Complete the corresponding expression in the numerator
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Complete the corresponding expression in the numerator
Examine the following problem:
Remember the fraction reduction operation,
In order for the fraction on the left side to be reducible all the terms in its numerator must have a common factor. Additionally the number 5 (which is in the denominator of the fraction on the right side) already exists in the denominator of the fraction on the left side, hence we don't want to reduce it,
We'll simply add that we want to obtain the number 2 that appears in the numerator of the fraction on the right side.
Now, we want to reduce the term from the denominator of the fraction on the left side given that it does not appear in the denominator on the right side and simultaneously obtain the term in the denominator of the fraction on the right side. Note that this term does not appear in the expression in the denominator of the fraction on the left side, therefore we will choose the expression:
Let's verify that from this choice we will obtain the expression on the right side. We will use the fact that multiplying a number by a fraction is actually multiplying the number by the fraction's numerator (in the first stage), and in fraction multiplication (in the second stage) in order to simplify the fraction resulting from this choice, then we'll reduce the simplified fraction:
Therefore this choice is indeed correct.
In other words - the correct answer is answer D.
Determine if the simplification shown below is correct:
\( \frac{7}{7\cdot8}=8 \)
The denominators are different! The left side has 5a while the right has 5b. You need to account for this difference by including variables in the numerator to make the fractions truly equal.
Look at what's different between the denominators. Since 5a needs to become 5b, you need 'a' in the numerator and 'b' in the denominator to cancel and transform properly.
In this case, both fractions already have 5 in the denominator, but they can't be simplified further because of the different variables. Cross multiplication is your best approach here.
Remember: you want to cancel out what's different. Since the left has 'a' and the right has 'b', put 'a' in the numerator (to cancel with denominator's 'a') and 'b' in the denominator.
Substitute your answer and simplify! after canceling the 'a' terms. It matches the right side! ✓
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