Parentheses in advanced Order of Operations: Exercises with fractions

Let's first address the fraction. We must begin by solving the exercise within the parentheses due to the rules of the order of arithmetic operations. Parentheses come before everything else:

$\frac{36-(20)}{8}-3\times2=$

Let's continue by simplifying the fraction, we subtract the exercise in the numerator and divide by 8:

$\frac{36-20}{8}=\frac{16}{8}=2$

We then arrange the exercise accordingly:

$2-3\times2=$

Finally we solve the multiplication exercise and then subtract:

$2-6=-4$

This simple example demonstrates the order of operations, which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations enclosed in parentheses precede all others,

Let's note that when a fraction (every fraction) is involved in a division operation, it means we can relate the numerator and the denominator to the fraction as whole numbers involved in multiplication, in other words, we can rewrite the given fraction and write it in the following form:

$\frac{(5-3)\cdot15+3}{5+6}-\frac{2\cdot8}{3+1}= \\ \downarrow\\ \big((5-3)\cdot15+3\big):(5+6)-(2\cdot8):(3+1)$We emphasize this by stating that fractions involved in the division and in their separate form , are actually found in multiplication,

Returning to the original fraction in the problem, in other words - in the given form, and simplifying, we separate the different fractions involved in the division operations and simplify them according to the order of operations mentioned, and in the given form:

$\frac{(5-3)\cdot15+3}{5+6}-\frac{2\cdot8}{3+1}= \\ \frac{2\cdot15+3}{11}-\frac{2\cdot8}{4}= \\ \frac{30+3}{11}-\frac{16}{4}=\\ \frac{33}{11}-\frac{16}{4}\\$$$In the first step, we simplified the fraction involved in the division from the left, in other words- we performed the multiplication operation in the division, in contrast, we performed the division operation involved in the fractions, in the next step we simplified the fraction involved in the division from the left and assumed that multiplication precedes division we started with the multiplication involved in this fraction and only then calculated the result of the division operation, in contrast, we performed the multiplication involved in the second division from the left,

We continue and simplify the fraction we received in the last step, this is done again according to the order of operations mentioned, in other words- we start with the division operation of the fractions (this is done by inverting the fractions) and in the next step calculate the result of the subtraction operation:

$\frac{33}{11}-\frac{16}{4}=\\ \frac{\not{33}}{\not{11}}-\frac{\not{16}}{\not{4}}=\\ 3-4=\\ -1$ We conclude the steps of simplifying the fraction, we found that:

$\frac{(5-3)\cdot15+3}{5+6}-\frac{2\cdot8}{3+1}= \\ \frac{2\cdot15+3}{11}-\frac{2\cdot8}{4}= \\ \frac{33}{11}-\frac{16}{4}=\\ 3-4=\\ -1$Therefore, the correct answer is answer d.

First, we solve the fraction expression.

Let's note that within the parentheses in the numerator there is a multiplication exercise, we will put it in parentheses to avoid confusion in the solution.

First we multiply and then we subtract:

$(5-(4\times3))=(5-12)=-7$

Now the exercise obtained in the numerator is:$-7:-7=1$

We arrange the exercise accordingly:

$\frac{1}{0}+3-2=$

Note that in the denominator of the fraction exercise, the number 0 appears.

Since according to the rules no number can be divided by 0, the exercise has no solution.

This simple rule is the foundation of the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations enclosed in parentheses precede all others,

First, we pay special attention to the given rule, the first break from the left is the number 0, remember that dividing the number 0 by any number always yields the result 0, (except dividing by the number 0 itself, which is generally forbidden, even though this simple rule that breaks in the given rule, in accordance with the order of operations mentioned, means that this break is worth nothing) therefore the value of this break is 0 and therefore we can simply omit it entirely (as if - the entire break) from the given rule, as this is a common practice that does not contribute anything in terms of numerical value,

$\frac{0}{5+4:2}-(5+3):4= \\ \downarrow\\ 0-(5+3):4= \\ -(5+3):4=$As usual we should not forget to keep the negative sign after the break, as this minus sign indicates multiplication by negative one,

We will continue and simplify this rule,

In accordance with the order of operations mentioned we will start with the multiplication and division operations, next we will calculate the result of the division operation:

$-(5+3):4=\\ -8:4=\\ -2$In the last step we did not forget that dividing a positive number by a negative number yields a negative result,

We received that the correct answer is answer c.

