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Exercises for the lesson “Solve Linear Systems by Graphing”
Skill Practice
1. A solution of a system of linear equations in two variables is an ordered pair that satisfies each equation in the system.
2. Graph both equations on the same coordinate plane. Find the point where the lines appear to intersect, substitute the x and y values from this point into each equation. If the ordered pair is a solution to each equation, it is a solution of the system.
3. x 1 y 5 2 2 x 1 5y 5 2
23 1 1 0 2 2 23 1 5(1) 0 2
22 5 22 ✓ 2 5 2 ✓
(23, 1) is a solution.
4. 2x 2 3y 5 4 2x 1 8y 5 11
2(5) 2 3(2) 0 4 2(5) 1 8(2) 0 11
4 5 4 ✓ 26 Þ 11
(5, 2) is not a solution.
5. 6x 1 5y 5 27 x 2 2y 5 0
6(22) 1 5(1) 0 27 22 2 2(1) 0 0
27 Þ 27 ✓ 24 Þ 0
(22, 1) is not a solution.
6. B; (0, 22)
x 1 y 5 22 7x 2 4y 5 8
22 1 0 0 22 7(22) 2 4(0) 0 8
22 5 22 ✓ 214 Þ 8
0 1 (22) 0 22 7(0) 2 4(22) 0 8
22 5 22 ✓ 8 5 8 ✓
7. B; (23, 6)
2x 1 3y 5 12 10x 1 3y 5 212
2(23) 1 3(3) 0 12 10(23) 1 3(3) 0 212
3 Þ 12 221 Þ 212
2(23) 1 3(6) 0 212 10(23) 1 3(6) 0 212
12 5 12 ✓ 212 5 212 ✓
8. (1, 23)
x 2 y 5 4 4x 1 y 5 1
1 2 (23) 0 4 4(1) 1 (23) 0 21
4 5 4 ✓ 1 5 1 ✓
(1, 23) is a solution.
9. (4, 2)
2x 1 y 5 22 2x 2 y 5 6
24 1 2 0 22 2(4) 2 2 0 6
22 5 22 ✓ 6 5 6 ✓
(4, 2) is a solution.
10. (3, 2)
x 1 y 5 5 22x 1 y 5 24
3 1 2 0 5 22(3) 1 2 0 24
5 5 5 ✓ 24 5 24 ✓
(3, 2) is a solution.
11. The error is that the y-intercept of equation 2 should be 21, not 23. When graphed correctly, the lines intersect at (23, 23).
x 2 3y 5 6 2x 2 3y 5 3
23 2 3(23) 0 6 2(23) 2 3(23) 0 3
6 5 6 ✓ 3 5 3 ✓
(23, 23) is a solution to the linear system.
12. y 5 2x 1 3 y 5 x 1 1
x
y
2
1
y 5 2x 1 3
(1, 2)
y 5 x 1 1
Test (1, 2).
2 0 21 1 3 2 0 1 1 1
2 5 2 ✓ 2 5 2 ✓
(1, 2) is a solution.
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It is important to check your solution because the coordinates of the solution may not be obvious by looking at the graph.
29. a. 2 1 } 4 x 1 6 5
1 } 2 x 1 3
6 5 3 } 4 x 1 3
3 5 3 } 4 x
4 5 x
b. y 5 2 1 } 4 x 1 6 y 5
1 } 2 x 1 3
x
y
1
1
(4, 5)
14y 5 2 x 1 6
12y 5 x 1 3
Test (4, 5).
5 0 2 1 } 4 (4) 1 6 5 0
1 } 2 (4) 1 3
5 5 5 ✓ 5 5 5 ✓
The solution is (4, 5).
c. The two equations from the system in part (b) were set equal to each other for the equation in part (a).
d. Set each side of the equation equal to y.
y 5 2 2 } 5 x 1 5 y 5
1 } 5 x 1 2
Graph both equations on the same coordinate plane. The point where the graphs intersect is the solution. The x-coordinate of the intersection point is the value of x in the given equation.
