If 2(2a+1)×3(3b+2)62=2x×3y
Chose the correct option:
x = 2a – 1, y = 3b
x = 2a + 1, y = 3b + 2
x = 2a + 1, y = 3b
2(2a+1)×3(3b+2)62=2(2a+1)×3(3b+2)(2×3)2 =2(2a+1)×3(3b+2)22×32 [∵(ab)m=am×bm]
=2(2a+1−2)×3(3b+2−2) [∵aman=am−n]
=2(2a−1)×3(3b)∵2(2a−1)×3(3b)=2x×3y
⇒ x=2a−1,y=3b
Find the values of x and y that satisfy the below given pair of equations, where x≠0 & y≠0. 2x+3y=13 5x−4y=−2