How do you simplify #(x+5)/(x-3)*(7x^2 - 21x) / (7x)#?

Answer 1

#x+5#

Completely factor out all expressions. You can factor out 7x from #(7x^2-21x)#.
#(x+5)/(x-3)*(7x^2-21x)/(7x)#
#=(x+5)/(x-3)*[(7x)(x-3)]/(7x)#
#=[(x+5)(7x)(x-3)]/[(7x)(x-3)]#
After you have factored everything out, cancel factors that appear in both the numerator and the denominator. In this problem, #7x# and #(x-3)# appear in both the numerator and denominator, so you can cancel them.
#[(x+5)cancel((7x))cancel((x-3))]/[cancel((7x))cancel((x-3))]#
#=(x+5)/1#
#=x+5#
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Answer 2

To simplify the expression (x+5)/(x-3)*(7x^2 - 21x) / (7x), we can cancel out common factors.

First, we can cancel out the common factor of (x) in the numerator and denominator of the first fraction, resulting in (x+5)/(x-3).

Next, we can cancel out the common factor of (7x) in the numerator and denominator of the second fraction, resulting in (7x^2 - 21x) / 1.

Combining these simplified fractions, we get (x+5)/(x-3) * (7x^2 - 21x).

Therefore, the simplified expression is (x+5)(7x^2 - 21x) / (x-3).

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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.

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.

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.

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.

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