Solve the following equations by trial and error method:
(i) 5p+2=17
(ii) 3m−14=4
In this method, we assume the value of variable and check whether it satisfies the equation or not.
(i) For, 5p+2=17
Let x=0
LHS: 5(0)+2=2≠17≠ RHS
Hence,x≠0
Let x=1
LHS: 5(1)+2=5+2=7≠17≠ RHS
Hence, x≠1
Let x=3
LHS: 5(3)+2=15+2=17= RHS
Hence, x=3 is the solution of the equation 5p+2=17.
(ii) For,3m−14=4
Let m=0
LHS: 3(0)−14=−14≠4≠ RHS
Hence, m≠0
Let m=1
LHS: 3(1)−14=3−14=−11≠4≠ RHS
Hence, m≠1
Let m=3
LHS: 3(3)−14=9−14=−5≠4≠ RHS
Hence, m≠3
Let m=6
LHS: 3(6)−14=18−14=4= RHS
Hence, m=6 is the solution of the equation 3m−14=4.