Introduction to Mathematical Thinking Solutions

This document provides solutions to exercises on mathematical thinking, including finding slopes, intercepts, and sketching graphs. It covers topics such as equations and graphs. Suitable for beginners.

Introduction to Mathematical Thinking Solutions
Q1.
a) For the points (2,4) and (5,6) ,
slope= dy
dx = 6−4
5−2 = 2
3
→ 2
3 = y−4
x−2
∴ y= 2
3 x + 8
3
b) For the points (4,3) and (-2,5) ,
slope= dy
dx = 5−3
−2−4 =−1
3
→− 1
3 = y−3
x−4
∴ y=−1
3 x + 13
3
Q2.
a) For the point (2,1) with gradient 3:
slope=3= y −1
x−2
∴ y=3 x−5
¿ the form of x
a + y
b =1, the equation becomes x
(5
3 )+ y
3 =1
, where x−intercept , a= 5
3 ∧ y −intercept , b=3
b) For the point (-2,1) with gradient 1
2 :
slope= 1
2 = y −1
x +2
∴ y= 1
2 x +2
¿ the form of x
a + y
b =1, the equation becomes x
−4 + y
2 =1
, where x−intercept , a=−4∧ y−intercept ,b=2

c) For the points (1,1) with y-intercept (0,3) ,
slope= dy
dx = 3−1
0−1 =−2
→−2= y −3
x−0
∴ y=−2 x+ 3
¿ the form of x
a + y
b =1, the equation becomes x
(3
2 )+ y
−2 =1
, where x−intercept , a= 3
2 ∧ y −intercept , b=−2
d) For the points (-1,3) with y-intercept (2,-1) ,
slope= dy
dx = 3+1
−1−2 =−4
3
→− 4
3 = y−3
x +1
∴ y=−4
3 x +5
3
¿ the form of x
a + y
b =1, the equation becomes x
( 5
4 )+ y
( 5
3 )=1
, where x−intercept , a= 5
4 ∧ y−intercept ,b= 5
3
Q3.
a) ¿ the form of x
a + y
b =1, the e quation is x
−1 + y
2 =1
, where x−intercept=a=−1∧ y−intercept =b=2
(as read¿the graph)
b) For the points (-2,1) with y-intercept (0,4) ,
slope= dy
dx = 4−1
0−−2 = 3
2
→ 3
2 = y−4
x−0
∴ y= 3
2 x +4

¿ the form of x
a + y
b =1, the equation becomes x
(−8
3 ) + y
4 =1
c) For y =2
3 x+2 ,¿ the form of x
a + y
b =1, the equation becomes x
(−3 ) + y
2 =1
d) For 2 x+ 3 y=6 ,
For x=0 , y −intercept=2 while for y=0 , x−intercept =3
∴∈the form of x
a + y
b =1 , the equationbecomes x
3 + y
2 =1
e) For y+1
x−2 = 1
2 ,
For x=0 , y −intercept=−2while for y=0 , x−intercept =4
∴∈the form of x
a + y
b =1 , the equationbecomes x
4 + y
−2 =1
Q4.
a) For x
2 + y
3 =1 ,the sketch of the graph is as shown below :
-6 -4 -2 0 2 4 6
-6
-4
-2
0
2
4
6
8
10
12
𝑥/2+ /3=1𝑦
y
Linear (y)
x−intercept ≅ 2∧ y−intercept ≅ 3
(as read¿the graph)

b) For y =2
3 x+1 , the sketch of the graph is as shownbelow :
-6 -4 -2 0 2 4 6
-3
-2
-1
0
1
2
3
4
5
𝑦=2/3 +1𝑥
y
Linear (y)
x−intercept ≅ −1.5∧ y−intercept ≅ 1
(as read¿the graph)
c) For 2 x+ 3 y=6 , the sketch of the graph is as shownbelow :

-6 -4 -2 0 2 4 6
-2
-1
0
1
2
3
4
5
6
2 +3 =6𝑥 𝑦
y
Linear (y)
x−intercept ≅ 3∧ y−intercept ≅ 2
(as read¿the graph)
d) For y+1
x−2 = 3+1
4−2 ,the sketch of the graph is as shown below :
-8 -6 -4 -2 0 2 4 6 8
-20
-15
-10
-5
0
5
10
( +1)/( −2)=(3+1)/(4−2)𝑦 𝑥
y
Linear (y)
x−intercept ≅ 2. 5∧ y−intercept ≅−5
(as read¿the graph)

