Question

How do you evaluate `(4y-2k)/(3h+k)` when `h=3` and `k=1`?

Answer 1

See the entire solution process below:

Explanation:

To evaluate this expression for `h = 3` and `k = 1` substitute `color(red)(3)` for each occurrence of `color(red)(h)` in the expression.

And, substitute `color(blue)(1)` for each occurrence of `color(blue)(k)` in the expression and perform the operations:

`(4y - 2color(blue)(k))/(3color(red)(h) + color(blue)(k))` becomes:

#(4y - (2 xx color(blue)(1)))/((3 xx color(red)(3)) + color(blue)(1)) -> (4y - 2)/(9 + 1) -> (4y -2)/10 -> (2(2y - 1))/10 -> (2y - 1)/5#

Or

`(2y)/5 - 1/5`