# 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$ $(p=0)$

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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.

Updated on 10-Oct-2022 13:33:38