Chapter 6
Exercises page 74
1
-x + 3y = -3
3y = -3 + x
y = (-3 + x)/3 = -1 + x/3
→ part (b) is the correct answer
2a- 4x + 5y2 = 2
5y2 = -4x + 2 → not linear
b- -3x + 2y = 7
2y = 3x + 7
y = 3/2x + 7/2 → linear
c- 2x2 + x – 3y2= x(2x – 2) + y(2 – 3y)
2x2 + x – 3y2 = 2x2 – 2x + 2y – 3y2
2x2 – 3y2 – 2x2 - 2y – 3y2 = -2x – x
-2y = -3x
y = 3/2x → linear
d- y = -6 → linear
3a- 3x + 2y = 10
 (1, 3) → 3(1) + 2(3) = 3 + 6 = 9 ≠ 10
→ doesn’t verify



(2, 2) → 3(2) + 2(2) = 6 + 4 = 10 true
→ verifies
(2, 3) → 3(2) + 2(3) = 6 + 6 = 12 ≠ 10
→ doesn’t verify
(10, -10) → 3(10) + 2(-10) = 30 - 20 = 10 true
→ verifies
b- 3x – y + 2 = 0
 (1, 5) → 3(1) – 5 + 2 = 3 – 5 + 2 = 0 true
→ verifies
 (5, 1) → 3(5) – 1 + 2 = 15 – 1 + 2 = 16 ≠ 0
→ doesn’t verify
 (2, 8) → 3(2) – 8 + 2 = 6 – 8 + 2 = 0 true
→ verifies
 (0, 2) → 3(0) – 2 + 2 = 0 – 2 + 2 = 0 true
→ verifies
c- x – 3y + 4 = 0
 (2, 2) → 2 – 3(2) + 4 = 2 – 6 + 4 = 0 true
→ verifies
 (-2, -2) → -2 – 3(-2) + 4 = -2 + 6 + 4 = 8 ≠ 0
→ doesn’t verify
 (5, 3) → 5 – 3(3) + 4 = 5 – 9 + 4 = 0 true
→ verifies
 (0, 1) → 0 – 3(1) + 4 = 0 – 3 + 4 = 1 ≠ 0
→ doesn’t verify
4a- 2x + 3y = 3
2(-3) + 3(3) = -6 + 9 = 3 true
→ (-3, 3) is a solution
b- (0, 1) is a solution since 2(0) + 3(1) = 3 true
(3, -1) is a solution since 2(3) +3(-1) = 6 – 3 = 3 true
5a- x – y = 3
3x + y = 17

(5, -5) is a solution if it verifies both equations
x – y = 3 → 5 – (-5) = 10 ≠ 3
so (5, -5) is not a solution

(-2, -3) is a solution if it verifies both equations
x – y = 3 → -2 – (-3) = 1 ≠ 3
so (-2, -3) is not a solution

(5, 2) is a solution if it verifies both equations
x – y = 3 → 5 – (2) = 3 true
3x + y = 17 → 3(5) + 2 = 17 true
so (5, 2) is a solution
b- 5x – 2y = 7
3x – 5y =8

(2, 3) is a solution if it verifies both equations
5x – 2y = 7 → 5(2) – 2(3) = 10 – 6 = 4 ≠ 7
→ (2, 3) is not a solution

(-2, -3) is a solution if it verifies both equations
5x – 2y = 7 → 5(-2) – 2(-3) = -10 + 6 = -4 ≠ 7
→ (-2, -3) is not a solution

(3, 2) is a solution if it verifies both equations
5x – 2y = 7 → 5(3) – 2(2) = 15 – 4 = 11 ≠ 7
→ (3, 2) is not a solution

(1, -1) is a solution if it verifies both equations
5x – 2y = 7 → 5(1) – 2(-1) = 5 + 2 = 7 true
3x – 5y = 8 → 3(1) – 5(-1) = 3 + 5 = 8 true
→ (1,-1) is a solution
c- y = 2
x+y=7

(2,2) is a solution if it verifies both equations
y = 2 true
x+y=7 → 2+2=4≠7
→ (2, 2) is not a solution

(5,2) is a solution if it verifies both equations
y = 2 true
x+y=7 → 5+2=7
→ (5, 2) is a solution

(0,2) is a solution if it verifies both equations
y = 2 true
x+y=7 → 0+2=2≠7
→ (0, 2) is not a solution

