Question

Solve $11+x=19$ by the trial-and-error method.

Answer 1

We have $11+x=19$

At $x=0$ LHS$=11+x=19\ne$RHS

At $x=1$ LHS$=11+1=12\ne$RHS

At $x=2$ LHS$=11+2=13\ne$RHS

At $x=3$ LHS$=11+3=14\ne$RHS

At $x=4$ LHS$=11+4=15\ne$RHS

At $x=5$ LHS$=11+5=16\ne$RHS

At $x=6$ LHS$=11+6=17\ne$RHS

At $x=7$ LHS$=11+7=19=$RHS

$\therefore x=7$ is a root of the equation