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Question

Solve the following equations by trial and error method:
(i) 5p+2=17
(ii) 3m14=4

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Solution

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=217 RHS

Hence,x0

Let x=1

LHS: 5(1)+2=5+2=717 RHS

Hence, x1

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,3m14=4

Let m=0

LHS: 3(0)14=144 RHS

Hence, m0

Let m=1

LHS: 3(1)14=314=114 RHS

Hence, m1

Let m=3

LHS: 3(3)14=914=54 RHS

Hence, m3

Let m=6

LHS: 3(6)14=1814=4= RHS

Hence, m=6 is the solution of the equation 3m14=4.


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