Add or subtract as indicated. Express your result in simplest form.

(5)/(4cd) Plus (11)/(4cd)

My answer: (4)/(cd)
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Add or subtract as indicated. Express your result in simplest form.

(x^2)/(x-3) Subtracted by (9)/(x-3)

MY answer is: x + 3
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Add or subtract as indicated. Express your result in simplest form.

(5a-12)/(a^2-8a+15) subtracted by (3a-2)/(a^2-8a+15)

My answer is: (2)/(a-3)

Good work! They are all correct.

Thanks for asking.

Looks OK!

Express the sum in the simplest form.

Great job! Your answers are correct for all three questions. Let's go through each question and see how you arrived at the correct solutions:

1) (5)/(4cd) Plus (11)/(4cd)

To add these fractions, you need to have a common denominator. In this case, the denominators are both 4cd, so you can simply add the numerators and keep the denominator the same:

(5 + 11)/(4cd) = 16/(4cd)

You can simplify this fraction by dividing both the numerator and denominator by 4:

16/(4cd) = 4/(cd)

So your answer is (4)/(cd).

2) (x^2)/(x-3) Subtracted by (9)/(x-3)

To subtract fractions, you also need a common denominator. In this case, the denominators are both x-3, so you can simply subtract the numerators while keeping the denominator the same:

(x^2 - 9)/(x-3)

The numerator can be simplified by recognizing that it is a difference of squares:

(x + 3)(x - 3)/(x-3)

Notice that the x-3 terms in the numerator and denominator cancel out:

(x + 3)

So your answer is x + 3.

3) (5a-12)/(a^2-8a+15) subtracted by (3a-2)/(a^2-8a+15)

Similar to the previous example, the denominators are the same, so you can subtract the numerators while keeping the denominator the same:

(5a - 12 - (3a - 2))/(a^2-8a+15)

Simplify the numerator by removing the parentheses:

5a - 12 - 3a + 2 = 2a - 10

The denominator remains the same: a^2-8a+15. Thus, the final expression is:

(2a - 10)/(a^2-8a+15)

This fraction cannot be simplified any further, so your answer is (2)/(a-3).

Well done on getting all the answers correct! If you have any more questions, feel free to ask.