prove to \( 48 \div 2(9+3) \) is 2

Topic Review

 The question \( 48 \div 2(9+3) \) has been a war on the internet for years now.
I will like you to follow my prove with a pen and note in your hand

Maths Is The Study Of Balance

You will probably think what does this guy mean my maths is the study of balance. Let take a look at weighing scale

Weighing scale

Weighing scale is a good example of left-hand side {LHS} and right-hand side {RHS} in our day to day mathematics. it also explains the value at the LHS is equal to the value at the RHS to make it stay balanced.

we have

• \( x^a \times x^b =  x^{ab} \) example \( 2^2 \times 2^3 = 2^{2+3} = 2^5 = 32 \)
• to bring our answer back to the first step we need to use this \( x^{ab} = x^a \times x^b \) example \( 32 = 2^5 = 2^{2+3} = 2^2 \times 2^3 \) which tells you ether you use the reverse way \( x^a \times x^b =  x^{ab} \) or \( x^{ab} = x^a \times x^b \) you are still use the rule of indices only in the reverse way

Word Example

I will take you back to your days in school

1.  From the days of addition and subtraction we have

     2 + 3 = ?    we all know the answer = 5

     3 - 1 = ?     ans = 2
note: 2 + 3 = 5  LHS = RHS which me for examine to set this as question they just need to remove 5 so you can work for it.

2.  To the days of Addition and Subtraction LHS = RHS

     2 + [  ]  = 5      [  ] = box

     2 + [3] = 5

here we have the LHS and RHS to enable you get the value to be input in the box

formed of question with 5

    5 = 2 + 3   where they replace 3 with box

3.  In the days of insert the correct number in the box LHS = RHS

     2 * [  ] = 6     or     2[  ] = 6
solution
     2 * [3] = 6     or     2[3] = 6
Form question with 6
     6 = 2 * 3      or replace the box      {where they replace 3 with box}

4.   To the days of unknown and variable LHS = RHS

     6 = 3y
solution

     \begin{align} \require{cancel} 6 & = 3y  \\ \frac{ \cancel 6^2}{ \cancel 3_1 } & = \frac{ ^1 \cancel 3y}{ \cancel 3_1 } \text{divide both sides by 3} \\ 2 & = y \end{align} or \begin{align} 6 & = 3y \\  \frac{ \cancel 3 \times 2 }{ \cancel 3} & = \frac{ \cancel 3y }{3} \text{divide both sides by 3} \\ 2 & = y \end{align}

Balance and check {scale}

Check and balance start here in maths i can also here is where Equation start UNKNOWN .  how to scale here is by replace our value of y and see if it will be equal to the RHS

\begin{align}& & 6 = 3y& \\ 6 & = 3(2) & \text{or} \qquad & 6 = 3 \times 2 \\ 6 & = 6 & \text{or} \qquad & 6 = 6 \end{align} 

form question with 6
     6 = 3 * 2     where they replace 2 with y

5.   To the days of adding & subtraction from LHS to RHS 

    2x + 2 = 8
solution
    2x + 2 = 8   subtract -2 from both side

    2x + 2 - 2 = 8 - 2

    2x = 6

   \( \require{cancel}  \frac{ ^1 \cancel 2x}{ \cancel 2_1 } = \frac{ \cancel 6^3}{ \cancel 2_1 } \)

    x = 3

Balance and check {scale}

\begin{align} 2x + 2 =8 \quad \text{or} \quad & 2x + 2 =8 \\ 2(3) + 2 = 8 \quad \text{or} \quad & 2 \times 3 + 2 = 8 \\ 6+2=8 \quad \text{or} \quad & 6+2=8 \\ 8=8 \quad \text{or} \quad & 8=8 \end{align}

form question with 3
\begin{align} 8 & = 6 + 2 \\ 8 & = (3 \times 2) + 2 \\ & \text{replace 3 with x} \\ 8 & = (x \times 2) + 2 \\ 8 & = 2x +2 \end{align}

6.  To the days of Multiplication and division from LHS to RHS

$$ 2x \div 2 = 3 $$
solution 
\begin{align} \require{cancel} 2x \div 2 & = 3 \\ \frac{ ^1 \cancel 2x }{ \cancel 2_1 } & = \frac{ ^3 \cancel 6 }{ \cancel 2_1 } \\ x & = 3 \end{align}

