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

Jul 30, 2017

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:

$\frac{- 1}{-} 1 \times \frac{5 k - 2}{-} 4 = \frac{2 - 5 k}{4}$

$\frac{- 1 \left(5 k - 2\right)}{- 1 \times - 4} = \frac{2 - 5 k}{4}$

$\frac{- 5 k + 2}{4} = \frac{2 - 5 k}{4}$

$\frac{2 - 5 k}{4} = \frac{2 - 5 k}{4}$

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:

$k = \left\{\mathbb{R}\right\}$

Jul 30, 2017

$k$ 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:

$4 \left(5 k - 2\right) = - 4 \left(2 - 5 k\right)$

$20 k - 8 = - 8 + 20 k$

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:

$\frac{5 k - 2}{-} 4$ can also be written as $\frac{- \left(5 k - 2\right)}{4}$, which leads to:

$\frac{- 5 k + 2}{4}$ which is the same as $\frac{2 - 5 k}{4}$

Now we have

$\frac{2 - 5 k}{4} = \frac{2 - 5 k}{4}$

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