This simple equation emphasizes the order of operations, indicating that exponentiation precedes multiplication and division, which come before addition and subtraction, and that operations within parentheses take precedence over all others,

Let's start by discussing the given equation, the first step from the left is division by the number 1, remember that dividing any number by 1 always yields the same number, so we can simply disregard the division by 1 operation, which essentially leaves the equation (with the division by 1 operation, or without it) unchanged, namely:

$\frac{(7-8)+3}{2}:1+(5-4)= \\ \downarrow\\ \frac{(7-8)+3}{2}+(5-4)=$

Continuing with this equation,

Let's note that both the numerator and the denominator in a fraction (every fraction) are equations (in their entirety) between which a division operation is performed, namely- they can be treated as the numerator and the denominator in a fraction as equations that are closed, thus we can rewrite the given equation and write it in the following form:

$\frac{(7-8)+3}{2}+(5-4)= \\ \downarrow\\ \big((7-8)+3\big):2+(5-4)$We highlight this to emphasize that fractions which are the numerator and similarly in its denominator should be treated separately, indeed as if they are closed,

Returning to the original equation, namely - in the given form, and simplifying, we simplify the equation that is in the numerator of the fraction and, this is done in accordance with the order of operations mentioned above and in a systematic manner:

$\frac{(7-8)+3}{2}+(5-4)= \\ \frac{-1+3}{2}+1= \\ \frac{2}{2}+1$In the first stage, we simplified the equation that is in the numerator of the fraction, this in accordance with the order of operations mentioned above hence we started with the equation that is closed, and only then did we perform the multiplication operation that is in the numerator of the fraction, in contrast, we simplified the equation that is in closed parentheses,

Continuing we simplify the equation in accordance with the order of operations mentioned above,thus the division operation of the fraction (this is done mechanically), and continuing we perform the multiplication operation:

$\frac{\not{2}}{\not{2}}+1 =\\ 1+1 =\\ 2$In this case, the simplification process is very short, hence we won't elaborate,

Therefore, the correct answer is option B.

We begin by addressing the numerator of the fraction.

First we solve the division exercise and the exercise within the parentheses:

$400:(-5)=-80$

$(93-61)=32$

We obtain the following:

$\frac{-80-(-2\times32)}{4}=$

We then solve the parentheses in the numerator of the fraction:

$\frac{-80-(-64)}{4}=$

Let's remember that a negative times a negative equals a positive:

$\frac{-80+64}{4}=$

$\frac{-16}{4}=-4$

$$This simple example illustrates the order of operations, which states that multiplication and division take precedence over addition and subtraction, and that operations within parentheses come first,

Let's say we have a fraction and a whole number (every whole number) between which a division operation takes place, meaning - we can relate to the fraction and the whole number as fractions in their simplest form, through which a division operation occurs, thus we can write the given fraction in the following form:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \downarrow\\ \big((2^2-3)^{15}+4^2\big):(15+2)-(3^2-2^2):5$We emphasize this by stating that we should relate to the fractions that are in the numerator and those in the denominator separately, as if they exist in their simplest form,

Let's return to the original fraction in question, meaning - in its given form, and simplify it, simplifying separately the different fractions that are in the numerator and those in the denominator (if simplification is needed), this is done in accordance with the order of operations mentioned above and in a systematic way,

