Special Cases (0 and 1, Inverse, Fraction Line): Exercises with fractions

Let's begin by multiplying the numerator:

$25+25=50$

We obtain the following fraction:

$\frac{50}{10}$

Finally let's reduce the numerator and denominator by 10 and we are left with the following result:

$\frac{5}{1}=5$

Let's focus on the fraction of the fraction.
According to the order of operations rules, we'll solve from left to right, since it only contains addition and subtraction operations:

$5+3=8$

$8-2=6$

Now we'll get the fraction:

$\frac{6}{3}$

We'll reduce the numerator and denominator by 3 and get:

$\frac{2}{1}=2$

First, let's solve the addition problem that appears in the denominator:

$18+36=54$

Note that in the resulting fraction (18:54), we can reduce both the numerator and denominator by 18.

$18:18=1$

$54:18=3$

Therefore, the result we get is: $\frac{1}{3}$

Let's rewrite the exercise in a more familiar equation form:

$9:(42+7)=$

First, let's solve the part in parentheses:

$42+7=49$

Now we should obtain the exercise:

$9:49$

Finally, let's write the exercise as a fraction:

$\frac{9}{49}$

Let's first solve the addition exercise in the numerator:

$100+1=101$

Note that the result we will get if we divide 25 by 100 will have a remainder.

Let's now check what the closest number to 101 is, by which we can then divide 25 without a remainder:

$\frac{100}{25}=4$

Finally we add the remainder to get the answer:

$4\frac{1}{25}$

According to the order of operations rules, we will solve the fraction numerator.

Since the exercise only has a subtraction operation, we will solve it from left to right:

$90-15=75$

$75-3=72$

Now we will get the fraction:

$\frac{72}{8}$

We will simplify the numerator and denominator by 8 and get:

$\frac{9}{1}=9$

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:

$2-6=-4$

According to the order of arithmetic operations, we first place the multiplication exercise inside parentheses:

$\frac{25+3-2}{13}+(5\cdot4)=$

We then solve the multiplication exercise:

$5\times4=20$

We obtain the exercise:

$\frac{25+3-2}{13}+20=$

Next we solve the exercise in the numerator of the fraction:

$25+3-2=28-2=26$

We obtain the fraction:

$\frac{26}{13}=2$

Lastly we obtain the following exercise:

$2+20=22$

First, we will solve the multiplication exercise that was broken:

$4\times5=20$

Now, we will solve the exercise that was broken:

$14+8-2=22-2=20$

We receive the solution:

$\frac{20}{20}=1$

Now, we receive the exercise:

$1\times2+3=$

According to the order of operations, we will first solve the multiplication exercise and then proceed:

$1\times2=2$

$2+3=5$

According to the order of arithmetic operations, first we place the multiplication exercises within parentheses:

$\frac{27-(5\cdot3)}{(6\cdot2)}+\frac{(15\cdot4)}{3}=$

We then solve the exercises within parentheses:

$5\times3=15$

$6\times2=12$

$15\times4=60$

Now we obtain the exercise:

$\frac{27-15}{12}+\frac{60}{3}=$

We solve the numerator of the fraction:

$27-15=12$

We obtain:

$\frac{12}{12}+\frac{60}{3}=$

We solve the fractions:

$\frac{12}{12}=1$

$60:3=20$

Finally we obtain the exercise:

$1+20=21$

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.

The first step is to resolve the problem in the numerator of the fraction:

$0.5+2=2.5$

Resulting in the following:

$\frac{2.5}{5}$

We then proceed to reduce the numerator and denominator by 2.5 in order to obtain the below fraction:

$\frac{1}{2}$

Let's write the exercise in a different form:

$(20-5):(7+3)=$

According to the order of operations in arithmetic, we'll first solve the expressions in parentheses:

$20-5=15$

$7+3=10$

Now we'll get the exercise:

$15:10=$

Let's write the exercise as a fraction:

$\frac{15}{10}$

Let's reduce the numerator and denominator by 10:

$\frac{1.5}{1}=1.5=1\frac{1}{2}$

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.

First, let's solve the exercise that was broken down:

$7+8-3=15-3=12$

We receive the breakdown:

$\frac{12}{2}=6$

Now, let's solve the exercise:

$6:3+4=$

According to the order of operations, we'll first solve the division exercise and then proceed:

$6:3=2$

$2+4=6$

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.

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.

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

Initially, we pay very close attention to the given rule, given that in the rule the existence of a number that is multiplied by 0, since multiplying any number by 0 always yields the result of 0, we disregard this multiplication, of course, meaning that it does not contribute anything, in contrast we focus on the second break from the left (as all the break from the right) and simplify the rule that is in it, this in accordance to the order of operations mentioned above, therefore we start with the multiplication that is in the break and continue to perform the division operation that is in this break:

$\frac{44-3\cdot0}{11}:4-\frac{3\cdot4+5}{17}= \\ \frac{44}{11}:4-\frac{12+5}{17}= \\ \frac{44}{11}:4-\frac{17}{17} \\$

We continue and simplify the rule we received in the last step, again, of course in accordance to the order of operations mentioned above, therefore we start with performing the division operation of the breaks, this is done sequentially, and continue to perform the division operation that is across the first, and finally perform the subtraction operation:

$\frac{\not{44}}{\not{11}}:4-\frac{\not{17}}{\not{17}}= \\ 4:4-1=\\ 1-1=\\ 0$ Simply put, this rule is short, therefore there is no need to elaborate,

We received whether the correct answer is answer c'.

Note:

Keep in mind that in the group of the last breaks in the solution to the problem, we can start recording the break and the division operation that is easy on it even without the break, but with the help of the division operation:

$\frac{44}{11}:4\\ \downarrow\\ 44:11:4$And in continuation we will calculate the division operation in the break and only after that we performed the division by the number 4, we emphasize that in total we simplified this rule in accordance to the natural order of operations, meaning we performed the operations one after the other from left to right, and this means that there is no precedence of one division operation in this rule over the other defined by the natural order of operations, meaning- in calculation from left to right, (Keep in mind in addition that the order of operations mentioned at the beginning, which states that exponentiation precedes multiplication and division, which precede addition and subtraction, and that parentheses precede all, it does not define precedence also between the multiplication and division operations, and therefore the rule between these two operations, in the absence of parentheses that constitute a different order, is in calculation from left to right).

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).