30. 23x 1 2y 5 1
2y 5 3x 1 1
y 5 3 } 2 x 1
1 } 2
2x 1 y 5 11
x
y
1
1
(5, 1)(1, 2)
(3, 5) y 5 22x 1 11
x 1 4y 5 9
4y 5 2x 1 9
y 5 2 1 } 4 x 1
9 } 4
Line 1 & 2: Test (3, 5).
23(3) 1 2(5) 0 1 2(3) 1 5 0 11
1 5 1 ✓ 11 5 11 ✓
Line 2 & 3: Test (5, 1).
2(5) 1 1 0 11 5 1 4(1) 0 9
11 5 11 ✓ 9 5 9 ✓
Line 3 & 1: Test (1, 2).
1 1 2(4) 0 9 23(1) 1 2(2) 0 1
9 5 9 ✓ 1 5 1 ✓
The vertices of the triangles are (3, 5), (5, 1), (1, 2).
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31. The percent who watch 1 hour or less will equal the percent who watch more than 1 hour 50 years after 1990, or in 2040.
32. B; y 5 2484x 1 17,424
y 5 2330x 1 15,840
33. Let x 5 number of small cards.
Let y 5 number of large cards.
x 1 y 5 25
3x 1 5y 5 95 x 1 y 5 25
3x 1 5y 5 95 y 5 2x 1 25
5y 5 23x 1 95
y 5 2 3 } 5 x 1 19
x
y
3
3
(15, 10)
x 1 y 5 25
3x 1 5y 5 95
Test (15, 10)
15 1 10 0 25 3(15) 1 5(10) 0 95
25 5 25 ✓ 95 5 95 ✓
She sold 15 small cards and 10 large cards.
34. a. Let x 5 minutes on stair machine.
Let y 5 minutes on stationary bike.
x 1 y 5 40 5x 1 6y 5 225
y 5 2x 1 40 6y 5 25x 1 225
y 5 2 5 } 6 x 1 37.5
y
5
5
(15, 25)
x 1 y 5 40
5x 1 6y 5 225
You should spend 15 minutes on the stair machine and 25 minutes on the stationary bike.
b. Let x 5 minutes on elliptical trainer.
Let y 5 minutes on stair machine.
x 1 y 5 30 8x 1 5y 5 225
y 5 2x 1 30 5y 5 28x 1 225
y 5 2 8 } 5 x 1 45
y
5
5
(25, 5)x 1 y 5 30
8x 1 5y 5 225
You should spend 25 minutes on the elliptical trainer and 5 minutes on the stair machine.
35. a. y 5 5x 1 15
y 5 8x
b. x y 5 5x 1 15 y 5 8x
1 20 8
2 25 16
3 30 24
4 35 32
5 40 40
c.
22
8
x
y
y 5 8x
y 5 5x 1 15
It makes sense to become a club member if you attend more than 5 movies a year. The graph shows that the y-values (total cost) are higher on the line representing a non-member when x (number of movies viewed) is greater than 5.
36. Let x 5 purchase price. y
x
25
50
(125, 100)y 5 0.8x
y 5 x 2 25 Let y 5 amount paid.
y 5 x 2 25
y 5 0.8x
You should choose $25 off if your purchase is less than $125, and you should choose 20% off if your purchase is more than $125, because 20% of amounts greater than $125 is more than $25.
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29. Once you have obtained a solution using substitution, graph both equations on the same coordinate plane. The point where the lines intersect should be the same as the solution you got using substitution.
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They rented 22 tubes for people and 4 “cooler” tubes.
33. x 1 y 5 9 → x 5 9 2 y
1.5x 5 1.2y
1.5(9 2 y) 5 1.2y
13.5 2 1.5y 5 1.2y
13.5 5 2.7y
5 5 y
x 5 9 2 5 5 4
The length from A to the string is represented by x, which is 4 inches. The length from the string to B is represented by y, which is 5 inches. The string should be placed 4 inches from point A.