(-3,2) is a solution if it verifies both equations
y = 2 true
x + y = 7 → -3 + 2 = -1 ≠ 7
→ (-3,2) is not a solution
6a- 4x + y = 10
5x – 2y = 6
using elimination method:
4x + y = 10 → (x2) → 8x + 2y = 20
5x – 2y = 6
add both equations → 13x = 26
x = 26/13 = 2
substitute x in equation (1):
4(2) + y = 10 → y = 10 – 8 = 2
then, (2, 2) is the solution
b- y = 3x + 7
y=x-4
using comparison method:
y = y → 3x + 7 = x - 4
3x – x = -4 – 7
2x = -11 → x = -11/2
substitute x in equation (2):
y = -11/2 – 4 = -19/2
then, (-11/2, -19/2) is the solution
c- x = 5y
2x – 3y = 10
using substitution method:
2(5y) – 3y = 10
10y – 3y = 10
7y = 10 → y = 10/7
substitute y in equation (1):
x = 5y = 5(10/7) = 50/7
then (50/7, 10/7) is the solution
d- 6x = 5y
2x – y = 12
→
x = 5/6y
using substitution method:
2( 5/6y) – y = 12
5y/3 – y = 12
2y/3 = 12 → y = 18
substitute y in equation (1):
x = 5/6(18) = 90/6 = 15
then, (15, 18) is the solution
7a- 2y + 4x = 3
2y = -4x + 3
y = -2x + 3/2
X
Y
b- y – 3 = 2x + 2
y = 2x + 2 + 3
y = 2x + 5
0
3/2
1
-1/2
X
Y
-1
3
-2
1
c- y/4x = 1/2
2y = 4x
y = 2x
X
Y
0
0
1
2
d- x + 2(y – 1) = y + 2(x + 1)
x + 2y – 2 = y + 2x + 2
2y – y = 2x – x + 2 + 2
y=x+4
X
Y
0
4
-1
3
e- x + 5 = y + x + 3
y=2
8a- y + x = 3
y–x=2

y + x = 3 → y = -x + 3
X
y

0
3
1
2
y–x=2 → y=x+2
x
Y
0
2
the two lines intersect in one point
→ the system has one solution
1
3
b- y + 2x = 2
3y + 6x = 4

y + 2x = 2 → y = -2x + 2
X
Y

0
2
1
0
3y + 6x = 4 → y = -2x + 4/3
X
Y
0
4/3
the two lines intersect at one point
→ one solution
1
-2/3
c- 2y – x = y + 3
y + 2x = 3x – 1

