Chapter 7
Exercises page 84
1a- x > 2
b- x ≤ -2
c- -0.5 < t < 1
d- -2 < y < 8
e- -2 < m < 0
2a- -9 < 8 < 10
true since -9 < 8 and 10 > 8
b- -3 > -10
true
c- -15 ≤ -21
false since -15 > 21
d- 6 > 0 > -11
true
e- (-1/2)2 ≥ 1/2 false since (-1/2)2 = 1/4 < 1/2
f- 23 < 32
true since 23 = 8 < 32 = 9
g- -2.5 < 0.5 < 1.5
true
3a- x – 2 ≥ 2
x≥2+2 → x≥4
b- m/6 < -0.5
m < -0.5 x 6 → m < -3
c- 3 ≤ -y/2
3 x 2 ≤ -y → -6 ≥ y
d- 2 – x > 0
-x > -2 → x < 2
e- 4 > y + 4
4–4>y → y<0
4a- 5(x – 3) > 3(x – 10)
5x – 15 > 3x – 30
2x > -15 → x > -15/2
Then 5 is a solution
b- 6(2x – 4) ≥ 0
12x – 24 ≥ 0 → x ≥ 2
Then 5 is a solution
c- 2(3 – x) < -2x + 7
6 – 2x < -2x + 7
0 < 1 → true for all values of x
Then 5 is a solution
d- 4/5 – 5x < 2
-5x < 2 – 4/5
-5x < 6/5
5x > -6/5 → x > -6/25
Then 5 is a solution
5a- 5(x – 3) > 3(x – 10)
5x – 15 > 3x – 30
2x > -15 → x > -15/2
b- -3x – 7 ≤ 4
-3x ≤ 4 + 7
x ≥ 11/3
c- -4(2x – 4) = -12
-8x + 16 ≥ -12
-8x ≥ -12 – 16
x ≤ 28/8 → x ≤ 7/2
d- 2(3x – 7) > 5x – 12
6x – 14 > 5x – 12
x>2
6a5 + x ≤ 4x – 1
-7x + 8 ≥ 6x – 20



2 ≤ -5/2
2≤0
2 ≤ -4
→
→
false
false
false
6≤x →
-13x ≥ 28
2≤x
x ≤ 28/3

2 ≤ 7/2 and 7/2 ≤ 28/3
→ 7/2 is a solution
b- x – 4 ≥ 2x – 6
3x – 8 ≥ x – 10




2≥x
x ≥ -1
x<7
x > -1
-5/2 < 7 true and -5/2 > -1 false
0 < 7 true and 0 > -1 true
→ 0 is a solution
7/2 < 7 true and 7/2 > -1 true
→ 7/2 is a solution
-4 < 7 true and -4 > -1 false
d- 2 + 8x ≥ 6x – 20
3x + 7 > 5x + 30




-4 + 6 ≥ 2x – x →
3x – x ≥ -10 + 8 →
2 ≥ -5/2 true and -5/2 ≥ -1 false
2 ≥ 0 true and 0 ≥ -1 true
→ 0 is a solution
c- x + 3 < 10
→
2(4x – 5) > x – 17 →


