
How do you solve the volume formula $R=4c+a$ , for a?
Answer
525k+ views
Hint: To solve the volume formula $R=4c+a$ , for “a”, we will use simple algebraic rules. First, let us make one side and all other terms on the other side. For this, we will follow the rule that all the positive terms (sum terms) when moved from LHS to RHS or vice-versa, become negative and all the negative terms (difference terms) when moved from LHS to RHS or vise-versa will be positive. This will yield the required answer.
Complete step-by-step solution:
We have to solve the volume formula $R=4c+a$ , for a. This is done using simple algebraic rules.
First, let us make one side and all other terms on the other side. For this, we will follow the rule that all the positive terms (sum terms) when moved from LHS to RHS or vice-versa, become negative and all the negative terms (difference terms) when moved from LHS to RHS or vise-versa will be positive. Hence we can write $R=4c+a$ as:
$R-4c=a$
We can write the above form as
$a=R-4c$
Hence the value of a will be $R-4c$ .
Note: Students have a chance of making mistakes when moving positive and negative terms from LHS to RHS. Similar to the sum and difference terms, we can also move a product and division terms. When we move a product term from LHS to RHS or vice-versa, it will be the divisor, that means, the terms on the opposite side will be divided by this term. Let us see the following example:
$\begin{align}
& px=6 \\
& \Rightarrow x=\dfrac{6}{p} \\
\end{align}$
Similarly, for a division term, the divisor will become the product when it is moved from LHS to RHS or vice-versa. Let us see the following example:
$\begin{align}
& x=\dfrac{6}{p} \\
& \Rightarrow px=6 \\
\end{align}$
Complete step-by-step solution:
We have to solve the volume formula $R=4c+a$ , for a. This is done using simple algebraic rules.
First, let us make one side and all other terms on the other side. For this, we will follow the rule that all the positive terms (sum terms) when moved from LHS to RHS or vice-versa, become negative and all the negative terms (difference terms) when moved from LHS to RHS or vise-versa will be positive. Hence we can write $R=4c+a$ as:
$R-4c=a$
We can write the above form as
$a=R-4c$
Hence the value of a will be $R-4c$ .
Note: Students have a chance of making mistakes when moving positive and negative terms from LHS to RHS. Similar to the sum and difference terms, we can also move a product and division terms. When we move a product term from LHS to RHS or vice-versa, it will be the divisor, that means, the terms on the opposite side will be divided by this term. Let us see the following example:
$\begin{align}
& px=6 \\
& \Rightarrow x=\dfrac{6}{p} \\
\end{align}$
Similarly, for a division term, the divisor will become the product when it is moved from LHS to RHS or vice-versa. Let us see the following example:
$\begin{align}
& x=\dfrac{6}{p} \\
& \Rightarrow px=6 \\
\end{align}$
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