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From Karnataka Open Educational Resources
Activities - progressions problems (view source)
Revision as of 10:22, 12 August 2014
, 10:22, 12 August 2014→Algorithm
10b = 2 ( 3b + 2 )<br>
10b = 2 ( 3b + 2 )<br>
10b = 6b + 4<br>
10b = 6b + 4<br>
10b – 6b = 4<br>
4b = 4<br>
b = 1<br>
Consider equation 2,
a + b = 10b
Put b = 1
a + (1) = 10 (1)
a + 1 = 10
a = 10-1
a = 9
