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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 – 2
2. 8 – 4z = 2z – 13
3. 8q – 12 = 3q + 23
Graph each pair of equations on the same coordinate plane.
4. y = 3x – 6
y = –x + 2
5. y = 6x + 1
y = 6x – 4
6. y = 2x – 5
6x – 3y = 15
7. y = x + 5
y = –3x + 5
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solutions
1.
2n + 3 = 5n – 2
2.
2n – 2n + 3 = 5n – 2n – 2
8 – 4z + 4z = 2z + 4z – 13
3 = 3n – 2
8 = 6z – 13
5 = 3n
21 = 6z
12 = n
1
3
3
3.
8 – 4z = 2z – 13
2
8q – 12 = 3q + 23
8q – 3q – 12 = 3q – 3q + 23
5q – 12 = 23
5q = 35
q=7
7-1
=z
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solutions (continued)
4. y = 3x – 6
5. y = 6x + 1
y = –x + 2
y = 6x – 4
6.
y = 2x – 5
7. y = x + 5
6x – 3y = 15
y = –3x + 5
–3y = –6x – 15
y = –6x  15
–3
y = 2x – 5
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solve by graphing. Check your solutions.
y = 2x + 1
y = 3x – 1
Graph both equations on the same coordinate plane.
y = 2x + 1
y = 3x – 1
The slope is 2. The y-intercept is 1.
The slope is 3. The y-intercept is –1.
7-1
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.
Check: See if (2, 5) makes both equations true.
y = 2x + 1
5 2(2) + 1
5 4+1
5=5
Substitute (2, 5) for (x, y).
7-1
y = 3x – 1
5 3(2) – 1
5 6–1
5=5
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Suppose you plan to start taking an aerobics class. Nonmembers 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 c
= number of classes.
Let T(c) = total cost of the classes.
Relate: cost
is
membership
fee
plus
Write: member T(c)
=
10
+
2 c
=
0
+
4 c
non-member T(c)
7-1
cost of classes
attended
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
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 + 10
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
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
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solve by graphing. 3x + 4y = 12
y=–3 x+3
4
Graph both equations on the same coordinate plane.
3x + 4y = 12
The y-intercept is 3.
The x-intercept is 4.
y = –3 x + 3
The slope is – 3 . The y-intercept is 3.
4
4
The graphs are the same line. The solutions are an
infinite number of ordered pairs (x, y), such that
y = – 3 x + 3.
4
7-1
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
equations true.
5. (0, 2);
6. (0, 0);
8. (1, 5);
9. (6, –1);
7. (1, 1);
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
10. (2, 2);
12. (2, 3);
13. a. 3 weeks
11. (4, 0);
16. no solution;
b. $35
17. infinitely many solutions;
14. 7 weeks
15. no solution;
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
18. no solution;
25. (20, 60);
19. no solution; same slope,
different y-int.
20. inf. many solutions;
equivalent equations
21. one solution; different slopes
22. inf. many solutions;
equivalent equations
23. A
24. 5 min
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
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
29. (2, 20)
31. (12, 30)
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.
30. (15, 40)
32. (–20, 0)
34. (–12, –16)
35. (–2, 10)
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
36. (–30, –2.5)
37. (–0.9, 1.6)
38. (2, 3)
39. a. c = 100 + 50t; c = 50 + 75t;
(2, 200);
b.
40. a.
b.
c.
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
The cost of renting either studio
different intercepts.
for 2 h is the same, $200.
[1] incorrect equation OR incorrect
no values
explanation
w =/ v
w=v
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
46. [4] a. y = 150 + 0.20x
y = 200 + 0.10x
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
48. It is translated 3 units left.
47. It is translated up 2 units.
52. 20% increase
49. It is translated 5 units up
and 2 units right.
50. 25% increase
51. 33 1 % decrease
3
53. 150% increase
54. 33 1 % decrease
3
55. 25% increase
56. 10% increase
57. 12.5% decrease
7-1
Solving Systems by Graphing
ALGEBRA 1 LESSON 7-1
Solve by graphing.
1. y = –x – 2
2. y = –x + 3
3. y = 3x + 2
y=2x+3
y = 2x – 6
6x – 2y = –4
(3, 1)
(3, 0)
Infinitely many
solutions
3
4. 2x – 3y = 9
5. –2x + 4y = 12
y=x–5
– 1 x + y = –3
(6, 1)
no solution
2
7-1
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 + 8
2. 4n = 9 – 2n
3. 1 t + 5 = 10
3
For each system, is the ordered pair a solution of both equations?
