14. a) Let x represent the number of sets of small (4 cubic
foot) boxes. Let y represent the number of sets of large (6
cubic foot) boxes.
16x + 24y 162
Graph the line 16x + 24y = 162.
x-intercept: x = 10.125
y-intercept: y = 6.75
The solution is all the whole number points to the left of
the line or on the line.
b) Use the larger boxes to minimize the number of boxes.
6 4 6 cubic feet = 144 cubic feet
1 4 4 cubic feet = 16 cubic feet
144 + 16 = 160
2 4 4 cubic feet = 32 cubic feet
144 + 32 = 176
7 4 6 cubic feet = 168 cubic feet
Use 7 sets of the larger boxes. They hold 168 cubic feet,
which is more than the 162 cubic feet Debbie and Gavin
need but a combination of large and small boxes results in
more boxes being used.
Les s on 6. 2: Ex pl ori ng G ra phs of S y st em s
of Li ne ar Equalit ie s, page 30 7
1. a) Part of the plane is shaded so the domain and range
are real numbers (x R, y R).
One boundary line is dashed so it is not included in the
domain and range. The other is solid so it is included.
b) Discrete points are shown in all quadrants so the
domain and range are integers, (x I, y I). The
boundary lines are included.
c) Discrete points are shown in two quadrants so the
domain and range are integers (x I, y I). The
boundary lines are included.
2. a) Graph the boundary lines.
–x + 2y = –4
x-intercept: x = 4
y-intercept: y = –2
y = x passes through (0, 0) and (5, 5).
Test the point (0, 0):
0 > –4 and 0 = 0 so (0, 0) is a solution of both inequalities.
Test the point (–4, 0):
–4 = –4 and –4 < 0 so the point is a solution.
Fundamentals of Mathematics 11 Solutions Manual
Draw solid lines and shade the area above both
lines:
b) Graph the boundary lines.
2x + 3y = 9
x-intercept: x = 4.5
y-intercept: y = 3
y – 6x = 1
x-intercept: x = –0.167
y-intercept: y = 1
Test the point (–4, 0):
–8 < 9 and 24 > 1 so the point is a solution.
Mark the discrete points in the solution region:
Les s on 6. 3: Gra phi ng t o S ol ve
Sy st em s of Li nea r E qualiti e s,
pa ge 31 7
1. a) Graph the solid line x + y = 2 and shade the
region for x + y 2.
Graph the dashed line for x = 4 and shade the
region for x < 4.
The overlapping area is the solution.
Check the point (0, 4) in the solution area.
0 + 4 2 and 0 < 4.
6-5
b) Graph the dashed line x + 2y = 6 and shade the region
for x + 2y < 6.
Graph the dashed line for x = y and shade the region for x
< y.
The overlapping area is the solution.
Check the point (0, 2) in the solution area.
0 + 4 < 6 and 0 < 2.
c) Graph the line 2x – 4 = y and shade the region for 2x –
4 y.
Graph the dashed line for 2y + 3x = 7 and shade the
region for 2y + 3x 7.
Mark the discrete points in the overlapping area and its
boundary lines.
The overlapping area is the solution.
Check the point (0, –6) in the solution area.
0 – 4 –6 and –12 + 0 7.
2. a) The solution set is the overlapping region with a
boundary along the line x = 6 and points along the
boundary are included. The other boundary is 3y – x = 6,
but points along it are not included.
iii) Test point (3, 2).
3(2) – 3 < 6
3<6
So (3, 2) is a solution.
iv) Test point (3, 3).
3(3) – 3 = 6
So (3, 3) is not a solution.
3. a) The boundary line y = –2x is included. The
boundary line –3 = x – y is not included so the
intersection is not included.
b) The boundary points with integer coordinates on
the boundary lines x + y –2 and 2y x are
included. The intersection point does not have
integer coordinates so it is not included.
c) The boundary points with integer coordinates on
the boundary line x + 3y 0 are included. The
boundary line x + y > 2 is not included, so the
intersection is not included.
