Question

Solve \[3(2p-1)=5-(3p-2)\] .

Answer 1

Hint: The given equation is a linear equation in one variable, which is \[p\]. Solving this equation is nothing but finding the value of \[p\]. To find that, we have to simplify the equation and group all the terms with \[p\] on one side and the remaining numbers on the other side.

Complete step by step answer:
In algebra, variables are quantities, whose values vary according to the problems under consideration. Mostly letters like \[a,b,x,y\] etc., are used to denote variables and numbers.Performing the operations of addition, subtraction, multiplication etc., on these letters and numbers give rise to algebraic expressions. If there are algebraic expressions on both sides of \[=\] sign, we get an algebraic equation. If the highest power of the variables involved is \[1\], then it is a linear equation. In our problem, \[p\] is the variable and its highest power is \[1\]. Hence it is a linear equation. It also has only one variable and hence with one equation we can solve for \[p\]. Let us start solving for \[p\].
The linear equation that is given to us is:
 \[3(2p-1)=5-(3p-2)\] …… ( \[1\] )
Simplifying And removing brackets in ( \[1\] )
 \[\Rightarrow 6p-3=5-3p+2\] …… ( \[2\] )
Let us group all terms with \[p\] on the left-hand side (LHS) and all the numbers on the right-hand side (RHS). When the terms change sides, positive terms become negative terms and negative terms become positive terms. So further simplification of ( \[2\] )
 \[\Rightarrow 6p+3p=5+2+3\]
 \[\Rightarrow 9p=10\] …… ( \[3\] )
Dividing on both sides of ( \[3\] ) by \[9\] gives,
 \[\therefore p=\dfrac{10}{9}\] which is our required solution.
In order to gain better understanding of the problem, let us verify whether this value of \[p\]
satisfies the given equation. To do that, we substitute the value of \[p\] on both the LHS and RHS of (\[1\] ).
Substituting \[p=\dfrac{10}{9}\] in LHS of (\[1\]),
 \[\Rightarrow LHS=3\left( 2\left( \dfrac{10}{9} \right)-1 \right)\]
Multiplying and dividing the second term, by \[9\],
 \[\begin{align}
& \Rightarrow LHS=3\left( \dfrac{20}{9}-\dfrac{9}{9} \right) \\
& =3\left( \dfrac{11}{9} \right) \\
& =\dfrac{33}{9} \\
\end{align}\]
Substituting \[p=\dfrac{10}{9}\] in RHS of (\[1\]),
 \[\Rightarrow RHS=5-\left( 3\left( \dfrac{10}{9} \right)-2 \right)\]
Multiplying and dividing the second term, by \[9\] ,
 \[\begin{align}
& \Rightarrow RHS=5-\left( \dfrac{30}{9}-2\left( \dfrac{9}{9} \right) \right) \\
& =5-\left( \dfrac{30}{9}-\dfrac{18}{9} \right) \\
& =5-\dfrac{12}{9} \\
\end{align}\]
Again, multiplying and dividing \[5\] by \[9\] ,
 \[\Rightarrow RHS=5\left( \dfrac{9}{9} \right)-\dfrac{12}{9}\]
 \[\begin{align}
& =\dfrac{45}{9}-\dfrac{12}{9} \\
& =\dfrac{33}{9} \\
\end{align}\]
Hence, we find that LHS \[=\] RHS for the \[p\] value we found. Thus our \[p\] value is correct.

Note:
Many real-life problems can be solved by converting those situations into algebraic equations. Depending on the number of variables and unknown quantities involved, we need to find that many equations or relationships that exist amongst them.The method for solving those equations, depends on the number of variables involved. Finding the value of two numbers whose difference is 2 and sum is 14 is an example involving two variables and two equations. When we take the two numbers as \[x,y\] then \[x-y=2\] and \[x+y=14\] are the required two equations. Now, solving these equations gives us the values of the two numbers. Simple addition and substitution within these equations, give the values as \[8\] and \[6\] respectively.