4. (5, 1)
y = –x + 4
5. (2, 2.4) 4x + 5y = 20
y=x–6
2x + 6y = 10
7-2
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solutions
1.
m – 6 = 4m + 8
m – m – 6 = 4m – m + 8
–6 = 3m + 8
2.
n = 11
–14 = 3m
–4 2 = m
2
3
3.
4n = 9 – 2n
4n + 2n = 9 – 2n + 2n
6n = 9
1 t + 5 = 10
3
1t =5
3
t = 15
7-2
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solutions (continued)
4. (5, 1) in y = –x + 4
1
–5 + 4
1 =/ –1
(5, 1) in y = x – 6
1
5–6
1 =/ –1
no
no
No, (5, 1) is not a solution of both equations.
5. (2, 2.4) in 4x + 5y = 20
4(2) + 5(2.4)
20
(2, 2.4) in 2x + 6y = 10
2(2) + 6(2.4) 10
8 + 12
20
20 = 20
yes
4 + 14.4 10
18.4 =/ 10
no
No, (2, 2.4) is not a solution of both equations.
7-2
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
–3x + 4 = 2x + 2
4 = 5x + 2
2 = 5x
0.4 = x
Start with one equation.
Substitute –3x + 4 for y in that equation.
Add 3x to each side.
Subtract 2 from each side.
Divide each side by 5.
Step 2: Solve for the other variable.
y = 2(0.4) + 2
y = 0.8 + 2
y = 2.8
Substitute 0.4 for x in either equation.
Simplify.
7-2
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.
2.8
–3(0.4) + 4
2.8
–1.2 + 4
2.8 = 2.8
Substitute (0.4, 2.8) for (x, y) in the equation.
7-2
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
4x + 2(2x –1) = 12
4x + 4x –2 = 12
8x = 14
x = 1.75
Start with the other equation.
Substitute 2x –1 for y in that equation.
Use the Distributive Property.
Combine like terms and add 2 to each side.
Divide each side by 8.
7-2
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Step 3: Solve for y in the other equation.
–2(1.75) + y = 1
–3.5 + y = –1
y = 2.5
Substitute 1.75 for x.
Simplify.
Add 3.5 to each side.
Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5).
7-2
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
Persons
v
7 v
+
+
c
5 c
= 5
= 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
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
7v – 5v + 25 = 31
2v + 25 = 31
2v = 6
v =3
Substitute –v + 5 for c in the second
equation.
Solve for v.
Step 3: Solve for c in either equation.
3+c=5
c=2
Substitute 3 for v in the first equation.
7-2
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 21 + 10, or 31. The answer is correct.
7-2
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
pages 349–352 Exercises
10. ( 3 , 9 3 )
4
8
22. 15 video rentals
1. D
11. (2, 0)
2. C
12. (7 7 , 11 8 )
17
17
23. 80 acres flax,
160 acres sunflowers
3. B
13. (6, –2)
24. 9 yr
4. A
14. (3, –2)
25. estimate: ( 1 , 1);
5–16. Coordinates given in 15. (8, –7)
alphabetical order.
16. (–3, 9.4)
5. (9, 28)
17. 4 cm by 13 cm
6. (– 1 , –4 1 )
18. 4 wk
2
2
7. (6 1 , – 1 )
19. (15, 15)
3
3
8. (2, 4 1 )
20. (9, 126)
2
9. (4, 20)
21. (–4, 4)
7-2
2
; ( 1 , 1)
2
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
28. estimate: (–3.5, –3.5);
26. estimate: (–2, 3);
; (–2, 3)
(– 10 , – 11 )
3
27. estimate: (–1, 1);
29. estimate: (– 3 , 4 3 );
4
4
2 14
)
3 3
(– ,
(–1, 1)
7-2
3
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
30. estimate: (–4, 0);
32. Answers may vary.
Sample: y = x and y = –3x + 2;
(1 ,1)
2
33. a. (x, y) such that y = 0.5x + 4
b.
(– 50 , – 2 )
11
11
31. a.
b.
c.