2y – x = y + 3
2y – y = x + 3
y=x+3
X
Y

0
3
1
4
y + 2x = 3x – 1
y = 3x – 2x – 2
y=x–2
X
Y
0
-2
1
-1
the two lines are parallel → no intersection → solution
9a- y = 4x – 3
y = 2x + 23
using comparison method:
y=y
4x – 3 = 2x + 23
4x – 2x = 23 + 3
2x = 26 → x = 13
substitute x in equation (1):
y = 4(13) – 3 = 49
then, (13, 49) is the solution
b- 2x + 4y = 6
2x – 4y = 4
using elimination method:
add both equations → 4x = 10
x = 10/4 = 5/2
substitute x in equation (1):
2(5/2)+ 4y = 6
4y = 6 – 5
y = 1/4
then, (5/2, 1/4) is the solution
c- 3x – 2y = 0
5x + 3y =17
using substitution method:
3x – 2y = 0 → 3x = 2y → x = 2/3y
substitute x
→ 5(2/3y) + 3y = 17
10/3y + 3y = 17
19/3y = 17 → y = 51/19
substitute y to find x:
x = 2/3y = 2/3(51/19) = 34/19
then, the solution is (34/19, 51/19)
d- 3x +14y = 17
5x – 14y = 3
using elimination method, add the two equations:
8x = 20 → x = 20/8 = 5/2
substitute x in equation (2):
5(5/2) – 14y = 3
-14y = 3 – 25/2
-14y = -19/2
y = 19/28
then, the solution is (5/2, 19/28)
10a- (x + y)/5 = 0.3 – (2x – y)/2 → (x 10) → 2x + 2y = 3 – 10x + 5y → y = 4x – 1
6y – 12x = 30 → (divide by 6) →
y – 2x = 5
→ y = 2x + 5
y = 4x – 1
X
Y
0
-1
1
3
y = 2x + 5
X
Y
0
5
-1
3
graphically, (3, 11) is the solution.
by calculation:
y = 4x – 1
y = 2x + 5
using comparison method:
4x – 1 = 2x + 5
4x – 2x = 5 + 1
2x = 6 → x = 3
substitute x in equation (1):
y = 4(3) – 1 = 12 – 1 = 11
→ (3, 11) is the solution.
b- 4x – y = 1
5x + 3y = 17
→
→
y = 4x – 1
y = -5/3x + 17/3
y = 4x – 1
X
Y
0
-1
1
3
0
17/3
1
4
y = -5/3x + 17/3
X
Y
graphically, the two lines intersect at (1.1, 3.7) which is the solution.
by calculation:
y = 4x – 1
y = -5/3x + 17/3
using comparison method:
4x – 1 = -5/3x + 17/3
4x + 5/3x = 17/3 + 1
17/3x = 20/3 → x = 20/17 = 1.17
substitute x in equation (1):
y = 4(20/17) – 1 = 80/17 – 1 = 3.7
→ the solution is (1.1, 3.7)
11(b) x = y – 5
x + y = 23
→
y=x+5
x + y 23
true
(c) x – y = 5
x + y = 23
→
x=y+5
x + y = 23
true
12a- sum of 2 numbers is 75 → x + y = 75
difference is 25 → x – y = 25
→ x + y = 75
x – y = 25
b- number exceeds another by 14 → x = y + 14
sum is 70 → x + y = 70
→ x = y + 14
x + y = 70
c- let x be the price Susan paid
let y be the price Jinan paid
both paid 24,000 → x + y = 24000
Susan paid twice as Jinan → x = 2y
→ x + y = 24000
x = 2y
d- let x be the number of apple trees
let y be the number of orange trees
46 trees in total → x + y = 46
apple tress three times as much as orange trees → x = 3y
→ x + y = 46
x = 3y
e- let x be the number of tickets Hani sold
let y be the number of tickets Sami sold
Hani sold twice as Sami → x = 2y
240 total tickets sold → x + y = 240
→ x = 2y
x + y = 240
f- difference is 26 → x – y = 26
quotient is 4/3 → x/y = 4/3 → 3x = 4y
→ x – y = 26
3x = 4y
g- difference is 50 → x – y = 50
one is double the other → x = 2y
→ x – y = 50
x = 2y
h- let x be the number of small bottles
let y be the number of large bottles
price of small bottles = 2000
price of large bottles = 3900
→ 2000x + 3900y = 21700
2000x = 3900y + 1700
13a- A(1, 1) and B(0, 5)
equation of (AB) is of the form y = ax + b
a = (yB – yA)/(xB – xA) = (5 – 1)/(0 – 1) = -4
A belongs to (AB) → yA = axA + b
1 = -4(1) + b
b=5
therefore (AB): y = 4x + 5
b- A(-1, 2) and B(1, 2)
(AB): y = ax + b
a = (yB – yA)/(xB – xA) = (2 – 2)/(1 + 1) = 0
→ (AB) is of the form y = b (parallel to x-axis)
A belongs to (AB) → yA = b → b = 2
therefore (AB): y = 2
c- A(2, 0) and B(0, 2)
(AB): y = ax + b
a = (yB – yA)/(xB – xA) = (2 – 0)/(0 - 2) = -1
A belongs to (AB) → yA = -1xA + b
0 = -1(2) + b
b=2
therefore (AB): y = -x + 2
d- A(-1, -1) and B(-1, 1)
(AB): y = ax + b