34. a. d 5 rt
d 5 1.9t ← Lane 2 swimmer
d 5 1.8(t 1 1.2)
1.9t 5 1.8(t 1 1.2)
1.9t 5 1.8t 1 2.16
0.1t 5 2.16
t 5 21.6
The swimmer in lane 2 will catch up after 21.6 seconds.
b. The race will end when the swimmers have gone 400 meters. If d is less than 400 when t 5 21.6 seconds, then the swimmer in lane 2 will catch up before the race ends.
d 5 1.9t
d 5 1.9(21.6)
d 5 41.04 meters
The swimmer in lane 2 will catch up to the swimmer in lane 1 before the race ends.
35. Let x 5 mL of 1% hydrochloric acid solution.
Let y 5 mL of 5% hydrochloric acid solution.
x 1 y 5 100 → x 5 100 2 y
0.01x 1 0.05y 5 0.03(100)
0.01(100 2 y) 1 0.05y 5 3
1 2 0.01y 1 0.05y 5 3
0.04y 5 2
y 5 50
x 5 100 2 50 5 50
You need to mix 50 mL of the 1% solution and 50 mL of the 5% solution.
36. Let x 5 number of dimes.
Let y 5 number of quarters.
x 5 y 1 3
0.1x 1 0.25y 5 4.50
0.1(y 1 3) 1 0.25y 5 4.50
0.1y 1 0.3 1 0.25y 5 450
0.35y 5 4.2
y 5 12
x 5 12 1 3 5 15
She has 12 quarters.
37. Let x 5 time cheetah runs.
Let y 5 time gazelle runs.
x 5 y
88x 2 73y 5 350
88x 2 73x 5 350
15x 5 350
x 5 23.3
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. The cheetah would catch up to the gazelle in 23.3
seconds, but since the cheetah can only sustain this speed for 20 seconds, the gazelle can stay ahead of the cheetah.
38. Let x 5 bushels of 100% vermiculite.
Let y 5 bushels of 60% vermiculite.
x 1 5 5 y
1x 1 0.5(5) 5 0.6y
x 1 2.5 5 0.6(x 1 5)
x 1 2.5 5 0.6x 1 3
0.4x 5 0.5
x 5 1.25
In order to make a mix that is 60% vermiculite and 40% peat moss, he would have to add 1.25 bushels of the 100% vermiculite which would give him 6.25 bushels total of the mixture. Since he only needs 6 bushels, he does have enough of the 50%/50% mix.
Quiz for the lessons “Solve Linear Systems by Graphing” and “Solve Linear Systems by Substitution”
1. x 1 y 5 22 2x 1 y 5 6
y 5 2x 2 2 y 5 x 1 6
x
y
1
1
(24, 2)
2x 1 y 5 6
x 1 y 5 22
Test (24, 2).
24 1 2 0 22 2(24) 1 2 0 6
22 5 22 ✓ 6 5 6 ✓
(24, 2) is a solution.
2. x 2 y 5 0 5x 1 2y 5 27
y 5 x 2y 5 25x 2 7
y 5 2 5 } 2 x 2
7 } 2
x
y
1
1(21, 21)
x 2 y 5 0
5x 1 2y 5 27
Test (21, 21). 5(21) 1 2(21) 0 27
21 2 (21) 0 0 27 5 27 ✓
0 5 0 ✓
(21, 21) is a solution.
3. x 2 2y 5 12 23x 1 y 5 21
22y 5 2x 1 12 y 5 3x 2 1
y 5 1 } 2 x 2 6
x
y
211
(22, 27)
23x 1 y 5 21
x 2 2y 5 12
Test (22, 27).
