Question

How do you solve `(5k-2)/-4=(2-5k)/4`?

Answer 1

See a solution process below:

Explanation:

First, both fractions need to be over common denominators. We can multiply the fraction on the left by the appropriate form of `1` giving:

`(-1)/-1 xx (5k - 2)/-4 = (2 - 5k)/4`

`(-1(5k - 2))/(-1 xx -4) = (2 - 5k)/4`

`(-5k + 2)/4 = (2 - 5k)/4`

`(2 - 5k)/4 = (2 - 5k)/4`

Because both sides of the equation are exactly the same `k` can be any value. Therefore there are infinite solutions. Or, `k` is the set of all Real numbers:

`k = {RR}`

Answer 2

`k` can have any value.

Explanation:

There is only one fraction on each side of the equal sign, so this means we can cross-multiply. However this gives an interesting result:

`4(5k-2) = -4(2-5k)`

`20k-8 = -8 +20k`

The two sides of the equation are identical.

If we continue to solve we will end up with

`0=0`

This is a true statement but there is no `k` to solve for.
This is the indication that it is an identity - an equation which will be true for any value of `k`

Look at the original equation again:

`(5k-2)/-4` can also be written as `(-(5k-2))/4`, which leads to:

`(-5k+2)/4` which is the same as `(2-5k)/4`

Now we have

`(2-5k)/4 =(2-5k)/4`

The two sides are identical and therefore cannot be solved for a unique value of `k`