a = (yB – yA)/(xB – xA) = (1 + 1)/(-1 + 1) = 2/0 impossible
→ (AB) is parallel to y-axis (no slope) of equation x = c
A belongs to (AB) → xA = c → c = -1
therefore (AB): x = -1
14a- y = x – 1
y = 3x + 3
using comparison method:
x – 1 = 3x + 3
x – 3x = 3 + 1
-2x = 4 → x = -2
substitute x:
y = x + 1 = -2 - 1 = -3
→ (-2, -3) is the point of intersection.
b- y = x + 1
0 = 2x – 1
0 = 2x – 1 → 2x = 1 → x = 1/2
y = x + 1 = 1/2 + 1 = 3/2
→ (1/2, 3/2) is the point of intersection.
c- y = 6x + 1
y = -2x + 3
using comparison method:
6x + 1 = -2x + 3
6x + 2x = 3 – 1
8x = 2 → x = 1/4
substitute x:
y = 6x + 1 = 6(1/4) + 1 = 5/2
→ (1/4, 5/2) Is the point of intersection.
d- y = 3
x=5
→(5, 3) is the point of intersection.
15a- To buy 8 crayons and a pen, we paid 10,000
and to buy 3 crayons and 2 pens, we paid 7000.
b- 8x + y = 10000
3x + 2y = 7000
using elimination method:
8x + y =10000 → (x -2)
-16x – 2y = -20000
3x + 2y = 7000
add both equations:
-13x = -13000
x = 1000
substitute x → 8(1000) + y = 10000
→ y = 2000
Chapter 7
Problems page 77
1a- A(1, 2)
B(-2, -1)
C(-1, 2)
D(1, -1)
b- (AB): y = ax + b
a = (yB – yA) / (xB – xA) = ((-1 – 2)/(-2 – 1) = -3/-3 = 1
A belongs to (AB) → yA = 1(xA) + b
2=1+b
b=1
therefore (AB): y = x + 1
(CD): y = ax + b
a = (yD – yC) / (xD – xC) = ((-1 – 2)/(1 + 1) = -3/ 2
C belongs to (CD) → yC = 1(xC) + b
2 = -1 + b
b=3
therefore (AB): y = x + 3
c- y = x + 1
y = -3/2x + 1/2
using comparison method:
x + 1 = -3/2x + 1/2
x + 3/2x = 1/2 – 1
5/2x = -1/2
x = -1/5
substitute x:
y = x + 1 = -1/5 + 1 = 4/5
therefore (-1/2, 4/5) is the point of intersection of (AB) and (CD)
3a- If you gave me 6 bills, I will have as much as you have.
If I give you 10 bills, you will have 2 times as I have.
b- a + 6 = t – 6
a + 10 = 2(t – 10)
→
→
a= t - 12
a = 2t – 30
using comparison method:
t – 12 = 2t – 30
t – 2t = -30 + 12
-t = - 18 → t = 18
substitute t:
→ a = t – 12 = 18 – 12 = 6
Therefore, Anthony has 6 bills and Thomas has 18 bills.
4a- u + v = 41
u–v=9
b- u > 0 and v > 0
c- u + v = 41
u–v=9
using elimination method, add both equations:
2u = 50 → u = 25
substsitute u:
25 + v = 41 → v = 41 – 25 = 16
u = x2 → x = +√25
x = +5
v = y2 → y = +√16
x = +4
or x = -√25
or x = -5
or x = -√16
or x = -4
5a- 2u + v = 5
u + 2v = 8
b- x ≠ 0 and y ≠ 0
c- 2u + v = 5
u + 2v = 8 → u = 8 – 2v
substitute u in equation (1):
2(8 – 2v) + v = 5
16 – 4v + v = 5
-3v = -11 → v = 11/3
substitute v:
u + 2(11/3) = 8
→ u = 2/3
u = 1/x → x = 1/u = 3/2
v = 1/y → y = 1/v = 3/11
6
let x be the number of 500 LL bills
let y be the number of 1000 LL bills
Ziad has 26 bills → x + y = 26
sum is 16500 LL → 500x + 1000y = 16500
x + y = 26
→
x = 26 - y
500x + 1000y = 16500
substitute x:
500(26 – y) + 1000y = 16500
13000 – 500y + 1000y = 16500
500y = 3500
y = 3500/500 = 7
substitute y to find x
→ x + 7 = 26
x = 10
then, there are 19 bills of 500LL and 7 bills of 1000LL.
7
let x be the number of chairs
let y be the number of tables
price of 1 chair = 40,000
price of 1 table = 16,000
total price is 1,880,000 LL → 40,000x + 16,000y = 1,880,000
chairs exceed tables by 5 → x = y + 5
then: 40000x + 16000y = 1880000
x=y+5
substitute x:
40000(y + 5) + 16000y = 1880000
40000y + 200000 + 16000y = 1880000
56000y = 1880000 - 200000
y = 1680000/56000 = 30
substitute y to find x
→ x = y + 5 = 30 + 5 = 35
then, there are 35 chairs and 30 tables.
8- rectangle : length = x
width = y
a- area of rectangle = length x width = x.y
b- width = y + 9
length = x – 2
same area = xy
→ length x width = (x – 2)(y + 9) = xy
xy + 9x – 2y – 18 = xy
9x – 2y = 18
c- length = x – 5
width = y + 2
same area = xy