22 2 2(27) 0 12 23(22) 1 (27) 0 21
12 5 12 ✓ 21 5 21 ✓
(22, 27) is a solution.
4. y 5 x 2 4
22x 1 y 5 18
22x 1 x 2 4 5 18
2x 5 22
x 5 222
y 5 222 2 4 5 226
The solution is (222, 226).
5. y 5 4 2 3x
5x 2 y 5 22
5x 2 (4 2 3x) 5 22
5x 2 4 1 3x 5 22
8x 5 26
x 5 3.25
y 5 4 2 3(3.25)
y 5 25.75
The solution is (3.25, 25.75).
6. x 5 y 1 9
5x 2 3y 5 7
5(y 1 9) 2 3y 5 7
5y 1 45 2 3y 5 7
2y 5 238
y 5 219
x 5 219 1 9 5 210
The solution is (210, 219).
7. 2y 1 x 5 24
x 5 22y 2 4
y 2 x 5 25
y 2 (22y 2 4) 5 25
y 1 2y 1 4 5 25
3y 5 29
y 5 23
x 5 22(23) 2 4 5 2
The solution is (2, 23).
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5. You cannot solve this problem using only the steps above because neither of the variables will cancel out after you add the equations. So you would still have 2 variables and be unable to solve for one of them in step 3.
Guided Practice for the lesson “Solve Linear Systems by Adding or Subtracting”
1. 4x 2 3y 5 5
22x 1 3y 5 27
2x 5 22
x 5 21
4(21) 2 3y 5 5
23y 5 9
y 5 23
The solution is (21, 23).
Check:
4(21) 2 3(23) 0 5 22(21) 1 3(23) 0 27
5 5 5 ✓ 27 5 27 ✓
2. 25x 2 6y 5 8
5x 1 2y 5 4
24y 5 12
y 5 23
5x 1 2(23) 5 4
5x 5 10
x 5 2
The solution is (2, 23).
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The original length is 5 inches, and the original width is 4 inches.
b. 3(5) 5 15
2(4) 5 8
The new length is 15 inches, and the new width is 8 inches.
34. ax 1 3y 5 2
4x 1 5y 5 6
If a 5 1 you can solve the system by substitution. You can solve the first equation for x and substitute 2 2 3y for x in the second equation. If a 5 0 you can solve the system by solving the first equation for y and substitute that value into the second equation. If a 5 4 or a 5 24, you can solve the system by addition or subtraction since the second equation contains the term 4x. That term could be eliminated without use of multiplication.
35. ax 2 by 5 4 x 5 4, y 5 2
bx 2 ay 5 10
4a 2 2b 5 4
4b 2 2a 5 10
4a 2 2b 5 4 3 2 → 8a 2 4b 5 8
22a 1 4b 5 10 22a 1 4b 5 10
6a 5 18
a 5 3
4(3) 2 2b 5 4
22b 5 28
b 5 4
So, a 5 3, b 5 4.
36. ax 2 by 5 4 x 5 2
bx 2 ay 5 10 y 5 1
2a 2 b 5 4
2b 2 a 5 10
2a 2 b 5 4 3 2 → 4a 2 2b 5 8
2a 1 2b 5 10 2a 1 2b 5 10
3a 5 18
a 5 6
2(6) 2 b 5 4
2b 5 28
b 5 8
So, a 5 6, b 5 8.
Problem Solving
37. Let x 5 number of hardcovers.
Let y 5 number of paperbacks.
x 1 y 5 8 3 2 → 2x 1 2y 5 16
4x 1 2y 5 26 4x 1 2y 5 26
22x 5 210
x 5 5
5 1 y 5 8
y 5 3
She purchased 5 hardcover books.
38. Let x 5 cost for 1 song.
Let y 5 cost for 1 album.
5x 1 y 5 14.94 3 2 → 10x 1 2y 5 29.88
3x 1 2y 5 22.95 3x 1 2y 5 22.95
7x 5 6.93
x 5 0.99
5(0.99) 1 y 5 14.94
y 5 9.99
The website charges $.99 to download a song and $9.99 to download an album.
