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

2 Answers
Jul 30, 2017

Answer:

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}#

Jul 30, 2017

Answer:

#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#