Addition and Subtraction of Algebraic Fractions

Let's examine the problem:

$\frac{12ab}{?}=1$Now let's think logically, and remember the known fact that dividing any number by itself always yields the result 1,

Therefore, in order to get the result 1 from dividing two numbers, the only way is to divide the number by itself, meaning-

The missing expression in the denominator of the fraction on the left side is the complete expression that appears in the numerator of the same fraction:

$12ab$.

Therefore- the correct answer is answer D.

Let's examine the problem, first we'll write the expression on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

$\frac{16ab}{?}=2b \\ \downarrow\\ \frac{16ab}{?}=\frac{2b}{1}$

Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all the terms in its denominator to have a common factor, additionally, we want to reduce the number 16 to get the number 2, and reduce the term $a$ from the fraction's denominator since in the expression on the right side it does not appear, therefore we will choose the expression:

$8a$

because:

$16=8\cdot 2$

Let's verify that with this choice we indeed get the expression on the right side:

$\frac{16ab}{?}=\frac{2b}{1} \\ \downarrow\\ \frac{\not{16}\not{a}b}{\textcolor{red}{\not{8}\not{a}}}\stackrel{?}{= }\frac{2b}{1} \\ \downarrow\\ \boxed{\frac{2b}{1}\stackrel{!}{= }\frac{2b}{1} }$

therefore this choice is indeed correct.

In other words - the correct answer is answer B.

Using the formula:

$\frac{x}{y}=\frac{z}{w}\xrightarrow{}x\cdot y=z\cdot y$

We first convert the 8 into a fraction, and multiply

$\frac{16ab}{?}=\frac{8}{1}$

$16ab\times1=8a$

$16ab=8a$

We then divide both sides by 8a:

$\frac{16ab}{8a}=\frac{8a}{8a}$

$2b$

Let's examine the problem, first we'll write down the expression on the right side as a fraction (using the fact that dividing a number by 1 doesn't change its value):

$\frac{19ab}{?}=a \\ \downarrow\\ \frac{19ab}{?}=\frac{a}{1}$
Now let's think logically, and remember the fraction reduction operation,

For the fraction on the left side to be reducible, we want all the terms in its denominator to have a common factor, additionally, we want to reduce the number 19 to get the number 1 and also reduce the term $b$ from the fraction's numerator since in the expression on the right side it doesn't appear, therefore we'll choose the expression:

$19b$

Let's verify that with this choice we indeed get the expression on the right side:

$\frac{19ab}{?}=\frac{a}{1} \\ \downarrow\\ \frac{\not{19}a\not{b}}{\textcolor{red}{\not{19}\not{b}}}\stackrel{?}{= }\frac{a}{1} \\ \downarrow\\ \boxed{\frac{a}{1}\stackrel{!}{= }\frac{a}{1} }$

therefore this choice is indeed correct.

In other words - the correct answer is answer D.

Let's examine the problem, first we'll write the expression on the right side as a fraction (using the fact that dividing a number by 1 does not change its value):

$\frac{27ab}{\text{?}}=3ab\\ \downarrow\\ \frac{27ab}{\text{?}}=\frac{3ab}{1}$

Now let's think logically, and remember the fraction reduction operation,

Note that both in the numerator of the expression on the right side and in the numerator of the expression on the left side exists the expression $ab$, therefore in the expression we are looking for there are no variables (since we are not interested in reducing them from the expression in the numerator on the left side),

Next, we ask which number was chosen to put in the denominator of the expression on the left side so that its reduction with the number 27 yields the number 3, the answer to this is of course - the number 9,

Because:

$27=9\cdot 3$

Let's verify that this choice indeed gives us the expression on the right side:

$\frac{27ab}{\text{?}}=\frac{3ab}{1} \\ \downarrow\\ \frac{\not{27}ab}{\textcolor{red}{\not{9}}}\stackrel{?}{= }\frac{3ab}{1} \\ \downarrow\\ \boxed{\frac{3ab}{1}\stackrel{!}{= }\frac{3ab}{1} }$

Therefore this choice is indeed correct.

In other words - the correct answer is answer A.