39. Let x 5 number of pies.
Let y 5 batches of applesauce.
5x 1 4y 5 169 5x 1 4y 5 169
3x 1 2y 5 95 3 2 → 6x 1 4y 5 190
2x 5 221
x 5 21
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. Graph the two equations. The point of intersection of the
graphs is the solution of the system.
The graphs intersect at (6, 23).
Step 2
Multiply Equation 2 by 23 so that you can eliminate the variable x by adding.
3x 1 5y 5 3 → 3x 1 5y 5 3
x 2 y 5 9 323 23x 1 3y 5 227
8y 5 224
y 5 23
Step 3
Graph the equations 3x 1 5y 5 3, 23x 1 3y 5 227, and y 5 23 on the same graphing calculator screen with the original equations.
Step 4
All of the equations intersect at (6, 23).
So, (6, 23) is the solution of the system.
7. x 2 2y 5 26 Equation 1
2x 1 y 5 8 Equation 2
Solve each equation for y.
x 2 2y 5 26 2x 1 y 5 8
22y 5 2x 2 6 y 5 22x + 8
y 5 1 }
2 x 1 3
Graph the two equations. The point of intersection of the graphs is the solution of the system.
The solution using the graphing calculator is (2, 4). Multiply Equation 1 by 22 so that you can eliminate the
variable x by adding.
x 2 2y 5 26 322 22x 1 4y 5 12
2x 1 y 5 8 → 2x 1 y 5 8
5y 5 20
y 5 4
Multiply Equation 2 by 2 so that you can eliminate the variable y by adding.
x 2 2y 5 26 → x 2 2y 5 26
2x 1 y 5 8 ×2 4x 1 2y 5 16
5x 5 10
x 5 2
The solution (2, 4) found using the graphing calculator has coordinates that are given by the equations resulting from using linear combinations to eliminate one of the variables from the original system.
Draw Conclusions
8. To check the solution of the system, you can add the equations and graph the sum. If the three lines intersect in one point, the solution is correct.
9. a. You get a false equation such as 5 5 3.
b. The lines are parallel.
c. When you add the equations, you don’t get an equation that can be graphed, so the method does not work.
Mixed Review of Problem Solving for the lessons ‘’Solve Linear Systems by Graphing”, “Solve Linear Systems by Substitution”, “Solve Linear Systems by Adding or Subtracting”, and “Solve Linear Systems by Multiplying First”
1. Let x 5 speed in still air.
Let y 5 speed in wind.
a. 15
} 0.25
5 60, 15
} 0.2
5 75
The average speed for the first flight is 60 km/h, and 75 km/h for the return.
b. x 2 y 5 60
x 1 y 5 75
c. 2x 5 135
x 5 67.5
67.5 1 y 5 75
y 5 7.5
The helicopter’s average speed in still air is 67.5 kilometers per hour. The speed of the wind is 7.5 kilometers per hour.
2. Let x 5 cost of a pound of potato salad.
Let y 5 cost of a pound of coleslaw.
1.8x 1 1.4y 5 9.70
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Two pounds of potato salad cost $6.50, and two pounds of coleslaw cost $5.50, a total of $12.
3. The point represented by the intersection of the graphs shows that the customers will have paid the same amount after 4 months.
4. Answers will vary.
5. d 5 rt 1 h
d 1 5 7200(0.5) 1 1705 5 5305
d 2 5 4000(0.5) 1 3940 5 5940
The balloon at Kirby Park will ascend to a height of 5305 feet. The balloon at Newman Park will ascend to a height of 5940 feet with regards to sea level. After about 42 minutes, the two balloons will be at the same distance from sea level.