2
Let n = number of nickels,
let d = number of dimes.
n + d = 28
0.05n + 0.10d = 2.05
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.
(15, 13)
7-2
c. Graphing shows only one line.
Substitution results in a true
equation with no variables.
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
34. a. no solution
b.
38. (2, 1 )
2
1
39. (– , 0)
2
40.
41.
42.
c. Graphing shows 2
parallel lines. Substitution
results in a false equation
with no variables.
36. (– 1 , – 1 )
37. (2, –4)
44.
45.
35. (2, 4)
2
43.
2
48. 4803
49. 520
50. [2] 7(–7) – 4(–2) 29
(4, –2)
–49 + 8 29
inf. many solutions
–41 =/ 29
No, (–2, –7) must
no solution
satisfy both
1 solution
equations to be a
solution of the
a. g = 23 b
22
system.
b. (b, g) = (572, 598)
[1] no explanation
c. 26
given
a. (t, d) = (9, 79.2)
b. yes
46. (r, s, t) = (7, 9, 4)
47. 29.8
7-2
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
51. (12, 10);
54.
52. (2, 1);
55.
57.
58.
56.
53. (4, 2);
59.
7-2
Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solve each system using substitution.
1. 5x + 4y = 5
2. 3x + y = 4
3. 6m – 2n = 7
y = 5x
2x – y = 6
3m + n = 4
(0.2, 1)
(2, 2)
(1.25, 0.25)
7-2
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
y = 2x + 13
2. y + 5x = 4
y = 7x – 20
7-3
3. y = –2x + 2
3x – 17 = 2y
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.
y = 2x + 13
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
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 – 20 + 5x = 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
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 = 2y
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
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
2x + 3(1) = 11
2x + 3 = 11
2x = 8
x=4
Choose the first equation.
Substitute 1 for y.
Solve for x.
7-3
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) 1
Substitute 4 for x and 1 for y into the
second equation.
–8 + 9
1
1 = 1
7-3
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
Write:
a
+ s
= 1139
total amount collected
5 a
Solve by elimination.
7-3
+ s = 3067
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
(continued)
Step 1: Eliminate one variable.
a + s = 1139
5a + s = 3067
–4a + 0 = –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.
482 + 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
482 + 657, or 1139. The money collected was $5(482), or $2410,
plus $1(657), or $657, which is $3067. The solution is correct.
7-3
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)
Step 2: Solve for x.
–12x = 48
x= 4
7-3
Add the equations to
eliminate y.
3x + 6y = –6
–15x – 6y = –42
–12x – 0 = –48
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.
12 + 6y = –6
Solve for y.
6y = –18
y = –3
The solution is (4, –3).
7-3
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:
5 p
p
+ n
= 240
7-3
+ 8 n = 1614
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
Step 2: Solve for n.
–3n = –414
n = 138
7-3
Subtract the equations
to eliminate p.
5p + 5n = 1200
5p + 8n = 1614
0 – 3n = –414
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 + 138 = 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
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)
Step 2: Solve for y.
4y = 20
y = 5
7-3
Subtract the equations
to eliminate x.
15x + 25y = 50
15x + 21y = 30
0 + 4y = 20
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 + 25 = 10
3x = –15
x = –5
The solution is (–5, 5).
7-3
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
pages 356–359 Exercises
11. (–2, – 5 )
2
1. (1, 3)
12. (1, 14)
2. (2, –2)
13. ( 1 , 1)
2
3. (5, –17)
14. (–2, 3)
4. (–3, 4)
15. a. 30w + = 17.65,
20w + 3 = 25.65
b. $.39 for a wallet size,
5. (–9, 1 )
2
1
6. (– , 10)
2
7. a.
b.
8. a.
b.
