Subtracting with unlike denominators

(9b)/(18b^3)-(4b)/(16b^2)

i know the answer is (-b+2)/4b^2
but i dont know how to get it

For each fractional term, we can reduce the common factors [First fraction - 2b;Second fraction - 4b] to get:

1/(2b²) - 1/(4b)

Then, since the LCD is 4b², multiply the top and bottom of the first fraction by 2 and the top and bottom of the second fraction by b to get:

1/(2b²) * 2/2 - 1/(4b) * b/b
= 2/(4b²) - b/(4b²)

Finally, combine both fractional terms to get:

(2 - b)/4b² or (-b + 2)/4b²

I hope this helps!

General rule:

(a / b) - (c / d) = (ad - bc) / (bd)

In other words, cross-multiply and keep the sign and then multiply denominators.

(a / b) + (c / d) = (ad + bc) / (bd)

(9b)/(18b^3)-(4b)/(16b^2)

= 1/2b^2 - 1/4b //// simplify each fraction

= 1/2b^2 * 2/2 - 1/4b * b/b

= 2/4b^2 - b/4b^2

= (2-b) / 4b^2

(9b)/(18b^3)-(4b)/(16b^2)=(72b-36b²)/144…‡
=36b(2-b)/144b³
=(2-b)/144b²