a(3-x^2)+b(4x^2-7)=7x^2-16,,,,a=?

expand LHS,

a(3-x^2)+b(4x^2-7)
= 3a - ax^2 + 4bx^2 - 7b
= x^2(4b-a) + 3a - 7b

We can see that
4b-a = 7
and
3a - 7b = -16

Now we may solve the equation with two unknown
4b-a = 7 -(1)
3a - 7b = -16 -(2)

From(1).
a= 4b-7 -(3)

By putting (3) into (2) ,we get
3(4b-7) - 7b = -16
12b - 21 - 7b = -16
5b = 5
b = 1

Sub b=1 into (3)
a = 4(1) - 7
a = 4-7
a = -3

therefore a = -3