$5.95 for an 8  10
x + y = 20, x – y = 4
16. a. x = burritos, y = tacos;
12 and 8
3x + 4y = 11.33, 9x + 5y = 23.56
b. $1.79 for a burrito, $1.49 for a taco
a + s = 456, 3.5a + s = 1131
17. (–1, –3)
270 adult, 186 student
9. (–5, 1)
18. (2.5, 1)
10. (11, –3)
19. (2, –2)
7-3
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
20. (10, 8)
21. (–1, 3 )
27. (5, 1); substitution;
one eq. solved for x
22. (1, 5)
28. ( 1 , 2 1 ); substitution;
2
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
3
3
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, 4 )
33. (10, –6)
34. (8, 12)
7-3
3
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
35. (–15, –1)
41. B1 = 3 volts; B2 = 1.5 volts
36. (20, 12)
42. (–3, 2)
37. (–2, 16)
43. ( c , 0) (a =/ 0, b =/ 1)
38. (33, –48)
44. (8, 13, 20)
39. 9
45. CD: $3.40, cassette: $1.80
a
40. Answers may vary. Sample:
46. a. 2.8 g of gold
You solve a system using the
b. about 2.7%
elimination method by adding
47. D
or subtracting the eqs. to
48. H
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
7-3
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
y – x = 13
7y + x = 11
8y = 24
y =3
3 – x = 13
x = –10; (–10, 3)
[1] no work shown
50. [4]
51–53. Coordinates given in
alphabetical order.
51. (6, 26)
49. [2]
52. (–1, 4)
53. (9, –5)
1
9
55. 2
9
56. 1
15
54.
1
h(b1 + b2)
2
= 1 (4)(4 + 2)
2
A =
= 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
7-3
57. 71
58. –18
59. –44
60. 22
61. 83
62. –6
Solving Systems Using Elimination
ALGEBRA 1 LESSON 7-3
Solve using elimination.
1. 3x – 4y = 7
2x + 4y = 8
(3, 0.5)
3. –6x + 5y = 4 (1, 2)
2. 5m + 3n = 22 (2, 4)
5m + 6n = 34
4. 7p + 5q = 2
8p – 9q = 17
3x + 4y = 11
7-3
(1, 1)
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 1 hour
2
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?
7-4
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
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
50% solution.
Let
b
= volume of the
25% solution.
Relate: volume of solution
amount of acid
Write:
0.5 a
a
+ b
= 10
+ 0.25 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
Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
(continued)
Step 2: Find b.
0.5a + 0.25b
= 0.4(10)
0.5(10 – b) + 0.25b = 0.4(10)
Substitute 10 – b for a.
Use parentheses.
5 – 0.5b + 0.25b = 0.4(10)
Use the Distributive Property.
5 – 0.25b
= 4
Simplify.
–0.25b
= –1
Subtract 5 from
each side.
b for
= b4 in either equation.
Divide each side by –0.25.
Step 3: Find a. Substitute 4
a + 4 = 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
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
expenses plus
computer purchase.
Write:
d = 0.5 p
Income is price
times pages typed.
+ 1750
d
7-4
= 5.5 p
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 =
5.5p =
5p =
p=
0.5p + 1750
0.5p + 1750
1750
350
Start with one equation.
Substitute 5.5p for d.
Solve for p.
To break even, you must type 350 pages.
7-4
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.
Relate:
Let W = the wind speed.
with tail wind
with head wind
(rate)(time) = distance
(rate)(time) = distance
(A + W) (time) = distance
Write:
(A + W)5.6 = 2800
(A + W) (time) = distance
(A + W)6.8 = 2800
7-4
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 = 2800
A + W = 500
Divide each side by 5.6.
(A – W)6.8 = 2800
A – W
412
Divide each side by 6.8.
Step 1: Eliminate W.
A + W = 500
A – W = 412
Add the equations to eliminate W.
2A + 0 = 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 = 500
Use the first equation.
456 + W = 500
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
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 + 0.08b = 1.2
c. 18 kg A, 6 kg B
4. a. at 8 wk
b. $160; $62
5. 600 games
7. a.
b.
c.
d.
s + c = 2.75
s – c = 1.5
2.125 mi/h
0.625 mi/h
8. a. (A + W )4.8 = 2100
(A – W )5.6 = 2100
b. 406.25 mi/h
c. 31.25 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.
6. 40 shirts
7-4
Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
18. 19 small mowers, 11 large mowers
19. a. 42 mi/h
b. 12 mi/h
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 = 0 + 2.5m;
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
20. a. g = 19 b
17
b. g + b= 1908
g = 19 b
17
901 boys, 1007 girls
21. a. 16 days
17. Answers may vary. Sample:
b. Answers may vary. Sample:
You have 10 coins, all dimes and
If you plan to ski for many years,
quarters. The value of the coins is
you should buy the equipment,
$1.75. How many dimes do you have?
since you will break even at
How many quarters do you have?
