Let
and
Find each of the following :
i. 2A – 3B
ii. B – 4C
iii. 3A – C
iv. 3A – 2B + 3C
(i) 2A=
=![]()
= 3B=
=![]()
= 2A-3B=
-
=![]()
= ![]()
Hence, 2A-3B=![]()
(ii) 4C=
=![]()
B-4C=
-![]()
=
=![]()
Hence, B-4C=![]()
(iii) 3A= ![]()
= 3A-C=
-![]()
=
=![]()
Hence, 3A-C=![]()
(iv) 3A=
=![]()
= 2A=
=![]()
= 3C=
=![]()
= 3A-2B+3C=
-
+![]()
=
=![]()
Hence, 3A-2B+3C=![]()
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