6. a. x 1 y 5 500
0.1x 1 0.3y 5 0.2(500)
b. x 5 500 2 y
0.1(500 2 y) 1 0.3y 5 100
50 2 0.1y 1 0.3y 5 100
0.2y 5 50
y 5 250
x 1 250 5 500
x 5 250
To make the 20% acid and 80% water mix, 250 milliliters of the 10% acid and 90% water mix is combined with 250 milliliters of the 30% acid and 70% water mix.
c. x 1 y 5 500
0.1x 1 0.3y 5 0.15(500)
0.1(500 2 y) 1 0.3y 5 75
50 2 0.1y 1 0.3y 5 75
0.2y 5 25
y 5 125
x 1 125 5 500
x 5 375
The chemist does need more of the 10% acid and 90% water mix because this new mix has less acid than the first mix created, so more of the less acidic solution must be added.
Lesson Solve Special Types of Linear Systems
Guided Practice for the lesson “Solve Special Types of Linear Systems”
1. 5x 1 3y 5 6
25x 2 3y 5 3
0 5 9
This is a false statement. Since the variables are eliminated and you are left with a false statement, regardless of the values of x and y. This tells you that the system has no solution.
2. y 5 2x 2 4
26x 1 3y 5 212
26x 1 3(2x 2 4) 5 212
26x 1 6x 2 12 5 212
212 5 212
The variables are eliminated and you are left with a statement that is true regardless of the values of x and y. This tells you that the system has infinitely many solutions.
3. x 2 3y 5 215 2x 2 3y 5 218
23y 5 2x 2 15 23y 5 22x 2 18
y 5 1 } 3 x 1 5 y 5
2 } 3 x 1 6
The system has one solution.
4. Let x 5 cost of regular print.
45x 1 30(x 1 3) 5 465
45x 1 30x 1 90 5 465
75x 5 375
x 5 5
5 1 3 5 8
A glossy print costs $8.00.
Exercises for the lesson “Solve Special Types of Linear Systems”
Skill Practice
1. A linear system with no solution is called an inconsistent system.
2. A linear system with infinitely many solutions is called a consistent dependent system.
3. The graph of a linear system with no solution is two parallel lines.
4. The graph of a linear system with infinitely many solutions is two lines that coincide producing a graph that appears to be a single line.
5. x 2 3y 5 29 x 2 y 5 21
23y 5 2x 2 9 2y 5 2x 2 1
y 5 1 } 3 x 1 3 y 5 x 1 1
Matches graph B. System has one solution.
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Matches graph A. System has infinitely many solutions.
8. x 1 y 5 22 y 5 2x 1 5
y 5 2x 2 2
1
x
y
y 5 2x 1 5
x 1 y 5 22
1
The system has no solution.
9. 3x 2 4y 5 12 y 5 3 } 4 x 2 3
24y 5 23x 1 12
y 5 3 } 4 x 2 3
21
1
x
y
3x 2 4y 5 12
y 5 x 2 334
The system has infinitely many solutions.
10. 3x 2 y 5 29 3x 1 5y 5 215
2y 5 23x 2 9 5y 5 23x 2 15
y 5 3x 1 9 y 5 2 3 } 5 x 2 3
3
x
y
3x 1 5y 5 215
3x 2 y 5 29
3
The system has one solution.
11. 22x 1 2 y 5 216 3x 2 6 y 5 30
2 y 5 2x 2 16 26 y 5 23x 1 30
y 5 x 2 8 y 5 1 } 2 x 2 5
22
2
x
y
3x 2 6y 5 30
22x 1 2y 5 216
The system has one solution.
12. 29x 1 6 y 5 18 6x 2 4y 5 212
6 y 5 9x 1 18 24y 5 26x 2 12
y 5 3 } 2 x 1 3 y 5
3 } 2 x 1 3
1
x
y
21
29x 1 6y 5 18
6x 2 4y 5 212
The system has infinitely many solutions.