16 days.
q + d = 10
22. x = 2
0.25q + 0.10d = 1.75
y=4
You have 5 dimes and 5 quarters.
7-4
Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
24. a. 2.50s + 4.00 = 10,000
29. (4, 1)
=5s
30. (–6, 7)
2
31. (3, 3 )
800 small, 2000 large
b. 560 h
c. $17.86/h
33. 1
25. C
34. 1
26. H
3
35. –2
27. B
28. [2]
2
32. 3
2
3V + C = 22
2V + 4C = 28
12V + 4C
2V + 4C
10V
V
= 88
= 28
= 60
= 60
6 people/van
[1]correct answer, no work shown
7-4
36. – 4
11
37. undefined
Applications of Linear Systems
ALGEBRA 1 LESSON 7-4
38. 6 < y < 10
39. –8 < n <
–3
40. 1 < k < 6
41. 1 <
–p<
–4
42. –5 < c <
–5
43. 5 > w > 1
7-4
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?
7.5 L of 10% solution; 12.5 L of 18% solution
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?
2500 disks
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.
still water: 3.1 mi/h; current: 0.6 mi/h
7-4
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 > –2
2. 8 <
– 8
3. 4n >
– n
Write each equation is slope-intercept form.
4. 2x – 3y = 9
5. y + 3x = 6
7-5
6. 4y – 3x = 1
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 8 <
– 8 is always true.
3. 4(1) >
– 1 is true and 4(–1) >
– –1 is false, so 4n >
– n is sometimes
true.
4. 2x – 3y = 9
–3y = –2x + 9
5. y + 3x = 6
y = –3x + 6
y = –2x + 9
–3
y = 2x – 3
3
6. 4y – 3x = 1
4y = 3x + 1
y = 3x  1
4
y = 3 x + 1
4
4
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
Graph y > –2x + 1.
First, graph the boundary line y = –2x + 1.
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
Substitute (0, 2) for (x, y).
2 > –2(0) + 1
2 > 1
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
Graph 4x – 3y >
– 6.
Solve 4x – 3y >
– 6 for y.
4x – 3y >
–6
–3y >
– –4x + 6
4
y<
– x – 2
3
Subtract 4x from each side.
Divide each side by –3. Reverse
the inequality symbol.
4
Graph y = 3 x – 2.
The coordinates of the points on the boundary
line make the inequality true. So, use a solid
line.
4
Since y <
– 3 x – 2, shade below the boundary
line.
7-5
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
streamers
plus
cost of
tablecloths
is less than
or equal to
Define: Let
s
= the number of rolls of streamers.
Let
t
= the number of rolls of streamers.
Write:
2 s
+
6
t
7-5
<
–
total budget
24
Linear Inequalities
ALGEBRA 1 LESSON 7-5
(continued)
Graph 2s + 6t <
– 24 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 <
– 24
2(0) + 6(0) <
– 24
0 <
– 24
Substitute (0, 0) for (s, t).
Since the inequality is true, (0, 0) is a solution.
7-5
Linear Inequalities
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
Linear Inequalities
ALGEBRA 1 LESSON 7-5
pages 373–376 Exercises
1. no
14.
17.
15.
18.
11.
2. yes
3. yes
4. yes
12.
5. no
6. yes
7. A
13.
8. B
16.
9. B
10. A
7-5
19. y < 2 x – 7 ;
–3
3
Linear Inequalities
ALGEBRA 1 LESSON 7-5
5 x – 2;
20. y >
–3
23. a. 3x + 5y <
– 48
b.
2
8
21. y <
– 3x– 3;
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.
3 x – 2;
22. y <
–
– 2
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
24. a. 3n + 10c > 250
b.
25.
28.
29.
26.
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.
30.
27.
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
31.
34. y > 2x – 1
35. x <
– –3
1x–2
36. y <
–
3
32.
37. a. 12x + 8y <
– 180
b.
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.
c. Yes; you can buy 8 CDs and 9 tapes.
d. 43
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
38. x > 0;
39. y < 0;
41. x < y;
45. a. 2w + 2 <
– 50;
42. It should be shaded above
the line, and the line should
be dashed.
b. Answers may vary.
43. y < x + 2
Sample: 10 ft by 10 ft,
44. a.
5 ft by 5 ft
40. y >
– 0;
c. No; (12, 15) is not in the
shaded region and is not
a sol. of the inequality.
