How do you solve #1/(c-2) + 1/c = 2/(c(c-2))#?

Question

Nathan Axel · Accepted Answer

I found no real value of #c#.

I would choose #c(c-2)# as common denominator and write: #(c+c-2)/(c(c-2))=2/(c(c-2))# where on the left I arranged the numerators to the new common denominator. So you can now simplify the two denominators: #(c+c-2)/cancel((c(c-2)))=2/cancel((c(c-2)))# #2c-2=2# #2c=4# #c=4/2=2# BUT if you use #c=2# into your original equation you get a division by zero that cannot be evaluated.

Abigail Allen · Answer

undefined To solve the equation 1/(c-2) + 1/c = 2/(c(c-2)), we can start by finding a common denominator for the fractions on the left side. The common denominator is c(c-2). Multiplying the first fraction by c/c and the second fraction by (c-2)/(c-2), we get (c + (c-2))/(c(c-2)) = 2/(c(c-2)). Simplifying the numerator, we have (2c - 2)/(c(c-2)) = 2/(c(c-2)). Now, we can cross-multiply to eliminate the denominators. This gives us 2(c(c-2)) - 2(c) = 2(c-2). Expanding and simplifying, we have 2c^2 - 4c - 2c + 4 = 2c - 4. Combining like terms, we get 2c^2 - 8c + 4 = 2c - 4. Rearranging the equation, we have 2c^2 - 10c + 8 = 0. Factoring the quadratic equation, we have (2c - 4)(c - 2) = 0. Setting each factor equal to zero, we find two possible solutions: c = 2 and c = 2. Therefore, the solution to the equation is c = 2.