Question

If:
(a) \[2p + 20\% {\text{ of }}\left( {2p - 1} \right) = 7\], find \[p\].
(b) \[3\left( {2x - 1} \right) + 25\% {\text{ of }}x = 97\], find \[x\].

Answer 1

Hint:
Here, we need to solve the given equations to find the required values. We will first convert the percentages to fractions, and then simplify the equation to find the required values of \[p\] and \[x\].

Complete step by step solution:
(a)
We will simplify the given equation to find the value of \[p\].
We know that \[20\% \] of \[\left( {2p - 1} \right)\] can be written as \[\dfrac{{20}}{{100}} \times \left( {2p - 1} \right)\].
Simplifying the expression, we get
\[ \Rightarrow \]\[20\% \] of \[\left( {2p - 1} \right)\]\[ = \dfrac{1}{5}\left( {2p - 1} \right)\]
Substituting \[20\% \] of \[\left( {2p - 1} \right)\]\[ = \dfrac{1}{5}\left( {2p - 1} \right)\] in the equation \[2p + 20\% {\text{ of }}\left( {2p - 1} \right) = 7\], we get
\[ \Rightarrow 2p + \dfrac{1}{5}\left( {2p - 1} \right) = 7\]
Multiplying \[\dfrac{1}{5}\] and \[\left( {2p - 1} \right)\] using the distributive law of multiplication, we get
\[ \Rightarrow 2p + \dfrac{{2p}}{5} - \dfrac{1}{5} = 7\]
Taking the L.C.M., we get
\[ \Rightarrow \dfrac{{10p + 2p - 1}}{5} = 7\]
Adding the like terms, we get
\[ \Rightarrow \dfrac{{12p - 1}}{5} = 7\]
Multiplying both sides of the equation by 5, we get
\[\begin{array}{l} \Rightarrow \dfrac{{12p - 1}}{5} \times 5 = 7 \times 5\\ \Rightarrow 12p - 1 = 35\end{array}\]
Adding 1 to both sides, we get
\[\begin{array}{l} \Rightarrow 12p - 1 + 1 = 35 + 1\\ \Rightarrow 12p = 36\end{array}\]
Finally, dividing both sides by 12, we get the value of \[p\] as
\[\begin{array}{l} \Rightarrow \dfrac{{12p}}{{12}} = \dfrac{{36}}{{12}}\\ \Rightarrow p = 3\end{array}\]
\[\therefore \] We get the value of \[p\] as 3.
(b)
We will simplify the given equation to find the value of \[x\].
We know that \[25\% \] of \[x\] can be written as \[\dfrac{{25}}{{100}} \times x\].
Simplifying the expression, we get
\[ \Rightarrow \]\[25\% \] of \[x\]\[ = \dfrac{x}{4}\]
Substituting \[25\% \] of \[x\]\[ = \dfrac{x}{4}\] in the equation \[3\left( {2x - 1} \right) + 25\% {\text{ of }}x = 97\], we get
\[ \Rightarrow 3\left( {2x - 1} \right) + \dfrac{x}{4} = 97\]
Multiplying 3 and \[\left( {2x - 1} \right)\] using the distributive law of multiplication, we get
\[ \Rightarrow 6x - 3 + \dfrac{x}{4} = 97\]
Taking the L.C.M., we get
\[ \Rightarrow \dfrac{{24x - 12 + x}}{4} = 97\]
Adding the like terms, we get
\[ \Rightarrow \dfrac{{25x - 12}}{4} = 97\]
Multiplying both sides by 4, we get
\[\begin{array}{l} \Rightarrow \dfrac{{25x - 12}}{4} \times 4 = 97 \times 4\\ \Rightarrow 25x - 12 = 388\end{array}\]
Adding 12 to both sides, we get
\[\begin{array}{l} \Rightarrow 25x - 12 + 12 = 388 + 12\\ \Rightarrow 25x = 400\end{array}\]
Finally, dividing both sides by 25, we get the value of \[x\] as
\[\begin{array}{l} \Rightarrow \dfrac{{25x}}{{25}} = \dfrac{{400}}{{25}}\\ \Rightarrow x = 16\end{array}\]

\[\therefore \] We get the value of \[x\] as 16.

Note:
The equations given are linear equations in one variable. A linear equation in one variable is an equation of the form \[ax + b = 0\], where \[a\] and \[b\] are integers. A linear equation of the form \[ax + b = 0\] has only one solution.
We have used the distributive law of multiplication to multiply \[\dfrac{1}{5}\] by \[\left( {2p - 1} \right)\], and 3 by \[\left( {2x - 1} \right)\]. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
We have added the like terms in one step of the solution. Like terms are the terms whose variables and their exponents are the same. For example, \[100x,150x,240x,600x\] all have the variable \[x\] raised to the exponent 1. Terms which are not like cannot be added together.