4 for x.
y = –6 Solve for y.
6y = –18
y = –3
The solution is (4, –3).
7-3

44. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Suppose the band sells cans of popcorn for $5 per can and cans of mixed nuts for $8 per can. The band sells a total of 240 cans and receives a total of $1614. Find the number of cans of popcorn and the number of cans of mixed nuts sold.
Define: Let p = number of cans of popcorn sold.
Let n = number of cans of nuts sold.
Relate: total number of cans total amount of sales
Write: p n = p n = 1614
7-3

45. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 1: Eliminate one variable.
Start with the given
system.
p + n = 240
5p + 8n = 1614
To prepare to eliminate p, multiply the first equation by 5.
5(p + n = 240)
5p + 8n = 1614
Subtract the equations to eliminate p.
5p + 5n = 1200
5p + 8n = 1614
0 – 3n = –414
Step 2: Solve for n.
–3n = –414
n = 138
7-3

46. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 3: Solve for the eliminated variable using either of the original equations.
p + n = 240 Choose the first equation.
p = 240 Substitute 138 for n.
p = 102 Solve for p.
The band sold 102 cans of popcorn and 138 cans of mixed nuts.
7-3

47. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve by elimination. 3x + 5y = 10
5x + 7y = 10
Step 1: Eliminate one variable.
Start with the given
system.
3x + 5y = 10
5x + 7y = 10
To prepare to eliminate x, multiply one equation by 5 and the other equation by 3.
5(3x + 5y = 10)
3(5x + 7y = 10)
Subtract the equations to eliminate x.
15x y = 50
15x y = 30
y = 20
Step 2: Solve for y.
4y = 20
y = 5
7-3

48. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 3: Solve for the eliminated variable x using either of the original equations.
3x + 5y = 10 Use the first equation.
3x + 5(5) = 10 Substitute 5 for y.
3x = 10
3x = –15
x = –5
The solution is (–5, 5).
7-3

49. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
pages 356–359  Exercises
1. (1, 3)
2. (2, –2)
3. (5, –17)
4. (–3, 4)
5. (–9, )
6. (– , 10)
7. a. x + y = 20, x – y = 4
b. 12 and 8
8. a. a + s = 456, 3.5a + s = 1131
b. 270 adult, 186 student
9. (–5, 1)
10. (11, –3)
11. (–2, – )
12. (1, 14)
13. ( , 1)
14. (–2, 3)
15. a. 30w + = 17.65,
20w = 25.65
b. $.39 for a wallet size,
$5.95 for an 8  10
16. a. x = burritos, y = tacos;
3x + 4y = 11.33, 9x + 5y = 23.56
b. $1.79 for a burrito, $1.49 for a taco
17. (–1, –3)
18. (2.5, 1)
19. (2, –2) 5 2 1 2 1 2 1 2
7-3

50. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
20. (10, 8)
21. (–1, )
22. (1, 5)
23–28. Methods may vary.
Samples are given.
23. (–1, –2); substitution;
both solved for y
24. (15, –10); elimination;
equations not solved for y
25. (10, 2); substitution;
one eq. solved for x
26. (–3, 11); elimination;
eqs. not solved for a variable
27. (5, 1); substitution;
one eq. solved for x
28. ( , 2 ); substitution;
eqs. solved for y
29. one night: $81.25;
one meal: $8.13
30. a. brass: $6; steel: $3
b. $99
31. She forgot to multiply –8 by 6.
32. Answers may vary.
Sample: 2x – 3y = 6, x + 3y = 9; (5, )
33. (10, –6)
34. (8, 12) 3 2 1 3 1 3 4 3
7-3

51. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
35. (–15, –1)
36. (20, 12)
37. (–2, 16)
38. (33, –48)
39. 9
40. Answers may vary. Sample:
You solve a system using the
elimination method by adding
or subtracting the eqs. to
eliminate one of the variables.
This sum or difference is one eq.
with one variable that can be solved.
Use addition: Use subtraction: Use multiplication:
3x + 2y = 6 5x + 3y = 15 4x + 5y = 20
–x – 2y = 4 5x – 2y = 10 2x – y = 10
41. B1 = 3 volts; B2 = 1.5 volts
42. (–3, 2)
43. ( , 0) (a = 0, b = 1)
44. (8, 13, 20)
45. CD: $3.40, cassette: $1.80
46. a. 2.8 g of gold
b. about 2.7%
47. D
48. H c a / /
7-3

52. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
49. [2]   y – x = 13
7y + x = 11
8y = 24
      y = 3
 3 – x = 13
     x = –10; (–10, 3)
[1] no work shown
50. [4]
A = h(b1 + b2)
= (4)(4 + 2)
= 2(6)
= 12
The area is 12 square units.
[3] graph and formula with one mathematical error
[2] graph but error in formula
[1] graph only
51–53. Coordinates given in alphabetical order.
51. (6, 26)
52. (–1, 4)
53. (9, –5)
54.
55.
56.
57. 71
58. –18
59. –44
60. 22
61. 83
62. –6 1 9 2 9 1 2 1 15 1 2
7-3

53. Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve using elimination.
1. 3x – 4y = m + 3n = 22
2x + 4y = m + 6n = 34
3. –6x + 5y = p + 5q = 2
3x + 4y = p – 9q = 17
(3, 0.5)
(2, 4)
(1, 2)
(1, 1)
7-3

54. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
(For help, go to Lesson 2-5.)
1. Two trains run on parallel tracks. The first train leaves a city hour before the second train. The first train travels at 55 mi/h. The second train travels at 65 mi/h. Find how long it takes for the second train to pass the first train.
2. Luis and Carl drive to the beach at an average speed of 50 mi/h. They return home on the same road at an average speed of 55 mi/h. The trip home takes 30 min. less. What is the distance from their home to the beach? 1 2
7-4

55. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
Solutions
1. 55t = 65(t – 0.5)
55t = 65t – 32.5
32.5 = 10t
3.25 = t
It takes 3.25 hours for the second train to pass the first train.
2. 50t = 55(t – 0.5)
50t = 55t – 27.5
27.5 = 5t
5.5 = t
50t = 50(5.5) = 275
It is 275 miles to the beach from their home.
7-4

56. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
A chemist has one solution that is 50% acid. She has another solution that is 25% acid. How many liters of each type of acid solution should she combine to get 10 liters of a 40% acid solution?
Define: Let a = volume of the Let b = volume of the
50% solution. 25% solution.
Relate: volume of solution amount of acid
Write: a b = a b = 0.4(10)
Step 1: Choose one of the equations and solve for a variable.
a + b = 10 Solve for a.
a = 10 – b Subtract b from each side.
7-4

57. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
(continued)
Step 2: Find b.
0.5a b = 0.4(10)
0.5(10 – b) b = 0.4(10) Substitute 10 – b for a.
Use parentheses.
5 – 0.5b b = 0.4(10) Use the Distributive Property.
5 – 0.25b = 4 Simplify.
–0.25b = –1 Subtract 5 from each side.
b = 4 Divide each side by –0.25.
Step 3: Find a. Substitute 4 for b in either equation.
a = 10
a = 10 – 4
a = 6
To make 10 L of 40% acid solution, you need 6 L of 50% solution and 4 L of 25% solution.
7-4

58. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
Suppose you have a typing service. You buy a personal computer for $1750 on which to do your typing. You charge $5.50 per page for typing. Expenses are $.50 per page for ink, paper, electricity, and other expenses. How many pages must you type to break even?
Define: Let p = the number of pages.
Let d = the amount of dollars of expenses or income.
Relate: Expenses are per-page Income is price
expenses plus times pages typed.
computer purchase.
Write: d = p d = p
7-4

59. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
(continued)
Choose a method to solve this system. Use substitution since it is easy to substitute for d with these equations.
d = 0.5p Start with one equation.
5.5p = 0.5p Substitute 5.5p for d.
5p = Solve for p.
p = 350
To break even, you must type 350 pages.
7-4

60. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
Suppose it takes you 6.8 hours to fly about 2800 miles from Miami, Florida to Seattle, Washington. At the same time, your friend flies from Seattle to Miami. His plane travels with the same average airspeed, but this flight only takes 5.6 hours. Find the average airspeed of the planes. Find the average wind speed.
Define: Let A = the airspeed. Let W = the wind speed.
Relate: with tail wind with head wind
(rate)(time) = distance (rate)(time) = distance
(A + W) (time) = distance (A + W) (time) = distance
Write: (A + W)5.6 = (A + W)6.8 = 2800
7-4

61. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
(continued)
Solve by elimination. First divide to get the variables on the left side of each equation with coefficients of 1 or –1.
(A + W)5.6 = A + W = 500 Divide each side by 5.6.
(A – W)6.8 = A – W Divide each side by 6.8.
Step 1: Eliminate W.
A + W = 500
A – W = 412 Add the equations to eliminate W.
2A = 912
Step 2: Solve for A.
A = 456 Divide each side by 2.
Step 3: Solve for W using either of the original equations.
A + W = Use the first equation.
W = Substitute 456 for A.
W = 44 Solve for W.
The average airspeed of the planes is 456 mi/h. The average wind speed is 44 mi/h.
7-4

62. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
pages 365–368  Exercises
1. a. 4a + 5b = 6.71
b. 5a + 3b = 7.12
c. pen: $1.19, pencil: $.39
2. D
3. a.
b. a + b = 24;
0.04a b = 1.2
c. 18 kg A, 6 kg B
4. a. at 8 wk
b. $160; $62
games
6. 40 shirts
7. a. s + c = 2.75
b. s – c = 1.5
c mi/h
d mi/h
8. a. (A + W )4.8 = 2100
(A – W )5.6 = 2100
b mi/h
c mi/h
9–14.  Answers may vary.
Samples are given.
9. Substitution; one eq. is solved for t.
10. Substitution; both eqs. are solved for y.
11. Elimination; subtract to eliminate m.
12. Substitution; both eqs. are solved for y.
7-4

63. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
13. Elimination; mult. first eq.
by 3 and add to elim. y.
14. Substitution; one eq. is
solved for u.
15. a. t = 99 – 3.5m; t = m;
t = 41.25°, m = 16.5 min
b. After 16.5 min, the temp.
of either piece will be 41.25°C.
16. 5 cm; 12 cm
17. Answers may vary. Sample:
You have 10 coins, all dimes and
quarters. The value of the coins is
$1.75. How many dimes do you have?
How many quarters do you have?
        q + d = 10
0.25q d = 1.75
You have 5 dimes and 5 quarters.
small mowers, 11 large mowers
19. a. 42 mi/h
b. 12 mi/h
20. a. g = b
b. g + b = 1908
     g = b
901 boys, 1007 girls
21. a. 16 days
b. Answers may vary. Sample:
If you plan to ski for many years,
you should buy the equipment,
since you will break even at
16 days.
22. x = 2
y = 4 19 17 19 17
7-4

64. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
24. a. 2.50s = 10,000
= s
800 small, 2000 large
b. 560 h
c. $17.86/h
25. C
26. H
27. B
28. [2] 3V + C = V + 4C = 88
2V + 4C =   2V + 4C = 28
10V = 60
V = 60
6 people/van
[1] correct answer, no work shown
29. (4, 1)
30. (–6, 7)
31. (3, )
32.
33. 1
34.
35. –2
36. –
37. undefined 5 2 3 2 3 2 1 3 4 11
7-4

65. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
38. 6 < y < 10  
39. –8 < n 3  
40. 1 < k < 6  
p 4  
42. –5 < c 5  
43. 5 > w > 1  
< –
< –
< –
< –
7-4

66. Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
1. One antifreeze solution is 10% alcohol. Another antifreeze solution is 18% alcohol. How many liters of each antifreeze solution should be combined to create 20 liters of antifreeze solution that is 15% alcohol?
2. A local band is planning to make a compact disk. It will cost $12,500 to record and produce a master copy, and an additional $2.50 to make each sale copy. If they plan to sell the final product for $7.50, how many disks must they sell to break even?
3. Suppose it takes you and a friend 3.2 hours to canoe 12 miles downstream (with the current). During the return trip, it takes you and your friend 4.8 hours to paddle upstream (against the current) to the original starting point. Find the average paddling speed in still water of you and your friend and the average speed of the current of the river. Round answers to the nearest tenth.
7.5 L of 10% solution; 12.5 L of 18% solution
2500 disks
still water: 3.1 mi/h; current: 0.6 mi/h
7-4

