
Solve $ 4+5(p-1)=34 $
Answer
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Hint: To solve the given equation first we will open the bracket by multiplying the terms. Then we will shift the variables at one side and constants at the other side of the equality. Then by solving the mathematical operators given in the question we will get the desired answer.
Complete step by step answer:
We have been given an equation $ 4+5(p-1)=34 $
Let us first simplify the bracket we get
$ \begin{align}
& 4+5(p-1)=34 \\
& \Rightarrow 4+5p-5=34 \\
\end{align} $
Now, we will solve the constant terms of the LHS we get
$ \begin{align}
& \Rightarrow 4-5+5p=34 \\
& \Rightarrow -1+5p=34 \\
\end{align} $
Now, shifting the constant term to the RHS we get
$ \Rightarrow 5p=34+1 $
Now, solving the sum in the RHS we get
$ \Rightarrow 5p=35 $
Now, divide the whole equation by 5 to make the coefficient of $ p=1 $ we get
\[\Rightarrow \dfrac{5p}{5}=\dfrac{35}{5}\]
Now, solving further we get
$ \Rightarrow p=7 $
So on solving the equation $ 4+5(p-1)=34 $ we get the value $ p=7 $
Note:
The given equation is a linear equation in one variable as there is only one unknown variable. The possibility of a mistake in this type of question is the calculation mistake. Be careful while doing calculations and while shifting the terms from LHS to RHS or vice-versa.
Complete step by step answer:
We have been given an equation $ 4+5(p-1)=34 $
Let us first simplify the bracket we get
$ \begin{align}
& 4+5(p-1)=34 \\
& \Rightarrow 4+5p-5=34 \\
\end{align} $
Now, we will solve the constant terms of the LHS we get
$ \begin{align}
& \Rightarrow 4-5+5p=34 \\
& \Rightarrow -1+5p=34 \\
\end{align} $
Now, shifting the constant term to the RHS we get
$ \Rightarrow 5p=34+1 $
Now, solving the sum in the RHS we get
$ \Rightarrow 5p=35 $
Now, divide the whole equation by 5 to make the coefficient of $ p=1 $ we get
\[\Rightarrow \dfrac{5p}{5}=\dfrac{35}{5}\]
Now, solving further we get
$ \Rightarrow p=7 $
So on solving the equation $ 4+5(p-1)=34 $ we get the value $ p=7 $
Note:
The given equation is a linear equation in one variable as there is only one unknown variable. The possibility of a mistake in this type of question is the calculation mistake. Be careful while doing calculations and while shifting the terms from LHS to RHS or vice-versa.
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