please help

Multiply the first equation by a^2:

$a^4+4a^3+21a^2=0$

We still have to clean up the a^3 term, so we multiply the first equation by -5a:

$-5a^3-20a^2-105a=0$

Add the two equations to get:

$a^4-a^3+a^2-105a=0$

Rearrange:

$a^4-a^3+a^2=105a$

The value of n is 105.

Hope this helps!

 simplfy the fist equation to get a^2+4a=-21 and it can also be (a+4)a=-21 so we get a=-3 since 3+4=7 and -3*7=-21 then we plug a=-3 into the second equation to get 81-(-27)+9=3n which is 117=3n so n=39

@Guest

There is a value of a

Your answer is incorrect Jimkey17 AND there is no real value of a.

I certainly do give you credit for having a go though