Question

How do you subtract `\frac { 9y z } { 4z ^ { 2} } - \frac { 2y + 3} { 2z }`?

Answer 1

See a solution process below:

Explanation:

To subtract fractions the must be over a common denominator. We can put the fraction on the right over the same, common, denominator as the fraction on the left by multiplying by the appropriate form of `1`:

`(9yz)/(4z^2) - ((2z)/(2z) xx (2y + 3)/(2z)) =>`

`(9yz)/(4z^2) - (2z(2y + 3))/(2z xx 2z) =>`

`(9yz)/(4z^2) - (4zy + 6z)/(4z^2) =>`

`(9yz)/(4z^2) - (4yz + 6z)/(4z^2)`

We can now subtract the numerators over the common denominator:

`(9yz - (4yz + 6z))/(4z^2) =>`

`(9yz - 4yz - 6z)/(4z^2) =>`

`((9 - 4)yz - 6z)/(4z^2) =>`

`(5yz - 6z)/(4z^2) =>`

`((5y - 6)z)/(4z^2) =>`

`(5y - 6)/(4z)`