13. 23x 1 4 y 5 12 23x 1 4y 5 24
4 y 5 3x 1 12 4y 5 3x 1 24
y 5 3 } 4 x 1 3 y 5
3 } 4 x 1 6
21
1
x
y
23x 1 4y 5 12
23x 1 4y 5 24
The system has no solution
14. The error is that, though the lines do not intersect in the graph shown, they are not parallel, so they will intersect at some point. Therefore the system has one solution.
15. 2x 1 5y 5 14 3 3 → 6x 1 15y 5 42
6x 1 7y 5 10 6x 1 7y 5 10
8y 5 32
y 5 4
2x 1 5(4) 5 14
2x 5 26
x 5 23
The solution is (23, 4).
16. 216x 1 2y 5 22
y 5 8x 2 1
216x 1 2(8x 2 1) 5 22
216x 1 16x 2 2 5 22
22 5 22
The system has infinitely many solutions.
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8. The parent absolute value funtion is defined by two
equations. For negative values of x, |x| 5 2x and
for nonnegative values of x, |x| 5 x; y 5 { 2x, if x < 0 x, if x ≥ 0 .
9.
laa111te-bkm-alA-a08
x
y
1
1
Yes; yes; the function is a piecewise function because, for every integer value of n, there is a unique equation that applies to the part of the domain defined by n ≤ x < n 1 1. The function is a step function because it is defined by a constant value over each part of its domain.
Lesson Solve Systems of Linear Inequalities
Guided Practice for the lesson “Solve Systems of Linear Inequalities”
1. y < x 2 4
1
x
y
21
y ≥ x 1 3
2. y ≥ 2x 1 2
1
x
y
21
y < 4
x < 3
3. y ≥ x 2 4
2
x
y
22
y < 5
y > 2x
4. x ≤ 3 5. x < 2
y > 2 }
3 x 2 1 y ≤ 4
6. x 2 y ≤ 8 → y ≥ x 2 8
10
x
y
210
x ≥ 26
x ≤ 34
y ≥ 0 Since the point does fall in the
shaded area, this bat can be used by a senior league player.
Exercises for the lesson “Solve Systems of Linear Inequalities”
Skill Practice
1. A solution of a system of linear inequalities is an ordered pair that is a solution of each inequality in the system.
2. Graph both inequalities in the same coordinate plane. Use a dashed line for Inequality 1 and a solid line for Inequality 2. Find the intersection of the half-planes and shade that area.
3. (1, 1) is not a solution. 4. (0, 6) is a solution.
33. No, the system has no solutions because no point makes both inequalities true.
34. x ≥ 2 y ≥ 1 x ≤ 6 y ≤ 4
35.
x
y
2
2
y ≥ 2 2 } 3 x 2 2
y ≥ 4 } 3 x 2 2
y ≤ 1 } 3 x 1 1
Problem Solving
36. x ≥ 0 music score
x ≤ 60 music score
y ≥ 0 visual score
y ≤ 40 visual score
10
x
y
210
37. x ≥ 0, y ≥ 0, x 1 y < 8,
1
1x
y
14x 1 7y < 70
7y < 214x 1 70
y < 22x 1 10
38. a. Let x 5 surfperch.
25
5
s
r
Let y 5 rockfish.
x ≥ 0, y ≥ 0
x ≤ 15, y ≤ 10
x 1 y ≤ 15
b. (11, 9)
You cannot catch 11 surfperch and 9 rockfish in
one day.
39. a. x ≥ 20
x ≤ 65
y ≥ 0.7(220 2 x) y ≤ 0.85(220 2 x)
y ≥ 154 2 0.7x y ≤ 187 2 0.85x
10 50 100
10
50
100
150
200
x
y
b. By finding the points on the graph, you can see that his heart rate does not stay in the suggested range. It falls below it.