→ length x width = (x – 5)(y + 2) = xy
xy + 2x – 5y – 10 = xy
-5y + 2x = 10
d- 9x – 2y = 18 → (x 5) →
-5y + 2x = 10 → (x -2) →
45x – 10y = 90
10y – 4x = -20
add both equations:
41x = 70 → x = 70/41
substitute x:
9(70/41) – 2y = 18
y = -54/49 impossible (negative value)
9
let x be the price of a rose
let y be the price of a daisy
rose costs 800 more than daisy → x = y + 800
7 roses and 5 daisies cost 10,400 → 7x + 5y = 10400
→ x = y + 800
7x + 5y = 10400
using substitution:
7(y + 800) + 5y = 10400
7y + 5600 + 5y = 10400
12y = 4800 → y = 4800/12 = 400
x = y + 800 = 400 + 800 = 1200
then, price of 1 rose is 1200 LL
and price of 1 daisy is 400 LL
10
let x be the numerator
let y be the numerator
→ (x + 10)/(y + 10) = 2/5
(x – 5)/(y-5) = 3/10
→
→
→ 5x + 50 = 2y + 20 →
10x – 50 = 3y – 15 →
5x – 2y = -30 → (x -2) →
10x – 3y = 35
5(x + 10) = 2(y + 10)
10(x – 5) = 3(y – 5)
add both equations:
y = 95
substitute y:
5x – 2(95) = -30
5x – 190 = -30
5x = 160
x = 160/5 = 32
then: x/y = 32/95
11
let x be the number of marbles of Samir
let y be the number of marbles of Jamil
x + y = 24
→
y + 2 = 2(x – 2) →
x + y = 24
y = 2x – 6
substitute y:
x + y = 24
x + 2x – 6 = 24
x = 30/3 = 10
substitute x:
x + y = 24
10 + y = 24 → y = 14
therefore, Samir has 10 marbles and Jamil has 14 marbles.
-10x + 4y = 60
10x – 3y = 35
12a- y = 2x – 7
X
Y
1
-5
2
-3
4
16
6
14
y = -x + 20
X
Y
Graphically, the solution is (9, 11).
b- y = 2x – 7
y = -x + 20
using comparison method:
2x – 7 = -x + 20
2x + x = 20 + 7
3x = 27
x = 27/3 = 9
substitute x:
y = 2x – 7 = 2(9) – 7 = 11
So, (9, 11) is the solution.
c- let x be the number of CDs of Anthony
let y be the number of CDs of Bernice
2x = y + 7
(x + 2) + (y + 2) = 24
d- 2x = y + 7
x + y = 20
→
→
2x = y + 7
x + y = 20
is equivalent to
y = 2x - 7
y = -x + 20
from part (b), we can deduce that x = 9 and y = 11
Therefore, Anthony has 9 CDs and Bernice has 11 CDs.
13a- let x be the price of bath towels
let y be the price of bath gloves
→ 4x + 5y = 110
6x + 4y = 172
add both equations:
14x = 750
x = 750/14 = 30
→ (x -4) →
→ (x 5) →
-16x – 20y = -440
30x + 20y = 860
substitute x:
4(30) + 5y = 110
120 + 5y = 110
5y = -10 → y = -2
b- Yes, since the price of the glove is negative which is impossible.
14a- x + y = 30
→
4x + 3y = 100
x = 30 - y
substitute x in the second equation:
4(30 – y) + 3y = 100
120 – 4y + 3y = 100
-y = -20
y = 20substitue y:
x = 30 – y = 30 – 20 = 10
b- let x be the number of books 4cm thick
let y be the number of books 3cm thick
x + y = 30
4x + 3y = 100
from part (a), we deduce that x = 10 and y = 20
Therefore, Walid has 10 books 4cm thick and 2o books 3 cm thick
15
let x be the cost covered by Tripoli
let y be the cost covered by Batroun
x + y = 150 000 000
x/280000 = y/70000 →
70000x = 280000y
→
x = 4y
substitute x:
4y + y = 150 000 000
5y = 150 000 000 → y = 30 000 000
substitute y:
x = 4y = 4(30 000 000) = 120 000 000
Therefore, Tripoli has to pay 120 000 000 LL and Batroun has to pay 30 000 000 LL.
16a- area of red shape = area of big square + area of small square
= (x.x) + (y.y)
= x2 + y2
area of blue shape =area of big square – area of small square
= x.x – y.y = x2 – y2
b- x2 + y2 = 13
x2 – y2 = 5
let u = x2
v = y2
u + v = 13
u–v=5
add both equations:
2u = 18 → u = 18/2 = 9
substitute u:
u + v = 13
9 + v = 13
v=4
x2 = u → x = +√9 or x = -√9
→ x = +3 or x = -3 (rejected)
y2 = v → y = +√4 or y = -√4
→ y = + 2 or y = -2 (rejected)
c- (x2 + y2) + (x2 – y2) = 18
x2 + y2 = 2.6(x2 – y2)
→
2x2 =18
→
x2 = 9
x2 + y2 = 2.6x2 – 2.6y2
substitute → 9 + y2 = 2.6(9) – 2.6y2
y2 + 2.6y2 = 23.4 – 9
y2 = 4
then x = 3 and y = 2 (deduced from part (b)).