
Is the following equation is true:
$\left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right)=\left( -45 \right)\div \left\{ \left( -15 \right)\div \left( -3 \right) \right\}$.
Answer
574.2k+ views
Hint: We first discuss the rules of preferences of operations. We will be using the BODMAS rule for the same. We have been given two parts of equality, so we will take one by one and perform the rules on them. Then, we will compare the two results and if they don’t match, then we can say that the given equality is wrong.
Complete step by step answer:
We will use the rule of BODMAS at the time of doing the binary operations. The rule gives us idea about the preferences of operations when more than one is present in one equation.
The first preference is brackets in the order of $\left( {} \right) > \left\{ {} \right\} > \left[ {} \right]$. Then comes ‘of’ which is another of saying multiplication. It’s rare to use the word between equations. Then comes the regular operations in the order of division-multiplication-addition-subtraction.
We have to check if both sides of the equation are equal or not. We first take the left-hand side equation $\left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right)$. We first perform the third bracket part.
$\begin{align}
& \left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right) \\
& =\left\{ \dfrac{-45}{-15} \right\}\div \left( -3 \right) \\
& =\left( 3 \right)\div \left( -3 \right) \\
\end{align}$
Now we perform the rest.
$\begin{align}
& \left( 3 \right)\div \left( -3 \right) \\
& =\dfrac{3}{-3} \\
& =-1 \\
\end{align}$
So, the answer is -1.
We now check the second part in the right-hand side equation $\left( -45 \right)\div \left\{ \left( -15 \right)\div \left( -3 \right) \right\}$. We first perform the third bracket part.
$\begin{align}
& \left( -45 \right)\div \left\{ \left( -15 \right)\div \left( -3 \right) \right\} \\
& =\left( -45 \right)\div \left\{ \dfrac{-15}{-3} \right\} \\
& =\left( -45 \right)\div \left( 5 \right) \\
\end{align}$
Now we perform the rest.
$\begin{align}
& \left( -45 \right)\div \left( 5 \right) \\
& =\dfrac{-45}{5} \\
& =-9 \\
\end{align}$
Therefore, the equation $\left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right)=\left( -45 \right)\div \left\{ \left( -15 \right)\div \left( -3 \right) \right\}$ is wrong.
Note: This problem shows how a slight change of brackets can change the solution of an equation. The rule just gives the preferences which have to strictly be followed. In case of more than one bracket of the same type, we start working from the start irrespective of anything else. For example: we will deal $\left( -45 \right)\div \left( -15 \right)\div \left( -3 \right)$ as $\left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right)$.
Complete step by step answer:
We will use the rule of BODMAS at the time of doing the binary operations. The rule gives us idea about the preferences of operations when more than one is present in one equation.
The first preference is brackets in the order of $\left( {} \right) > \left\{ {} \right\} > \left[ {} \right]$. Then comes ‘of’ which is another of saying multiplication. It’s rare to use the word between equations. Then comes the regular operations in the order of division-multiplication-addition-subtraction.
We have to check if both sides of the equation are equal or not. We first take the left-hand side equation $\left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right)$. We first perform the third bracket part.
$\begin{align}
& \left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right) \\
& =\left\{ \dfrac{-45}{-15} \right\}\div \left( -3 \right) \\
& =\left( 3 \right)\div \left( -3 \right) \\
\end{align}$
Now we perform the rest.
$\begin{align}
& \left( 3 \right)\div \left( -3 \right) \\
& =\dfrac{3}{-3} \\
& =-1 \\
\end{align}$
So, the answer is -1.
We now check the second part in the right-hand side equation $\left( -45 \right)\div \left\{ \left( -15 \right)\div \left( -3 \right) \right\}$. We first perform the third bracket part.
$\begin{align}
& \left( -45 \right)\div \left\{ \left( -15 \right)\div \left( -3 \right) \right\} \\
& =\left( -45 \right)\div \left\{ \dfrac{-15}{-3} \right\} \\
& =\left( -45 \right)\div \left( 5 \right) \\
\end{align}$
Now we perform the rest.
$\begin{align}
& \left( -45 \right)\div \left( 5 \right) \\
& =\dfrac{-45}{5} \\
& =-9 \\
\end{align}$
Therefore, the equation $\left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right)=\left( -45 \right)\div \left\{ \left( -15 \right)\div \left( -3 \right) \right\}$ is wrong.
Note: This problem shows how a slight change of brackets can change the solution of an equation. The rule just gives the preferences which have to strictly be followed. In case of more than one bracket of the same type, we start working from the start irrespective of anything else. For example: we will deal $\left( -45 \right)\div \left( -15 \right)\div \left( -3 \right)$ as $\left\{ \left( -45 \right)\div \left( -15 \right) \right\}\div \left( -3 \right)$.
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