4. a) Graph the solid line x + y = 3 and shade the
region for x + y 3.
Graph the dashed line for y = 2 and shade the
region for y > 2.
The overlapping area, not including the
intersection point, is the solution.
Check the point (–2, 3) in the solution area.
–2 + 3 3 and 3 > 2.
b) 2x + y > 0 for all whole number points except (0,
0).
Graph the dashed line for y = x and mark all the
discrete whole number points above the line.
Check the point (2, 6) in the solution area.
2(4) + 6 = 14 > 0 and 6 > 2
b) i) Test point (6, 4).
3(4) – 6 = 6
So (6, 4) is not a solution.
ii) Test point (8, 2).
8>6
So (8, 2) is not a solution.
6-6
Chapter 6: Quadratic Functions
c) Graph the solid line 3y – 2x = 6 and shade the region
for 3y – 2x 6.
Graph the solid line for 2y – 3x = 6 and shade the region
for 2y – 3x 6. Mark all the natural number points in the
shaded area and on the lower boundary line.
Check the point (8, 2) in the solution area.
3(2) – 2(8) = –10 6 and 2(2) – 3(8) = –20 6.
d) Graph the solid line y – x = 3 and shade the region for
y – x 3.
Graph the solid line for y + 2 = x and shade the region for
y + 2 x.
The two lines are parallel and the solution regions do not
overlap. So there is no solution.
5. i) a) Graph the dashed line x + y = 5 and shade the
region for x + y > 5.
Graph the solid line for y = 4 and shade the region for
y 4.
ii) a) The inequality x + y 1 is true for all natural
numbers.
Graph the dashed line for 2 = x – 2y and shade the
region for 2 > x – 2y. Mark all the points with
natural number coordinates above the line.
b) The solution is all the points with natural number
coordinates above the line 2 = x – 2y and on and
above the line x + y 1.
c) Check the point (2, 2) in the solution area.
2 + 2 1 and 2 > 2 – 2(2) = –2
iii) a) Graph the dashed line 4x + 11y = 44 and
shade the region for 4x + 11y > 44.
Graph the solid line for 2x – 6 = 3y and shade the
region for 2x – 6 3y. Mark the points with whole
number coordinates in the overlapping region
above both lines and on the boundary line
2x – 6 = 3y.
b) The solution is all the points with whole number
coordinates above the line 4x + 11y > 44, and
above or on the line 2x – 6 = 3y,
c) Check the point (6, 4) in the solution area.
4(6) + 11(4) = 68 > 44
2(6) – 6 = 6 3(4) = 12
6. a) Let x represent the number of egg salad
sandwiches. Let y represent the number of ham
and cheese sandwiches.
{(x, y) | x + y 450, x W, y W}
{(x, y) | y 2x, x W, y W}
b) The variables must be whole numbers (x W,
y W) because the cafeteria would not sell parts
of a sandwich.
b) The overlapping area below the boundary line y = 4
and above the boundary line x + y = 5 is the solution. It
includes the line y = 4 but not the line x + y = 5.
c) Check the point (6, 3) in the solution area.
6 + 3 > 5 and 3 4.
Fundamentals of Mathematics 11 Solutions Manual
6-7
c) Graph the solid line x + y = 450 and shade the region
for x + y 450.
Graph the solid line for y = 2x and shade the region for y 2x. Mark the points with whole number coordinates in the
overlapping region and on the boundaries.
The points are solutions if they lie within the
overlapping area or on the boundary line 3x – 6y = 18.
8. a) Let x represent the number of school friends.
Let y represent the number of rugby friends.
{(x, y) | x + y 500, x W, y W}
{(x, y) | x 3y, x W, y W}
b) The variables must be whole numbers (x W, y
W) because she cannot count parts of people.
c) Graph the line x + y = 500 and shade the region
for x + y 500.
Graph the line for x = 3y and shade the region for x
3y. Mark the points with whole number
coordinates in the overlapping area and its
boundaries.
d) e.g., 50 egg salad and 300 ham and cheese; 100 egg
salad and 200 ham and cheese
7. a) Graph the dashed line 9x + 18y = 18 and shade the
region for 9x + 18y < 18.