Solutions to Tutorial Exercises
Ex1.
a) For the points (1,2) and (-3,5) ,
slope= dy
dx = 5−2
−3−1 =−3
4
→− 3
4 = y−2
x−1
∴ y=−3
4 x + 11
4
b) For the points (1,4) and (3,1) ,
slope= dy
dx = 4−1
1−3 =−3
2
→− 3
2 = y−3
x−4
∴ y=−3
2 x +9
Ex2.
a) For the point (2,1) with gradient −3
2 :

slope=−3
2 = y−1
x−2
∴ y=−3
2 x + 4
¿ the form of x
a + y
b =1, the equation becomes x
( 8
3 ) + y
4 =1
, where x−inter cept , a= 8
3 ∧ y−intercept , b=4
b) For the point (2,1) with x-intercept (-1,0) ,
slope= dy
dx =1−0
2+1 = 1
3
→ 1
3 = y−0
x +1
∴ y= 1
3 x + 1
3
¿ the form of x
a + y
b =1, the equation becomes x
−1 + y
( 1
3 )=1
, where x−intercept , a=−1∧ y−intercept , b= 1
3
c) For the points (-1,1) with y-intercept (3,-2) ,
slope= dy
dx = 1+2
−1−3 =−3
4
→− 3
4 = y−1
x +1
∴ y=−3
4 x + 1
4
¿ the form of x
a + y
b =1, the equation becomes x
(1
3 )+ y
( 1
4 )=1
, where x−intercept , a= 1
3 ∧ y −intercept , b= 1
4
Ex3.
a) ¿ the form of x
a + y
b =1, the equation is x
−2 + y
4 =1
, where x−intercept=a=−2∧ y−intercept =b=4

b) For the points (3,1) with y-intercept (6,3) ,
slope= dy
dx = 3−1
6−3 = 2
3
→ 2
3 = y−1
x −3
∴ y= 2
3 x−1
¿ the form of x
a + y
b =1, the equation becomes x
(3
2 )+ y
−1 =1
c) For 6 x + y=2 , For x=0 , y −intercept=2 while for y=0 , x−intercept =1
3
∴∈the form of x
a + y
b =1 , the equationbecomes x
( 1
3 ) + y
2 =1
d) For−5 x+ 2 y =2 ,
For x=0 , y −intercept=1 while for y=0 , x−intercept =−2
5
∴∈the form of x
a + y
b =1 , the equationbecomes x
(−2
5 )+ y
1 =1
e) For y−2
x−6 = 2
3 ,
For x=0 , y −intercept=−2while for y=0 , x−intercept =3
∴∈the form of x
a + y
b =1 , the equationbecomes x
3 + y
−2 =1
Ex4.
a) For x
−3 + y
4 =1 , the sketch of the graph is as shownbelow :

-6 -4 -2 0 2 4 6
-4
-2
0
2
4
6
8
10
12
𝑥/(−3)+ /4=1𝑦
y
Linear (y)
x−intercept ≅ −3∧ y−intercept ≅ 4
(as read¿the graph)
b) For y =−1
4 x+2 , the sketch of the graph is as shownbelow :

-15 -10 -5 0 5 10 15
-1
0
1
2
3
4
5
𝑦=−1/4 +2𝑥
y
Linear (y)
x−intercept ≅ 8∧ y−intercept ≅ 2
(as read¿the graph)
c) For−x +3 y=3 , the sketch of the graph is as shown below :
-6 -4 -2 0 2 4 6
-1
-0.5
0
0.5
1
1.5
2
2.5
3
−𝑥+3 =3𝑦
y
Linear (y)
x−intercept ≅−3∧ y−intercept ≅ 1
(as read¿the graph)
d) For y−2
x+3 = 1−2
−4 +3 ,the sketchof the graphis as shown below :

-8 -6 -4 -2 0 2 4 6 8
-4
-2
0
2
4
6
8
10
12
14
( −2)/( +3)=(1−2)/(−4+3)𝑦 𝑥
y
Linear (y)
x−intercept ≅−5∧ y−intercept ≅ 5
(as read¿the graph)