Question

What will be the value for P in \[S = B + \left( {\dfrac{1}{2}} \right)Pl\] ?
\[
A) P = \dfrac{{2(S - B)}}{l} \\
B) P = \dfrac{{2(S - B)}}{{{l^2}}} \\
C) P = \dfrac{{2{{(S - B)}^2}}}{l} \\
D) P = \dfrac{{(S - B)}}{l} \\
\]

Answer 1

Hint:The key for solving this problem easily is to isolate P as we need to find the value of P in the given equation. The elimination method can be used for solving systems of linear equations and to solve the value for P here we can add the same value to each side of the equation while keeping the equation balanced.

Complete step by step solution:
In order to isolate the term containing \[P\] while keeping the equation balanced, we need to subtract \[B\] from each side of the equation.
\[
\Rightarrow S - B = - B + B + \left( {\dfrac{1}{2}} \right)Pl \\
\Rightarrow S - B = 0 + \left( {\dfrac{1}{2}} \right)Pl \\
\Rightarrow S - B = \left( {\dfrac{1}{2}} \right)Pl \\
\]
Now we need to multiply both the sides of equations by 2 for eliminating the fraction by keeping the equation balanced
\[
\Rightarrow 2(S - B) = 2 \times \left( {\dfrac{1}{2}} \right)Pl \\
\Rightarrow 2(S - B) = Pl \\
\]
To solve for \[P\] we need to divide each side of the equation by \[l\] while keeping the equation balanced
\[
\Rightarrow \dfrac{{2(S - B)}}{l} = \dfrac{{Pl}}{l} \\
\Rightarrow \dfrac{{2(S - B)}}{l} = P \\
\Rightarrow P = \dfrac{{2(S - B)}}{l} \\
\]
So the value for \[P\] comes out to be \[\dfrac{{2(S - B)}}{l}\] which means option A is correct.

Additional Information: To solve this problem in a better way it is important to simplify each side of the equation by removing parentheses and combining like terms. Addition or subtraction is used to isolate variable terms on one side of the equation while to solve the variable multiplication or subtraction is used.

Notes: Remember we need to isolate the variable by dividing each side by factor that is not containing the variable. The value of \[P\] can further be written in more simplified way as
\[P = \left( {\dfrac{{2S}}{l}} \right) - \left( {\dfrac{{2B}}{l}} \right)\]
We can remove fractions by multiplying each side of the equation with the common denominator. Remember that the solution of the linear equation is not affected if the same number is added or subtracted from both sides of the equation and if both the sides of the equation is multiplied or divided by the same non-zero number.