Balance and check {scale}

Am not going to balance this because the World don't know the relationship between muiltipication, division and bracket.

form question with 3
\begin{align} \require{cancel} 3 & = 6 \div 2 \\ 3 & = 2(3) \div 2 \\ 3 & = 2x \div 2 \end{align}
 same to
\begin{align} \require{cancel} 2 \div 2x & = 3 \\ \frac{ ^1 \cancel 2}{  \cancel 2_1x } & = 3 \\ \frac{1}{x} & = 3  \quad \text{ multiply both side by x }  \\ \frac{1}{x}  \times x & = 3 \times x  \quad \text{ divide both side by x } \\ \frac{1}{3} & = \frac{3x}{3} \quad \text{ rearrange }  \\ x & = \frac{1}{3} \end{align}

Balance and check {scale}

Am not going to balnce this because the World don't know the relationship between muiltipication, division and bracket.

form question with 3
\begin{align} 3 & = 2  \div 6 \\ & \text{replace 3 with x} \\ 3 & = 2 \div 2 \times 3 \end{align}Note:- you need replace 3 with x to form this question \begin{align} 3 & = 2 \div 2 \times x \\ 3 & = 2 \div 2x \quad \text{ rearrange } \\ 2 & \div 2x = 3 \end{align}

prove that \( 48 \div 2(9+3) \) is \( \neq \) 288 it just a misunderstanding of of peranthsis 

Let look at this question \begin{align} \text{THE SAME OR DIFFERENT} \\ \text{Discover by computing!} \\ 44 \div 2(9 +3) = \text{?} \\ 44 \div (2(9 +3)) = \text{?} \end{align} check HERE to see peoples comment on FACEBOOK

Working

1. 48 \( \div \) 2(9+3)
    = 48 \( \div \) 2 × 12
    = 24 × 12
    = 288
where some people say the working is this
       48 \( \div \) 2(9+3)
    =48 \( \div \) 2(12)
    =48 \( \div \) 24
    =2

Also they all agreed this is correct
2.   48 \( \div \) (2(9+3))
  = 48 \( \div \) (2(12))
  = 48 \( \div \) 24
  = 2

  Proves 

\( 48 \div 2(9+3) \) is \( \neq \) 288
let view this question with my number 1 example where I say
[ Note: 2 + 3 = 5,  LHS = RHS which mean the examine will set this question by remove 5 so you can work for it. ]
and you can see this question have 2 answer \( 48 \div 2(9+3) \) = ?. the first working solution said the answer is 288 now let put 288 to the LHS of the question to see if this question is balance
[ Note: in example 5 and 6 I replace 3 with x.  the reason for doing this is make you work for the x which mean x must be equal to the value of number remove ]
set of people that interpreted this \begin{align} 48 \div 2(9+3) & = 48 \div 2 × 12 \\ & = \frac{48}{2} × 12 \end{align} let check if \( 48 \div 2(9+3) = 288 \) by applying the logic of example 1, 5 and 6
I will like to replace 3 with x
\( 48 \div 2( 9 + x) = 288 \)

Solution

\begin{align} \require{cancel} 48 \div 2( 9 + x) & = 288 \\  44 \div 18 + 2x & = 288 \\ \frac{44}{18} + 2x & = 288 \\ \frac{ ^{22} \cancel{44}}{ \cancel{18}_9} + 2x & = 228 \\ \frac{22}{9} + 2x & = 288 \\ \text{LCM} \\ \frac{ 22 + 18x } {9} & = 288 \\ \text{ cross multiply} \\ 22 + 18x & =288 \times 9 \\ 22 + 18x & = 2592 \\ 18x & = 2592 - 22 \\ 18x & = 2570 \\ x & = \frac{2570}{18} \simeq 142.78 \end{align} you can clear see our answer is not what we use to replace it which is 3

