Compare Complex Expressions: (36-6·2)÷((√16+2)·1/6) vs Multiple Operations

For solving a problem involving addition or subtraction each of the terms that come into play is treated separately,

This is done within the framework of the order of operations, which states that multiplication precedes addition and subtraction, and that the preceding operations are performed before division and subtraction, and that the preceding operations are performed before all others,

A. We will start with the term that appears on the left in the given problem:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)$In this term, a division operation is established between two terms that in their sum give a specific result, we take the term that in their sum gives the left, remembering that multiplication precedes subtraction, therefore the multiplication in these terms is performed first and then the subtraction operation, in contrast- the term that in their sum gives the right (the denominators) is considered as a multiplication of a number by the term that in their sum (the numerators) therefore the multiplication in this term, this is within that we remember that multiplication precedes division and that the root cause (the definition of the root as strong) is strong for everything, therefore we will calculate its numerical value and then perform the division operation that in this term:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ (36-12):\big((4+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big) \\$We will continue and note that in the term that was received in the last stage the multiplication operation is found in the terms and accordingly precedence is given to the division operation that precedes them,the multiplication is performed within that we remember that the multiplication in the break means the multiplication by the break, in continuation the division operation of the break is performed, this by summarizing:

$24:\frac{6\cdot1}{6}=\\ 24:\frac{\not{6}}{\not{6}}=\\ 24:1=\\ 24$In the last stage we performed the remaining division operation, this within that we remember that dividing any number by the number 1 will yield the number itself,

$$We will conclude the simplification of the term that appears on the left in the given problem, we will summarize the simplification stages:

We received that:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big) =\\ 24:\big(6\cdot\frac{1}{6}\big)= \\ 24$

B. We will continue and simplify the term that appears on the right in the given problem:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13}$In this part, in the first section, we simplify the term within the framework of the order of operations,

In this term, a division operation is established between two terms that in their sum, in this part, in the first section, we simplify that two terms in contrast, the term that in their sum gives the left is simplified within performing the division and subtraction operations, in contrast we simplify the term that in their sum gives the right, given that multiplication and division precede subtraction we start from the second simplification stage in these terms , and given that the order of operations does not define precedence to one of the multiplication or division operations is performed one after the other according to the order from left to right (which is the natural order of operations) , in continuation we will calculate the result of the subtraction operation that in these terms:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:(17-4)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\$We will continue and simplify the term that was received in the last stage, in this part, first the division and multiplication operations are performed one after the other from left to right:

$17:13\cdot\frac{1}{13} =\\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17\cdot1}{13\cdot13}=\\ \frac{17}{13^2}$In the first stage, given that the result of the division operation is a result that is not whole we mark it as a break (a break from above- given that the numerator is larger than the denominator) in continuation we perform the multiplication of the breaks within that we remember that when we multiply two breaks we multiply numerator by numerator and denominator by denominator and keep the essence of the original break.

We will conclude the simplification of the term that appears on the right in the given problem, we will summarize the simplification stages:

We received that:

$(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} =\\ 17:(17-20:5)\cdot\frac{1}{13} =\\ 17:13\cdot\frac{1}{13} \\ \frac{17}{13}\cdot\frac{1}{13} =\\ \frac{17}{13^2}$We will return to the original problem, and we will present the results of the simplification of the terms that were reported in A and B:

$(36-6\cdot2):\big((\sqrt{16}+2)\cdot\frac{1}{6}\big)\textcolor{red}{☐}(16-3+4):(17-5\cdot4:5)\cdot\frac{1}{13} \downarrow\\ 24 \textcolor{red}{☐}\frac{17}{13^2}$As a result, the correct answer here is answer B.