→
→
true
→
→
2x ≥ -22
-2x >23
→
→
x ≥ -11
x < -23/2
-5/2 ≥ -11 true and -5/2 < -23/2 false
0 ≥ -11 true and 0 < -23/2 false
-4 ≥ -11 true and -4 < -23/2 false
7/2 ≥ -11 true and 7/2 < -23/2 false
7a- x + 3 > 5x
→
2x + 7 < x – 4 →
3 > 4x
x < -11
→
→
x < 3/4 false
x < -11
b- 2x – 4 < -3(x – 7)
2(x + 1) < -2 + x
→
→
2x – 4 < -3x + 21
2x + 2 < -2 + x
c- x + 2 > 3x
→
-x – 3 > x + 18 →
2x < 1 →
-2x > 21
a- 3x + 4 > 2(x – 4)
2x + 10 < x + 7
→
→
→
→
x<5
x < -4
x < 1/2
x < 21/2
8-
→ -12 < x < -3
x > -12
x < -3
x belongs to ]-12, -3[
b- 4/3x + 8 ≤ 3/2x + 10 →
4(5 – x) ≤ x + 17
→
→ x ≥ 3/5
3x + 4 > 2x – 8 →
2x – x < 7 – 10 →
→
4/3x – 3/2x ≤ 10 – 8 →
20 – 4x ≤ x + 17
→
x belongs to [3/5, +∞[
x ≥ -12
x ≥ 3/5
c- 3x + 4/5 > 2x – 1
-2/3 + 5x < 3x + 7
→
→
x > -1 -4/5
2x > 7 + 2/3
→
→
x > -9/5
x > 23/6
→
→
x≥0
x ≤ 12/7
→ x > 23/6 or x belongs to ]23/6, +∞[
d- x ≥ 0
→
4 + 3/2x ≤ x/3 + 6
x≥0 →
x≥ 0
3/2x – x/3 ≤ 6 – 4
x belongs to [0, 12/7]
e- x – 4 ≤ 5/2x + 13/2
2x- 23/2 ≥ x/2 + 5
→
→
x – 5/2x ≤ 13/2 + 4
2x – x/2 ≥ 5 + 23/2
→
→
x ≥ -7
x ≥ 11
→ x ≥ 11 or x belongs to [11, +∞[
f- x – 4 ≤ 5x/2 →
2x – 3/2 ≥ x/2 + 5
-3x/2 ≤ 4
→
3x/2 ≥ 5 + 3/2 →
→ x ≥ 13/3 or x belongs to [13/3, +∞[
9a- integer that precedes n : n – 1
integer that follows n : n + 1
b- n – 1 < n < n + 1
c- n + 1 > n > n – 1
x ≥ -8/3
x ≥13/3
d- (n – 1) + (n) + (n + 1) > 13
(n – 1) + (n + 1) < 15
→n=5
So, n – 1, n, n + 1
→
→
→
4, 5, 6
3n > 13
2n < 15
→
→
n > 13/3
n < 15/2
Chapter 7
Problems page 86
1a- Sum = 360 degrees
Angle CLU is obtuse → CLU > 180
We have sum of all angles = 360
→ CLU +LCB + CBU + BUL = 360
CLU + 3x – 6 + 2x + x = 360
CLU = 360 – 6x + 6
CLU = 354 – 6x
Then 354 – 6x > 18
-6x > 18 – 354
-6x > -336
x < 56
BCL < 80
3x – 6 < 80
3x < 86
x < 86/3
Values of x are the common values (shaded part)
→ x < 86/3
3a- Using comparison method:
y = 2m – 3x
y = 11 – 2x
→ 2m – 3x = 11 – 2x
2m – 11 = x → x = 2m – 11
Substitute in equation (2):
→ 2(2m – 11) + y = 11
4m – 22 + y = 11
y = 33 – 4m
b- x > 0
y>0
→
→
2m – 11 > 0
33 – 4m > 0
→
→
m > 11/2
m < 33/4
The intersection is the solution (shaded part)
→ 11/2 < m < 33/4
4a- age of ziad = 45
age of wife = 38
age of son = 12
let x be the number of years
→ 45 + x > 3(12 + x)
45 + x > 36 + 3x
-2x > -9
x > 9/2
age of ziad less than twice his wife
→ 45 + x < 2(38 + x)
45 + x < 76 + 2x
-x < 31
x > -31
shaded part is the solution
x belongs to ]-31, 9/2[
5a- AB = 8
BC = 6
CF = 2
Dl = x
b- Area of trapezoid ABCL = [(AB + LC) x BC] / 2
= (8 + 8 – x) x 6 / 2
= 48 – 3x
Area of triangle ADL = (AD x DL) / 2
= 6x/2 = 3x
Area of rectangle CDEF = CF x CD
= 2 x 8 = 16
Area of ABCL ≤ area of ADL
48 – 3x ≤ 3x
48 ≤ 6x → x ≥ 8
Area of ABCL ≥ area of CDEF
48 – 3x ≥ 16
48 – 16 ≥ 3x
32 ≥ 3x → x ≤ 32/3
Solution: 8 ≤ x ≤ 32/3
6a- ABCD is a right trapezoid at A and D, with DC = 6, AB = 4, and AD = 10.
Let M be a point on AB such that AM = x.
b- Area of BMC = area of ABCD – (area of DMC + area of ABM)
= (AB + CD)AD/2 – (DM.DC/2 + AM.AB/2)
= (4x6)10/2 – ((10 – x)6/2 + 4x/2)
= 50 – (30 – 3x + 2x)
= 50 – 30 + x
= 20 + x
Area of DMC = DM.DC/2 = (10 – x)(6)/2 = 30 – 3x
Area of AMB = AM.AB/2 = 4x/2 = 2x


area of BMC < area of DMC
→ 20 + x < 30 – 3x
→
x < 10/4
area of BMC > area of AMB
20 + x > 2x
x < 20
solution: x < 10/4
7a- x = 2cm
b- Area(AMBG) = l.w = 3x
Area(FECG) = l.w = (6 – x)(2) = 12 – 2x
Area(MDEF) = l.w = (6 – x)(1) = 6 – x
→ Area(AMBG) < Area(FECG)
Area(AMBG) > Area(MDGF)
→
x belongs to ]3/2, 12/5[
8a- (u): y = 2x
X
Y
0
0
1
2
0
2
1
3
(v): y = x + 2
X
Y
3x < 12 – 2x
3x > 6 – x
→
x < 12/5
x > 3/2
(w): y = 2/3x + 4
X
Y
0
4
3
6
b- ‘A’ point of intersection of (u) and (v)
→ y = 2x
y=x+2
by comparison → 2x = x + 2
x=2
substitute → y = 2x = 2(2) = 4
then A(2, 4)
‘B’ point of intersection of (u) and (w)
→ y = 2x
y = 2/3x + 4
by comparison → 2x = 2/3x + 4
x=3
substitute → y = 2x = 2(3) = 6
then B(3, 6)
‘C’ is the point of intersection of (v) and (w)
→y=x+2
y = 2/3x + 4
by comparison → x + 2 = 2/3x + 4
1/3x = 2 → x = 6
substitute → y = x + 2 = 6 + 2 = 8
then C(6, 8)
c- 2x ≤ x + 2 ≤ 2/3x + 4
2x ≤ x + 2
x + 2 ≤ 2/3x + 4
→
x≤2
x≤6
then x ≤ 2 or x belongs to ]-∞, 2[
d- x + 2 ≤ 2x ≤ 2/3x + 4
x + 2 ≤ 2x
2x ≤ 2/3x + 4
→
x≥2
x≤3
e- x + 2 ≤ 2/3x + 4
2/3x + 4 ≤ 2x
→
x – 2/3x ≤ 4 – 2
2/3x – 2x ≤ -4
→
-1/3x ≤ -2
-x ≤ -2
x≥6
x≥2
then x belongs to [3, 6]
f- 2/3x + 4 ≤ x + 2 ≤ 2x
2/3x + 4 ≤ x + 2
x + 2 ≤ 2x
→
→
x≤6
x≥3
then x ≥ 6 or x belongs to [6, +∞[