67. Describe each statement as always, sometimes, or never true.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
(For help, go to Lesson 3-1 and 6-3.)
Describe each statement as always, sometimes, or never true.
1. –3 > – n n
Write each equation is slope-intercept form.
4. 2x – 3y = y + 3x = y – 3x = 1
< –
> –
7-5

68. 1. –3 < –2 is true, so –3 > –2 is never true.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
Solutions
1. –3 < –2 is true, so –3 > –2 is never true.
2. 8 = 8 is true, so is always true.
3. 4(1) is true and 4(–1) –1 is false, so 4n n is sometimes true.
4. 2x – 3y = y + 3x = 6
–3y = –2x y = –3x + 6
y =
y = x – 3
6. 4y – 3x = 1
4y = 3x + 1
y = x +
< –
> –
> –
> –
–2x + 9 –3 2 3
3x  1 4 3 4 1 4
7-5

69. First, graph the boundary line y = –2x + 1.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
Graph y > –2x + 1.
First, graph the boundary line y = –2x
The coordinates of the points on the boundary line do not make the inequality true. So, use a dashed line.
Shade above the boundary line.
Check The point (0, 2) is in the region of the graph of the inequality.
See if (0, 2) satisfies the inequality.
y > –2x + 1
2 > –2(0) + 1
2 > 1
Substitute (0, 2) for (x, y).
7-5

70. –3y –4x + 6 Subtract 4x from each side.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
Graph 4x – 3y
> –
Solve 4x – 3y for y.
> –
4x – 3y 6
–3y –4x Subtract 4x from each side.
y x – 2 Divide each side by –3. Reverse the inequality symbol. 4 3
< –
>
Graph y = x – 2.
The coordinates of the points on the boundary line make the inequality true. So, use a solid line. 4 3
Since y x – 2, shade below the boundary line. 4 3
< –
7-5

71. Relate: cost of plus cost of is less than total budget
Linear Inequalities
ALGEBRA 1 LESSON 7-5
Suppose your budget allows you to spend no more than $24 for decorations for a party. Streamers cost $2 a roll and tablecloths cost $6 each. Use intercepts to graph the inequality that represents the situation. Find three possible combinations of streamers and tablecloths you can buy.
Relate: cost of plus cost of is less than total budget
streamers tablecloths or equal to
Define: Let s = the number of rolls of streamers.
Let t = the number of rolls of streamers.
Write: s t
< –
7-5

72. Graph 2s + 6t 24 by graphing the intercepts (12, 0) and (0, 4).
Linear Inequalities
ALGEBRA 1 LESSON 7-5
(continued)
Graph 2s + 6t by graphing the intercepts (12, 0) and (0, 4).
The coordinates of the points on the boundary line make the inequality true. So, use a solid line.
Graph only in Quadrant I, since you cannot buy a negative amount of decorations.
< –
Test the point (0, 0).
2s + 6t
2(0) + 6(0) Substitute (0, 0) for (s, t).
Since the inequality is true, (0, 0) is a solution.
< –
7-5

73. Linear Inequalities (continued)
ALGEBRA 1 LESSON 7-5
(continued)
Shade the region containing (0, 0). The graph below shows all the possible solutions of the problem.
Since the boundary line is included in the graph, the intercepts are also solutions to the inequality.
The solution (9, 1) means that if you buy 9 rolls of streamers, you can buy 1 tablecloth. Three solutions are
(9, 1), (6, 2), and (3, 3).
7-5

74. Linear Inequalities pages 373–376 Exercises 1. no 2. yes 3. yes 4. yes
ALGEBRA 1 LESSON 7-5
pages 373–376  Exercises
1. no
2. yes
3. yes
4. yes
5. no
6. yes
7. A
8. B
9. B
10. A
14.
15.
16.
17.
18.
19. y x – ;
11.
12.
13. 2 3 7 3
< –
7-5