We start with the first numerator from the left in the given fraction, noting that in this case it changes the fraction in the denominators that are in multiplication, therefore, we start with this fraction, this in accordance with the aforementioned order of operations, noting further that in this fraction in the denominators (which are in multiplication of 15) there exists a multiplication, therefore, we start calculating its numerical value in multiplication and then perform the subtraction operation that is in the denominators:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{(4-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5} \\$ We continue with the fraction we received in the previous step and simplify the numerators and the denominators in the fraction, this is done in accordance with the order of operations mentioned above, therefore, we start calculating their numerical values in multiplication and then perform the division and subtraction operations that are in the numerators and in the denominators:

$\frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{1+16}{17}-\frac{9-4}{5}= \\ \frac{17}{17}-\frac{5}{5}\\$ We continue and simplify the fraction we received in the previous step, again, in accordance with the order of operations mentioned above, therefore, we perform the division operation of the denominators, this is done systematically, and then perform the subtraction operation:

$\frac{17}{17}-\frac{5}{5}=\\ \frac{\not{17}}{\not{17}}-\frac{\not{5}}{\not{5}}=\\ 1-1=\\ 0$

We conclude with this, the steps of simplifying the given fraction, we received that:

$\frac{(2^2-3)^{15}+4^2}{15+2}-\frac{3^2-2^2}{5}= \\ \frac{1^{15}+4^2}{15+2}-\frac{3^2-2^2}{5} =\\ \frac{1+16}{17}-\frac{9-4}{5}= \\ 0$Therefore, the correct answer is answer D.

This simple rule is the emphasis on the order of operations which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others,

Let's consider that the numerator is the whole and the denominator is the part which breaks (every break) into whole pieces (in their entirety) among which division operation is performed, meaning- we can relate the numerator and the denominator of the break as whole pieces in closures, thus we can express the given fraction and write it in the following form:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}= \\ \downarrow\\ \big((25-2-16)^2+3\big):(8+5):\sqrt{9}$We highlight this by noting that fractions in the numerator of the break and in its denominator are considered separately, as if they are in closures,

Let's return to the original fraction in question, meaning - in the given form, and simplify, separately, the fraction in the numerator of the break which causes it and the fraction in its denominator, this is done in accordance to the order of operations mentioned and in a systematic way,

Let's consider that in the numerator of the break the fraction we get changes into a fraction in closures which indicates strength, therefore we will start simplifying this fraction, given that this fraction includes only addition and subtraction operations, perform the operations in accordance to the natural order of operations, meaning- from left to right, simplifying the fraction in the numerator of the break:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}=\\ \frac{7^2+3}{13}:\sqrt{9}\\$We will continue and simplify the fraction we received in the previous step, this of course, in accordance to the natural order of operations (which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others), therefore we will start from calculating the numerical values of the exponents in strength (while we remember that in defining the root as strength, the root itself is strength for everything), and then perform the division operation which is in the numerator of the break:

$\frac{7^2+3}{13}:\sqrt{9}=\\ \frac{49+3}{13}:3=\\ \frac{52}{13}:3\\$We will continue and simplify the fraction we received in the previous step, starting with performing the division operation of the break, this is done by approximation, and then perform the remaining division operation:

$\frac{\not{52}}{\not{13}}:3=\\ 4:3=\\ \frac{4}{3}$In the previous step, given that the outcome of the division operation is different from a whole (greater than whole for the numerator, given that the divisor is greater than the dividend) we marked its outcome as a fraction in approximation (where the numerator is greater than the denominator),

We conclude the steps of simplifying the given fraction, we found that:

$\frac{(25-2-16)^2+3}{8+5}:\sqrt{9}=\\ \frac{7^2+3}{13}:\sqrt{9}=\\ \frac{52}{13}:3=\\ \frac{4}{3}$Therefore, the correct answer is answer b'.

