You want the exponent of x in the first term of ...
[tex]\dfrac{4xy^3+8x^2y^2}{2xy^2}=\frac{4}{2}x^{(1-1)}y^{(3-2)}+\frac{8}{2}x^{(2-1)}y^{(2-2)}\\\\=2x^0y^1+4x^1y^0[/tex]
The exponent a is 0.
The value of a is: 0
We are given a expression by:
Quantity 4 times x times y cubed plus 8 times x squared times y to the fifth power all over 2 times x times y squared.
which is mathematically written as:
[tex]=\dfrac{4xy^3+8x^2y^5}{2xy^2}[/tex]
Now, this expression could also be written as:
[tex]=\dfrac{4xy^3}{2xy^2}+\dfrac{8x^2y^5}{2xy^2}[/tex]
Since,
[tex]\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}[/tex] )
Now, on further simplifying we have:
[tex]=2x^{1-1}y^{3-2}+4x^{2-1}y^{5-2}[/tex]
since,
[tex]\dfrac{x^m}{x^n}=x^{m-n}[/tex]
Hence, we have :
[tex]=2x^0y^1+4x^1y^3[/tex]
Hence, on comparing it with:
[tex]2x^ay^b+4x^cy^d[/tex]
we have:
[tex]a=0,\ b=1\ ,\ c=1\ and\ d=3[/tex]