Graph the solid line for 3x – 6y = 18 and shade the region
for 3x – 6y 18.
e.g., The points (200, 50) and (350, 100) are in the
solution region so she could have 200 school
friends and 50 rugby friends or 350 school friends
and 100 rugby friends.
9. Graph the line 3x + y = 2 and shade the region
for 3x + y 2.
Graph the line for 2y + 3x = 1 and shade the region
for 2y + 3x > 1. Mark the points with whole number
coordinates in the overlapping area and on the
boundary line 3x + y = 2.
Test the point (0, –3) in the solution area.
9(0) + 18(–3) = –54 < 18
3(0) – 6(–3) = 18 18
b) i) Test point (4, –1).
9(4) + 18(-1) = 18
So (4, –1) is not a solution.
ii) Test point (–2, 2).
9(–2) + 18(2) = 18
So (–2, 2) is not a solution.
iii) Test point (–4, –2).
9(–4) + 18(–2) = –72 < 18
So (–4, –2) is a solution.
iv) Test point (9, 1).
9(9) + 18(1) = 99 > 18
So (9, 1) is not a solution.
v) Test point (–2.5, –1.5).
9(–2.5) + 18(–1.5) = –49.5 < 18
So (–2.5, –1.5) is a solution.
vi) Test point (2, –2).
9(2) + 18(–2) = –18 < 18
So (2, –2) is a solution.
6-8
e.g., The points (–3, 8) and (–4, 10) are in the
solution area.
Chapter 6: Quadratic Functions
10. Let x represent the number of songs for young
listeners. Let y represent the number of songs for older
listeners.
{(x, y) | x + y 36, x W, y W}
{(x, y) | x 2y, x W, y W}
e.g., 28 songs for young listeners and 4 songs for older
listeners; 18 songs for young listeners and 8 songs for
older listeners; 24 songs for young listeners and 6 songs
for older listeners
11. e.g.,
a) For the boundaries to be included in the solution, I
need to use inequalities with or . For all points in the
solution region to be valid solutions, they must be real
numbers.
{(x, y) | x + y 3, x R, y R}
{(x, y) | 2x – y 4, x R, y R}
b) For the boundaries to be excluded from the solution, I
need to use inequalities with > or <. For only whole
number points in the solution region to be valid solutions,
they must be whole numbers.
{(x, y) | x + y > 7, x W, y W}
{(x, y) | 2x – y < 4, x W, y W}
12. Graph the line 2y + 8 = x and shade the region for
2y + 8 x.
Graph the line –4 + y + x = 0 and shade the region for
–4 + y + x < 0.
Simplify 3x + y + 3 = 3(1 + x) to y = 0. Graph the line y = 0
and shade the region for y > 0.
There is no overlapping area for all three regions so there
is no solution.
Appl yi ng P robl em- S olvi ng
Strat e gi es, pa ge 3 20
A. The black digits were given. The grey digits are
the unknowns.
B. I could see four 5s, so I knew only two more 5s
were possible. I saw that one 5 was needed in the
second column and another 5 was needed in the
fifth column. When I looked at the inequality signs,
I knew that the 5 in the second column was
needed in the fourth row and the 5 in the fifth
column was needed in the third row.
I also saw that four more 6s were needed. The
cells with greater-than signs were logical places for
the 6s.
C. I used deductive reasoning through a process of
elimination. For example, when I decided where
the missing 5s should go, I used what I knew about
rows and columns, along with what I knew about
inequality signs, to identify the only places where
the remaining 5s could go.
D.–F. Answers will vary, e.g.,
Cha pt e r 6 : Mi d- C hapte r Re vi e w,
pa ge 32 3
1. a) Let x represent the number of kilograms of
white chocolates. Let y represent the number of
kilograms of dark chocolates.
{(x, y) | x + y 70, x W, y W}
b) x-intercept: x = 70
y-intercept: y = 70
Fundamentals of Mathematics 11 Solutions Manual
6-9