Understanding Engineering Mathematics

(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