For does that say \( 48 \div 2(9+3) = 2 \) let balance it \begin{align} \require{cancel} 48 \div 2(9+x) & = 2 \\ 28 \div 18 + 2x & = 2 \\ \frac{ ^{24} \cancel{48}}{ \cancel{18}_9 } + 2x & = 2 \\ \frac{24}{9} + 2x & = 2 \\ \frac{24 + 18x}{9} & = 2 \\ 24 + 18x & = 2 \times 9 \\ 18x & = 18 - 24 \\ 18x & = -6 \\ x & = \frac{-6}{18} \\ x & = \frac{ - \cancel 6^1}{ \cancel{18}_3 } \\ x & = \frac{1}{3} \simeq 0.33 \end{align}alternate solution \begin{align} \require{cancel} 48 \div 2(9+x) & = 2 \\ 24 (9+x) & =2 \\ 216 +24x & = 2  \\ 24x & = 2-216 \\ \frac{24x}{24} & = \frac{-214}{24} \\ \frac{ ^1 \cancel{24}x}{ \cancel{24}_1 } & = \frac{ - \cancel{214}^{107}}{ \cancel{24}_{12}} \\ x & = \frac{-107}{12} \simeq -8.92 \end{align} another posibility \begin{align} \require{cancel} 48 \div 2(9+x) & = 2 \\ 28(9+x) & = 2 \\ \frac{ \cancel{28}(9+x}{ \cancel{24} } & = \frac{ \cancel{2}^1}{ \cancel{26}_{14}} \\ 9 + x & = \frac{1}{14} \\x & = \frac{1}{14} - 9 \\ x & = \frac{1-9}{14} \\ x & = \frac{-8}{14} \\x & = \frac{- \cancel 8^4}{ \cancel {14}_7} \\ x & = \frac{-4}{7} \simeq 0.57 \end{align} the both prove is wrong because we use the wrong interpretation of parenthesis  Note there is a close relationship between multiplication and division that why this answer is the opposite of our replacement

set of people that interpreted this  \begin{align} 48 \div 2(9+3) & = 48 \div 2(12) \\ & =  48 \div 24 \end{align}let check if \( 48 \div 2(9+3) = 2 \) by applying the logic of example 1, 5 and 6
I will like to replace 3 with x
\( 48 \div 2( 9 + x) = 2 \)

Solution

 \begin{align} 48 \div 2(9+x) & = 2 \\ 48 \div 18 +2x & = 2 \\  48 \div 24 +2x & = 2 \\  \frac{ ^{24} \cancel{48}}{ \cancel{18}_9 } + 2x & = 2 \\ \frac{24}{9} + 2x & = 2 \\ \frac{24 + 18x}{9} & = 2 \\ 24 + 18x & = 2 \times 9 \\ 18x & = 18 - 24 \\ 18x & = -6 \\ x & = \frac{-6}{18} \\ x & = \frac{ - \cancel 6^1}{ \cancel{18}_3 } \\ x & = \frac{1}{3} \simeq 0.33 \end{align} 

 the above prove is wrong because we use the wrong interpretation parenthesis  Note there is a close relationship between multiplication and division that why this answer is the opposite of our replacement.

you will properly thing where is the point in this. if you notice you will see that the answer of this prove is equal to the above set of people.  please read till the end to get the point 

Calebprof Prove 

am using my above examples of FORMING QUESTION to prove this 

let check if \( 48 \div 2(9+3) = 2 \) if 2 can form the expression

\begin{align} 2 & =48 \div 24 \\ 2 & = 48 \div 2(12) \\ 2 & = 48 \div 2(9+3) \end{align} proved

using the other way order of operation say we should solve this
let check if \( 48 \div 2(9+3) = 2 \) if 2 can form the expression \begin{align} 48 \div 2(9+3) & = 2 \\ 24(12) = 2 \\ 288 = 2 \end{align} this tell us 288 is correct because it is work with the wrong us of perenthsis
and the other way of working\begin{align} \require{cancel} 48 \div 2(9+3) & = 2 \\ 48 \div 18+6 & = 2 \\ \frac{48}{18} + 6 & = 2 \\ \frac{ \cancel{48}^{24}}{ \cancel{18}} + 6 = 2 \\ \frac{24+54}{9} = 2 \\ 2= \frac{78}{9} \simeq 8.67 \end{align}

note: if you are confused with this try to understand example 5 and 6 forming

Next week Thursday I'm going to explain the use of parenthesis
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