(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