46. y < x – 3
b.
y>x
c.
5
12
47. y >
– 2x – 2
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
48. a. Answers may vary.
Sample: 2x + y > 3
49. a. yes
b. yes
c. Answers may vary.
Sample: (2, 3)
d.
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
50. A
51. H
52. D
53. F
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
54. [2] First, graph the solid line y = 3x – 4.
Then shade below the line.
59. 11
60. –6
61. 8, 18
62. –6.7, –2.1
63. 12
64. 28
65. 11
[1] correct graph given, no explanation
66. 16 2
55. about 775 games
3
67. – 1
3
56. 0.75 mi/h, 3.25 mi/h
68. 4
57. 5
69. 13
58. 7
70. 6.5
7-5
Linear Inequalities
ALGEBRA 1 LESSON 7-5
1. Determine whether (4, 1) is a solution of 3x + 2y >
– 10.
Graph each inequality.
2. x > –2
3. 5x – 2y > 10
7-5
yes
4. 2x + 6y <
– 0
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 – 6
2. y = – 1 x + 4
y = –x + 2
y = –1x + 3
2
2
3. x + y = 4
2x – y = 8
Graph each inequality.
4. y > 5
2 x – 1
5. y <
– 3
7-6
6. 4x – 8y >
– 4
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solutions
1. y = 3x – 6
2. y = – 1 x + 4
y = –x + 2
y = –1x + 3
The solution is (2, 0)
There is no solution.
2
2
3. x + y = 4
4.
2x – y = 8
Solve equations
for y:
y = –x + 4
y = 2x – 8
The solution is (4, 0).
7-6
y > 5
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solutions (continued)
2
5. y <
– 3x – 1
6. 4x – 8y >
– 4
–8y >
– –4x + 4
4x  4
y <
–
8
y < 1x – 1
– 2
2
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solve by graphing.
y < –x + 3
–2x + 4y >
– 0
Graph y < –x + 3 and –2x + 4y >
– 0.
Check
The point (–1, 1) is in the region graphed by both inequalities.
See if (–1, 1) satisfies both inequalities.
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
(continued)
y < –x + 3
1 < –(–1) + 3
Substitute (–1, 1) for (x, y).
1 < 4
–2x + 4y >
– 0
–2(–1) + 4(1) >
– 0
2 + 4 >
– 0
The coordinates of the points in the [lavender] region where the graphs of
the two inequalities overlap are solutions of the system.
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Write a system of inequalities for each shaded region below.
[red] region
[blue] region
boundary: y = – 1 x + 2
boundary: y = 4
The region lies below the
boundary line, so the
inequality is
The region lies below the
boundary line, so the
inequality is
y < – 1 x + 2.
y < 4.
2
2
System for the [lavender] region:
y < –1 x + 2
2
y < 4
7-6
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
Let
= length of the dog run.
w
Relate: The
length
= width of the dog run.
is no
more than
Write:
Solve by graphing.
<
–
60 ft.
The
perimeter
60
2
<
– 60
2 + 2w <
– 136
7-6
+ 2 w
is no
more than
<
–
136 ft.
136
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
(continued)
2
<
– 60
m = 0:
b = 60
+ 2w <
– 136
Graph the intercepts
(68, 0) and (0, 68).
Shade below
= 60.
Test (0, 0). 2(0) + 2(0) <
– 136
0 <
– 136
So shade below
2 + 2w = 136
The solutions are the coordinates of the points that lie in the region shaded
lavender and on the lines
= 60 and 2 + 2w = 136.
7-6
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
of hours
worked
Write:
b
+
s
is less
20.
than or
equal to
<
–
20
7-6
The amount
earned
5 b
+ 6 s
is at
least
90.
>
–
90
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
(continued)
Solve by graphing.
b + s <
– 20
5b + 6s >
– 90
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
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
pages 380–384 Exercises
9.
6.
1. no
2. yes
3. no
7.
4.
10.
8.
11.
5.
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
12.
16. x > 5 and y >
– –x + 3
13.
5
4
1
2
19. y <
– 5x – 3
– – 3 x – 4 and y >
22. a. 5.99x + 9.99y <
– 50,
x>
1
1
– 0, y >
–1
17. y >
– – 2 x – 2 and y <
– 2x + 2
b.
