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Check whether the value given in the brackets is a solution to the given equation or not:
$(a).\ n+5=19$ $(n=1)$
$(b).\ 7n+5=19$ $(n=-2)$
$(c) 7n+5=19$ $(n=2)$
$(d).\ 4p-3=13$ $(p=1)$
$(e).\ 4p-3=13$ $(p=-4)$
$(f).\ 4p-3=13$ $(p=0)$
To do:
We have to check whether the value given in the brackets is a solution to the given equation or not.
Solution:
(a) Here,
L.H.S $=n+5$
R.H.S $=19$
And $n=1$ (given)
By putting, $n=1$,
L.H.S $=1+5=6$
$≠$R.H.S
L.H.S $≠$ R.H.S, so $n=1$ is not a solution of the equation.
(b) Here,
L.H.S $=7n+5$
R.H.S $=19$
And $n=-2$ (given)
By putting, $n=-2$,
L.H.S $=7 (-2)+5=-9$
$≠$R.H.S
L.H.S $≠$ R.H.S, so $n=-2$ is not a solution of the equation.
(c) Here,
L.H.S$=7n+5$
R.H.S $=19$
And $n=2$ (given)
By putting, $n=2$,
L.H.S $=7 (2)+5=19$
$=$R.H.S
L.H.S$=$R.H.S, so $n=2$ is a solution of the equation.
(d) Here,
L.H.S$=4p-3$
R.H.S$=13$
And $p=1$ (given)
By putting, $p=1$,
L.H.S$=4 (1)-3=1$
$≠$R.H.S
L.H.S$≠$R.H.S, so $p=1$ is not a solution of the equation.
(e) Here,
L.H.S$=4p-3$
R.H.S$=13$
And $p=-4$ (given)
By putting, $p=-4$,
L.H.S$=4 (-4)-3=-19$
$≠$R.H.S
L.H.S$≠$R.H.S, so $p=-4$ is not a solution.
(f) Here,
L.H.S$=4p-3$
R.H.S$=13$
And $p=0$ (given)
By putting, $p=0$,
L.H.S$=4(0)-3=-3$
$≠$R.H.S
L.H.S$≠$R.H.S, so $p=0$ is not a solution of the equation.
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