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

Question

Julia Adams · Accepted Answer

See a solution process below:

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}#

Ethan Adkins · Answer

#k# can have any value.

We can cross-multiply because there is only one fraction on either side of the equal sign, but the outcome is intriguing: #4(5k-2) = -4(2-5k)# #20k-8 = -8 +20k# The equation's two sides are the same. If we carry on solving, we'll arrive at #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# Examine the initial equation once more: #(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# Currently, we have #(2-5k)/4 =(2-5k)/4# The two sides are identical and therefore cannot be solved for a unique value of #k#

Eva Adamson · Answer

undefined To solve the equation (5k - 2) / -4 = (2 - 5k) / 4, you can follow these steps:

1. Cross multiply to eliminate the denominators: 
   (-4) * (2 - 5k) = (4) * (5k - 2)
   -8 + 20k = 20k - 8

2. Rearrange the equation by moving the variables to one side and constants to the other:
   -8 + 20k - 20k + 8 = 0
   0 = 0

3. Since the equation simplifies to 0 = 0, this means that it is an identity, indicating that the equation holds true for all real values of k. Therefore, the solution is all real numbers.