1 x + 1 and y – 3 x + 3
18. y >
–
>
–
–
20. a. 1.5f + 2.5c < 9.50, f + c > 4
b.
14.
15.
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
21.
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
23. x + y >
– 30, 1.25x + 3y <
– 60
24. a. x + y >
– 12, 6x + 4y <
– 60
25. x <
– 3, x >
– –3, y <
– 3, y >
– –3
26. y >
– 2, x < 5, y <
–x
2 x – 2, y < 2 x + 2
27. y >
–3
3
28. y >
– –x – 3, y <
–x – 3
– –x + 3, y <
– x + 3, y >
29. Answers may vary.
Sample: x >
– –2, x <
– 4, y <
– 1, y >
– –2
30. a. –1
b. 8
31. a. triangle
b. Answers may vary. Sample:
b. (2, 2), (–4, –1), (–4, 2)
(8, 3), (9, 1), (10, 0)
c. 9 units2
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
32. a.
b.
c.
33. a.
b.
c.
34. a.
b.
c.
35. a.
square
(1, –1), (5, –1), (1, 3), (5, 3)
16 units2
trapezoid
(0, –4), (0, 2), (2, –4), (2, 0)
10 units2
triangle
(2, –3), (2, 2), (7, –3)
12.5 units2
x>
– 1, 10.99x + 4.99y <
– 45
36. a.
b. No; they are parallel.
c. no
d. no
37. a.
b. No; they are parallel.
c. It is a strip between the lines.
b. (3, 0), (3, 1), (3, 2), (4, 0)
7-6
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
44.
45. Answers may vary.
y > 1 x, y < 2x
2
46. Answers may vary.
y > x + 1, y > –x + 1
47. a. h >
– 2 + 0.400a
b.
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 + 0.15s < 60
c. Answers may vary.
Sample: 5 hits, 6 at-bats
b. Answers may vary. Sample:
(8, 4.5); 12.5g; gold: $45.00, silver: $1.20
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
48. a. 180x + 240y >
– 2700
x+y<
– 17
x>y
y>
–4
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.
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
[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
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
53.
59. 5
56.
2
60. –3
61. –8
62. – 1
54.
63.
57.
64.
65.
58.
66.
55.
67.
4
1
–
5
– 8
3
10
9
6
13
15
4
68. ƒ(x) = 7x
69. ƒ(x) = x + 6
70. ƒ(x) = x 2
7-6
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solve each system by graphing.
1. x >
– 0
y < 3
2. 2x + 3y > 12
2x + 2y < 12
4. Write a system of inequalities for the following graph.
7-6
2
3. y = x – 3
3
2x – 3y >
– –9
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
Solve each system by graphing.
1. x >
– 0
y < 3
2. 2x + 3y > 12
2x + 2y < 12
7-6
2
3. y >
– 3 x – 3
2x – 3y >
– –9
Systems of Linear Inequalities
ALGEBRA 1 LESSON 7-6
4. Write a system of inequalities for the following graph.
y < x + 3
y > –1 x – 2
2
7-6
Systems of Equations and Inequalities
ALGEBRA 1 CHAPTER 7
1. (2, –1)
9.
(8, 25)
19. q + n = 15,
10. (3, –7)
0.25q + 0.05n = 2.75;
11. (2, –9)
10 quarters, 5 nickels
12. (3.5, 1)
2. (0, 4)
13.
14.
15.
3. one
16.
4.
5.
6.
7.
8.
17.
one
none
one
one
inf. many
18.
20. Answers may vary.
Sample: You solve a
(8, –24)
system of linear equations
2
(11, – )
by finding a single point that
5
satisfies all the equations in
(6, 4)
the system. You solve a
(–5, –2)
system of linear inequalities
by graphing a region that
b + m = 35,
contains points that satisfy
b + 2m = 45; $25; $10
all the inequalities in the
n = 5 , n + p = 24;
system.
p 3
15 novelists, 9 poets 21. C
7-A
Systems of Equations and Inequalities
ALGEBRA 1 CHAPTER 7
22.
28. a. w <
– 30, 2 + 2w <
– 180
25.
b.
23.
26. Answers may vary.
Sample: y = x + 1,
y = 3x – 5; (3, 4)
27. a. 0.10d + 0.25q < 5.00;
24.
b. 49 items
c. 19 items
7-A
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