Understanding Engineering Mathematics

(やまだぃちぅ) #1
(iii)

7
3

×

4
7

=

4
3

(iv)

7
5

×

3
14

=

1
5

×

3
2
(cancelling 7 from top and bottom, as in (iii))

=

1 × 3
5 × 2

=

3
10

(v)

3
4

÷

4
5

=

3
4

×

5
4

=

15
16

(vi)

1
2

+

1
3

=

3
2 × 3

+

2
2 × 3
on multiplying top and bottom appropriately to get the common
denominator in both fractions,

=

3
6

+

2
6

=

3 + 2
6

=

5
6

(vii)

1
2


1
3

=

3
6


2
6

=

3 − 2
6

=

1
6

(viii)


4
15


7
3

=

4
15


5 × 7
15

=

4 − 35
15

=−

31
15

(ix) 1+

1
2

+

1
3

=

6
6

+

3
6

+

2
6

=

11
6
Here we found the LCM of 2 and 3 (6) and put everything over
this, including the 1.

(x)

2
3


3
4

+

1
8
We want the LCM of 3, 4, 8. This is 24, so
2
3


3
4

+

1
8

=

2 × 8
24


3 × 6
24

+

3
24

=

16 − 18 + 3
24

=

1
24
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