Question

How do you solve for a in \[SA=2ab+2ac+2bc\]?

Answer 1

Hint: From the question, we were given to find the value of a in \[SA=2ab+2ac+2bc\]. Let us consider this equation (1). Now we have to perform operations and transpose the terms such that we have all the terms in ‘a’ on one side and the remaining terms on the other side of the equation. In this way, we can find the value of a in equation (1).

Complete step by step answer:
From the given question, it is given that to find the value of a in \[SA=2ab+2ac+2bc\].
Let us consider this equation as equation (1),
\[\Rightarrow SA=2ab+2ac+2bc.....(1)\]
So, it is clear that we should find the value of ‘a’ from equation (1).
So, we need all the terms in ‘a’ in LHS of equation (1) and the remaining terms in RHS of equation (1).
By, using the above concept, we will transpose 2bc from RHS to LHS
\[\Rightarrow SA-2bc=2ab+2ac\]
Now in RHS of the above equation, we will have 2a as common. So, we will take out the term 2a as common and we will get
\[\Rightarrow SA-2bc=2a\left( b+c \right)\]
Now we have to keep a on the RHS. So, by using cross multiplication we get
\[\Rightarrow \dfrac{SA-2bc}{2\left( b+c \right)}=a\]
Now by rewriting the equation, we get
\[\Rightarrow a=\dfrac{SA-2bc}{2\left( b+c \right)}\]
So, from the above equation it is clear that the value of ‘a’ is equal to \[\dfrac{SA-2bc}{2\left( b+c \right)}\] from the given equation \[SA=2ab+2ac+2bc\].

Note:
Students should understand the given question in the correct manner. If students assume ‘ac’ as ‘bc’ or ‘ab’ as ‘bc’ or any other miss-assumption will interrupt the final answer. So, students should read the question carefully. Students have to know about the constants and variables concept. Students should also avoid calculation mistakes while solving this problem.