75. c. Answers may vary. Sample: 8 blue
Linear Inequalities
ALGEBRA 1 LESSON 7-5 5 3
20. y x – 2;
21. y x – ;
22. y – x – 2;
23. a. 3x + 5y 48 b.
c. Answers may vary. Sample: 8 blue
and 4 gold, 2 blue and 8 gold, 12 blue
and 2 gold
d. No; you cannot buy –2 rolls of paper.
> –
< – 2 3 8 3
< – 3 2
< –
7-5

76. c. Answers may vary. Sample: 30 canvas and 10 nylon,
Linear Inequalities
ALGEBRA 1 LESSON 7-5
25.
26.
27.
24. a. 3n + 10c > 250 b.
c. Answers may vary. Sample:
30 canvas and 10 nylon,
26 canvas and 20 nylon,
35 canvas and 10 nylon
d. Domain and range values must
be positive integers, since you
cannot buy portions of packs or
a negative number of packs.
28.
29.
30.
7-5

77. c. Yes; you can buy 8 CDs and 9 tapes. d. 43 < – 31.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
34. y > 2x – 1
35. x –3
36. y x – 2
37. a. 12x + 8y b.
c. Yes; you can buy 8 CDs and 9 tapes.
d. 43 1 3
< –
31.
32.
33. For an inequality written in the
form y < or y , shade below the
boundary line. For an inequality
written in the form y > or y ,
shade above the boundary line.
> –
<
7-5

78. 42. It should be shaded above the line, and the line should be dashed.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
38. x > 0;
39. y < 0;
40. y 0;
41. x < y;
42. It should be shaded above
the line, and the line should
be dashed.
43. y < x + 2
44. a.
b. y > x c.
45. a. 2w ;
b. Answers may vary.
Sample: 10 ft by 10 ft,
5 ft by 5 ft
c. No; (12, 15) is not in the
shaded region and is not
a sol. of the inequality.
46. y < x – 3
47. y 2x – 2
< –
> –
> – 5 12
7-5

79. c. Answers may vary. Sample: (2, 3) d.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
48. a. Answers may vary.
Sample: 2x + y > 3
b. Answers may vary.
Sample: 3x + y < 1
c. If an inequality is in
standard form, where
A, B, and C are all
positive, you shade
above the line for >
or and below the
line for < or .
d. yes
49. a. yes
b. yes
c. Answers may vary. Sample: (2, 3) d.
50. A
51. H
52. D
53. F
> –
< –
7-5

80. 54. [2] First, graph the solid line y = 3x – 4.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
54. [2] First, graph the solid line y = 3x – 4.
Then shade below the line.
[1] correct graph given, no explanation
55. about 775 games
mi/h, 3.25 mi/h
57. 5
58. 7
59. 11
60. –6
61. 8, 18
62. –6.7, –2.1
63. 12
64. 28
65. 11
66. 16
67. –
68. 4
69. 13 2 3 1 3
7-5

81. 1. Determine whether (4, 1) is a solution of 3x + 2y 10.
Linear Inequalities
ALGEBRA 1 LESSON 7-5
1. Determine whether (4, 1) is a solution of 3x + 2y
Graph each inequality.
2. x > – x – 2y > x + 6y
> –
yes
< –
7-5

82. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
(For help, go to Lessons 7-1 and 7-5.)
Solve each system by graphing.
1. y = 3x – y = – x x + y = 4
y = –x y = – x x – y = 8
Graph each inequality.
4. y > y x – x – 8y 1 2 1 2 2 3
< –
> –
7-6

83. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solutions
1. y = 3x – y = – x + 4
y = –x y = – x + 3
The solution is (2, 0) There is no solution.
3. x + y = y > 5
2x – y = 8
Solve equations
for y:
y = –x + 4
y = 2x – 8
The solution is (4, 0). 1 2 1 2
7-6

84. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solutions (continued)
5. y x – 1
6. 4x – 8y
–8y –4x + 4 y
y x – 2 3
< –
> –
> –
4x  4 8
< – 1 2 1 2
< –
7-6

85. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solve by graphing. y < –x + 3
–2x + 4y
> –
Graph y < –x and –2x + 4y
> –
Check The point (–1, 1) is in the region graphed by both inequalities. See if (–1, 1) satisfies both inequalities.
7-6

86. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
(continued)
y < –x –2x + 4y
1 < –(–1) Substitute (–1, 1) for (x, y). –2(–1) + 4(1)
1 <
> –
The coordinates of the points in the [lavender] region where the graphs of the two inequalities overlap are solutions of the system.
7-6

87. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Write a system of inequalities for each shaded region below.
[red] region
boundary: y = – x + 2 1 2
[blue] region
boundary: y = 4
The region lies below the boundary line, so the inequality is
y < – x 1 2
The region lies below the boundary line, so the inequality is
y < 4.
System for the [lavender] region: y < – x + 2
y < 4 1 2
7-6

88. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
You need to make a fence for a dog run. The length of the run can be no more than 60 ft, and you have 136 feet of fencing that you can use. What are the possible dimensions of the dog run?
Define: Let = length of the dog run.
Let w = width of the dog run.
Relate: The is no 60 ft. The is no 136 ft.
length more than perimeter more than
Write: w
< –
Solve by graphing w
< –
7-6

89. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
(continued) 60
m = 0:
b = 60
Shade below
= 60.
< – w
Graph the intercepts
(68, 0) and (0, 68).
< –
Test (0, 0). 2(0) + 2(0)
So shade below
w = 136
< –
The solutions are the coordinates of the points that lie in the region shaded lavender and on the lines = 60 and w = 136.
7-6

90. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Suppose you have two jobs, babysitting that pays $5 per hour and sacking groceries that pays $6 per hour. You can work no more than 20 hours each week, but you need to earn at least $90 per week. How many hours can you work at each job?
Define: Let b = hours of babysitting.
Let s = hours of sacking groceries.
Relate: The number is less 20. The amount is at 90.
of hours than or earned least
worked equal to
Write: b s b s
> –
<
7-6

91. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
(continued)
Solve by graphing. b + s
5b + 6s
< –
> –
The solutions are all the coordinates of the points that are nonnegative integers that lie in the region shaded lavender and on the lines b + s = 20 and 5b + 6s = 90.
7-6

92. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
pages 380–384  Exercises
1. no
2. yes
3. no 4. 5. 9.
10.
11. 6. 7. 8.
7-6

93. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
16. x > 5 and y –x + 3
17. y – x – 2 and y x + 2
18. y – x + 1 and y – x + 3
19. y – x – 4 and y x – 3
20. a. 1.5f + 2.5c < 9.50, f + c > 4 b.
21.
12.
13.
14.
15.
> –
22. a. 5.99x y 50,
x 0, y 1 b.
c. 2 books and 6 CDs;
no, (2, 6) is not in the
shaded region.
d. Answers may vary.
Sample: 3 books and
3 CDs for $47.94
< – 1 2 1 2
> –
> –
> –
< – 1 5 3 4
> –
> – 2 3 1 5
< –
> –
7-6

94. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
25. x 3, x –3, y 3, y –3
26. y 2, x < 5, y x
27. y x – 2, y < x + 2
28. y –x – 3, y –x + 3, y x + 3, y x – 3
29. Answers may vary.
Sample: x –2, x 4, y 1, y –2
30. a. –1     
b. 8
31. a. triangle     
b. (2, 2), (–4, –1), (–4, 2)     
c. 9 units2
23. x + y 30, 1.25x + 3y 60
24. a. x + y 12, 6x + 4y 60
b. Answers may vary. Sample:
(8, 3), (9, 1), (10, 0)
> –
< –
< –
> –
< –
> –
> –
< – 2 3 2 3
> –
> –
< –
< –
> –
> –
< –
> –
< –
< –
> –
7-6

95. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
32. a. square
b. (1, –1), (5, –1), (1, 3), (5, 3)
c. 16 units2
33. a. trapezoid
b. (0, –4), (0, 2), (2, –4), (2, 0)
c. 10 units2
34. a. triangle
b. (2, –3), (2, 2), (7, –3)
c units2
35. a. x 1, 10.99x y 45
b. (3, 0), (3, 1), (3, 2), (4, 0)
36. a.
b. No; they are parallel.
c. no
d. no
37. a.
c. It is a strip between the lines.
> –
< –
7-6

96. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
38–42. Answers may vary. Samples are given.
38. x 1 and y 2
39. x < 0 and y > 0
40. y > 5 and y < 3
41. x < 2 and y < 5
42. x > 0 and y < 0
43. a. s + d > 10, s + d < 20,
d 3, 10d s < 60
b. Answers may vary. Sample:
(8, 4.5); 12.5g; gold: $45.00, silver: $1.20
44.
45. Answers may vary.
y > x, y < 2x
46. Answers may vary.
y > x + 1, y > –x + 1
47. a. h a b.
c. Answers may vary.
Sample: 5 hits, 6 at-bats
< –
< – 1 2
> –
> –
7-6

97. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
48. a. 180x + 240y
x + y 17
x > y
y 4
b. Answers may vary. Sample:
ten 14-in. drums and
six 18-in. drums
49. D
50. G
51. [2] all points on the line 3x + 4y = 12
[1] incorrect description given
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52. [4] a. Let x = number of toppings,
and y = cost of pizza.
Maria’s: y = 0.50x + 8
Tony’s: y = 0.75x + 7
b. (4, 10)  With 4 toppings, the cost is
$10 at either Tony’s or Maria’s.
c. Answers may vary. Sample: Since
I prefer more than 4 toppings, I will
go to Maria’s, because the pizza will
be less expensive.
[3] (a) and (b) only done correctly
[2] (a) done correctly, but student makes a computational error in (b)
[1] error in (a), but system solved correctly
< –
> –
7-6

98. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
59.
60. –3
61. –8
62. –
63. –
64. –
65.
66.
67.
68. ƒ(x) = 7x
69. ƒ(x) = x + 6
70. ƒ(x) = x 2 5 2
53.
54.
55.
56.
57.
58. 1 4 1 5 8 3 10 9 6 13 15 4
7-6

99. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solve each system by graphing.
1. x x + 3y > y = x – 3
y < x + 2y < x – 3y –9
4. Write a system of inequalities for the following graph. 2 3
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7-6

100. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solve each system by graphing.
1. x x + 3y > y x – 3
y < x + 2y < x – 3y –9 2 3
> –
> –
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7-6

101. Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
4. Write a system of inequalities for the following graph.
y < x + 3
y > – x – 2 1 2
7-6

102. Systems of Equations and Inequalities
ALGEBRA 1 CHAPTER 7
9. (8, 25)
10. (3, –7)
11. (2, –9)
12. (3.5, 1)
13. (8, –24)
14. (11, – )
15. (6, 4)
16. (–5, –2)
17. b + m = 35,
b + 2m = 45; $25; $10
= , n + p = 24;
15 novelists, 9 poets
1. (2, –1)
2. (0, 4)
3. one
4. one
5. none
6. one
7. one
8. inf. many
19. q + n = 15,
0.25q n = 2.75;
10 quarters, 5 nickels
20. Answers may vary.
Sample: You solve a
system of linear equations
by finding a single point that
satisfies all the equations in
the system. You solve a
system of linear inequalities
by graphing a region that
contains points that satisfy
all the inequalities in the
system.
21. C 2 5 n p 5 3
7-A

103. Systems of Equations and Inequalities
ALGEBRA 1 CHAPTER 7
22.
23.
24.
25.
26. Answers may vary.
Sample: y = x + 1,
y = 3x – 5; (3, 4)
27. a. 0.10d q < 5.00;
b. 49 items
c. 19 items
28. a. w 30, w b.
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7-A

104. Systems of Equations and Inequalities
ALGEBRA 1 CHAPTER 7
29. a.
b. x + y = 200, 0.3x + 0.5y = 84; 120 liters of
50% insecticide and 80 liters of 30% insecticide
7-A