Note:

Let's consider that in the group of the previous steps in solving the problem, we can start recording the break and the division operation that affects it even without the break, but with the help of the division operation:

$\frac{52}{13}:3\\ \downarrow\\ 52:13:3$And from here on we will start calculating the division operation in the break and only after that we performed the division in number 3, we emphasize that in general we simplify this fraction in accordance to the natural order of operations, meaning we perform the operations one after the other from left to right, and this means that there is no precedence of one division operation in the given fraction over the other except as defined by the natural order of operations, meaning- in calculating from left to right, (Let's consider additionally that defining the order of operations mentioned at the beginning of the solution, which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that operations within parentheses precede all others, does not define precedence even among multiplication and division, and therefore the judgment between these two operations, in different closures, is in a different order, it is in calculating from left to right).

Before solving the exercise, let's start by simplifying the power and the root:

$7^2=7\times7=49$

$\sqrt{36}=\sqrt{6^2}=6$

Now, we arrange the exercise accordingly:

$\frac{49-6:6}{3+3}\times(5+2)=$

According to the rules of the order of operations, parentheses are solved first:

$\frac{49-6:6}{3+3}\times(7)=$

Now we focus on the fraction, we start with the division exercise in the numerator, then we add and subtract as appropriate:

$\frac{49-1}{3+3}\times(7)=\frac{48}{6}\times(7)=$

We solve the exercise from left to right, first the division exercise and finally we multiply:

$8\times7=56$

This simple example demonstrates the order of operations, which states that multiplication and division take precedence over addition and subtraction, and that operations within parentheses take precedence over all others,

Let's consider the numerator and the denominator separately (each separately) which between them performs a division operation, meaning- we can relate to the numerator and the denominator separately as fractions in their own right, thus we can write the given fraction and write it in the following form:

$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \downarrow\\ \big((25-2-16)^2+3\big):\big((\sqrt{9}\cdot7):\sqrt{3} \big)$We emphasize this by noting that the fractions in the numerator and the denominator should be treated separately, indeed as if they are in their own parentheses,

Let's consider additionally that the division operation between the parentheses implies that we are dividing by the value of the denominator (meaning the denominator as a whole, it is the result of the division between the numerator and the denominator) and therefore in the given fraction to form a division that we marked for attention, the denominator being in parentheses is additionally important,

Returning to the original fraction problem, meaning - in the given form, and proceed simply,

We will start and simplify the fraction in the numerator (meaning- the numerator fraction that we are dividing by), this is done in accordance with the order of operations mentioned above, therefore we will start by calculating the numerical values of the fraction that takes precedence (this within the context of setting the root as a priority, the root being strong for everything) and then proceed with the multiplication which is in the numerator, in contrast let's consider within the parentheses that are left, those parts in the denominator are divided by the whole, they are fractions in the stronger parentheses, therefore we will also simplify this fraction, this in accordance with the order of operations mentioned above:

$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{3\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{21}{3}\\$We will continue and simplify the fraction we received in the previous step, continue simply the fraction found within the parentheses that are divided by the whole, they are the parentheses that are left, remembering that multiplication takes precedence over addition and subtraction, therefore we will start by calculating their numerical values that take precedence in those parentheses and then proceed with the subtraction operation, in the next step the division operation of the whole (and not the division operation in the whole) takes place, and in the last step the remaining division operation takes place:

$\big(5^2-2^2\big):\frac{21}{3}=\\ \big(25-4\big):\frac{21}{3}=\\ 21:\frac{21}{3}=\\ 21:\frac{\not{21}}{\not{3}}=\\ 21:7=\\ 3$Let's consider that we advanced the division operation of the whole over the division operation in the whole itself, and this means that the number 21 in the fraction we discussed is divided by its numerical values of the whole (in its entirety)- which is the result of the division of the numerator by the denominator, therefore it was necessary to complete first the calculation of the numerical values of the whole and only then to divide the number 21 in this value,

We will conclude thus with the steps of simplifying the given fraction:
$\big((3-2+4)^2-2^2\big):\frac{\sqrt{9}\cdot7}{3}= \\ \big(5^2-2^2\big):\frac{21}{3}=\\ 21:7=\\ 3$Therefore, the correct answer is answer d'.