If any number is in the denominator than it become difficult to solve.
s
-
t
= 3 ;
(ii)
s
3
+
t
2
= 6
Soln.
s
3
+
t
2
= 6
... (i)
Substituting equation (iii) in equation (ii) ;
Substituting t = 6 in equation (iii) ;
s
= 6 + 3
s
= 9
Solution s = 9, t = 6
Now how to solve such equations?
By Substitution Method
There are two denominator
3 & 2
Their LCM = 6
Multiplying throughout by 6,
s
3
+
t
2
= 6 × 6
6
6
3
2
2s
= 36
+
3t
2
= 36
(t + 3)
+ 3t
2t
+
6
3t
= 36
+
5t
+
6
= 36
5t
=
36
- 6
5t
=
30
t
=
30
5
t = 6
Constant → R.H.S
... (ii)
s
-
t
= 3
s
= t + 3
... (iii)
How to get the value of s ?
We have to substitute
t = 6
Either eqn (i), eqn (ii) or eqn (iii)
Let us substitute eqn (iii)
Number the equation as (i)
Number the equation as (ii)
Consider one of the two equations
Let us Consider equation no. (i)
Consider (i)
Which equation is to be considered
You can consider either of the two equations
It is better to consider simpler of the two equations
s
-
t
= 3
Number the equation as (iii)
What is the name of the method ?
SUBSTITUTION Method
So we need to substitute something
substitute what ?
Substitute eqn. (iii)
Where ?
In the equation which was not considered
Write their eqn.
either s = something
or t = something
Remove the denominator
6