40. Let x 5 3 in. by 5 in. pictures.
Let y 5 4 in. by 6 in. pictures.
a. x 1 y ≥ 16 → y ≥ 2x 1 16
8 }
4 x 1
8 } 2 y ≤ 48
x
y
4
4
x 1 y 5 16
2x 1 4y 5 48(12, 6)
4y ≤ 22x 1 48
y ≤ 2 1 } 2 x 1 12
b. Because the point falls within the solution region, you are able to buy 12 pictures that are 3 inches by 5 inches and 6 pictures that are 4 inches by 6 inches.
220Algebra 1Worked-Out Solution Key
CS10_CC_A1_MEWO710631_C6.indd 220 5/14/11 11:53:01 AM
.Chapter Review for the chapter "Systems of Equations and Inequalities"
1. A system of linear inequalities consists of two or more linear inequalities in the same variables.
2. A system of linear equations consists of two or more linear equations in the same variables.
3. Graph each of the inequalities on the same coordinate plane. Shade the intersection of the two half-planes. To check your solution, choose a point in the shaded region and substitute its x and y-values into each inequality.
4. Answers will vary.
5. y 5 23x 1 1
y 5 x 2 7
x
y
211
y 5 23x 1 1
y 5 x 2 7
(2, 25)
Check (2, 25).
25 0 23(2) 1 1 25 0 2 2 7
25 5 25 ✓ 25 5 25 ✓
The solution is (2, 25).
6. y 5 3x 1 4
y 5 22x 2 1
x
y
1
1
(21, 1)
y 5 22x 2 1
y 5 3x 1 4
Check (21, 1).
1 0 3(21) 1 4 1 0 22(21) 2 1
1 5 1 ✓ 1 5 1 ✓
The solution is (21, 1).
7. x 1 y 5 3 x 2 y 5 5
y 5 2x 1 3 y 5 x 2 5
x
y
1
1(4, 21)
y 5 x 2 5
y 5 2x 1 3
Check (4, 21).
4 1 (21) 0 3 4 2 (21) 0 5
3 5 3 ✓ 5 5 5 ✓
The solution is (4, 21).
8. y 5 2x 2 7 9. x 1 4y 5 9
x 1 2y 5 1 x 2 y 5 4
x 1 2(2x 2 7) 5 1 x 5 y 1 4
x 1 4x 2 14 5 1 y 1 4 1 4y 5 9
5x 5 15 5y 5 5
x 5 3 y 5 1
y 5 2(3) 2 7 5 21 x 5 1 1 4 5 5
The solution is (3, 21). The solution is (5, 1).
10. 2x 1 y 5 215
y 2 5x 5 6
y 5 5x 1 6
2x 1 5x 1 6 5 215
7x 5 221
x 5 23
y 5 5(23) 1 6 5 29
The solution is (23, 29).
11. Let x 5 tubes of paint.
Let y 5 brushes.
3x 1 0.5y 5 16
y 5 2x
3x 1 0.5(2x) 5 16
3x 1 x 5 16
4x 5 16
x 5 4
y 5 2(4) 5 8
She purchases 4 tubes of paint and 8 brushes.
12. x 1 2y 5 13 13. 4x 2 5y 5 14
x 2 2y 5 27 24x 1 y 5 26
2x 5 6 24y 5 8
x 5 3 y 5 22
3 1 2y 5 13 4x 2 5(22) 5 14
2y 5 10 4x 5 4
y 5 5 x 5 1
The solution is (3, 5). The solution is (1, 22).
14. x 1 7y 5 12 15. 9x 2 2y 5 34
22x 1 7y 5 18 5x 2 2y 5 10
3x 5 26 4x 5 24
x 5 22 x 5 6
22 1 7y 5 12 9(6) 2 2y 5 34
7y 5 14 22y 5 220
y 5 2 y 5 10
The solution is (22, 2). The solution is (6, 10).
223Algebra 1
Worked-Out Solution Key
CS10_CC_A1_MEWO710631_C6.indd 223 5/14/11 11:53:04 AM