How do you combine #c/(7-c)+(2c-7)/(c-7)#?

Question

Genesis Allen · Accepted Answer

#1#

There is a very useful way of changing the signs of an expression around.

"Multiplying by a negative, changes the signs"

# - (x-y) = (-x+y)= (y-x)#

OR: #(2p - q) = -(q-2p)" "#Notice the "switch-rounds"

We can apply this in the second fraction to make the denominators the same:

#c/((7-c))color(red)(-)(2c-7)/color(red)((7-c)) " "rArr " same denominators"#

=#(c color(blue)(-)(2c-7))/(7-c) " notice the multiplying by a negative"#

#(c -2c+7)/(7-c)#

=#color(teal)(-c+7)/(7-c) = color(teal)(7-c)/(7-c) " " color(teal)(" by the commutative law")#

#(cancel(7-c))/(cancel(7-c)) = 1#

Eva Abramson · Answer

undefined To combine the given expression, c/(7-c) + (2c-7)/(c-7), we need to find a common denominator and then add the fractions together. The common denominator is (7-c)(c-7).

Multiplying the first fraction by (c-7)/(c-7) and the second fraction by (7-c)/(7-c), we get:

(c(c-7))/((7-c)(c-7)) + ((2c-7)(7-c))/((7-c)(c-7))

Expanding and simplifying the numerators, we have:

(c^2 - 7c + 14c - 7)/((7-c)(c-7))

Combining like terms in the numerator, we get:

(c^2 + 7c - 7)/((7-c)(c-7))

Therefore, the combined expression is (c^2 + 7c - 7)/((7-c)(c-7)).