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How do you simplify #(4z)/(z - 4) + (z + 4)/(z + 1)#?

Answer 1

Savannah Anderson

#=(5z^2+4z-16)/((z-4)(z+1))#

#(4z)/(z-4) + (z+4)/(z+1)" "larr# find the LCD

Make equivalent fractions:

#=(4z)/(z-4) xx color(red)((z+1)/(z+1)) + (z+4)/(z+1) xx color(red)((z-4)/(z-4))#

#=(4z(z+1) +(z+4)(z-4))/((z-4)(z+1))#

#=(4z^2+4z+z^2-16)/((z-4)(z+1))#

#=(5z^2+4z-16)/((z-4)(z+1))#

NOTE:

#color(red)((z-4)/(z-4)) and color(red)((z+1)/(z+1))# are both equal to #1#.

Multiplying by #1# does not change the value.

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Answer 2

Eva Abramson

To simplify the expression (4z)/(z - 4) + (z + 4)/(z + 1), we need to find a common denominator and combine the fractions. The common denominator is (z - 4)(z + 1).

Multiplying the first fraction by (z + 1)/(z + 1) and the second fraction by (z - 4)/(z - 4), we get:

[(4z)(z + 1) + (z + 4)(z - 4)] / [(z - 4)(z + 1)]

Expanding and simplifying the numerator, we have:

(4z^2 + 4z + z^2 - 16) / [(z - 4)(z + 1)]

Combining like terms in the numerator, we get:

(5z^2 + 4z - 16) / [(z - 4)(z + 1)]

Therefore, the simplified expression is (5z^2 + 4z - 16) / [(z - 4)(z + 1)].

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.