Question

Hey Friends!! Good Morning.. :)
Give one example for:-
i) Substitution Method
ii) Elimination Method
iii) Cross Multiplication Method
ALL THE VERY BEST FOR TODAY'S EXAM FRIENDS.. :)

Answer 1

substitution method: ex s —t = 3


Answer :
Given equations are
s – t = 3 …(i)
 …(ii)
From eqn (i), we get
⇒ s = 3 + t …(iii)
On substituting s = 3 + t in eqn (ii), we get


⇒ 6 + 2t + 3t = 6×6
⇒ 5t = 36 – 6

Now, putting the t = 6 in eqn (iii), we get
⇒ s = 3 + 6
⇒ s = 9
Thus, s = 9 andt = 6 is the required solution

Elimination method: 


Answer :
Given pair of linear equations is
 …(i)
And  …(ii)
On multiplying Eq. (i) by 5 and Eq. (ii) by 2 to make the coefficients of  equal, we get the equation as
 …(iii)
 …(iv)
On subtracting Eq. (iii) from Eq. (iv), we get





On putting  in Eq. (ii), we get




Hence,  and  , which is the required 

cross multiplication method: 

Answer :
The given equations are
 –  = – 2 … (1)
 +  =  … (2)
By taking LCM,
(1) becomes 9x – 10y + 12 = 0
(2) becomes 2x + 3y – 13 = 0
For cross multiplication method, we write the coefficients as

Hence, we get  =  = 
⇒  =  = 
⇒  =  = 
⇒ x =  = 2
⇒ y =  = 3
∴ (2, 3) is the solution to the given system.

Answer 2

hope it's help u.....