Solve the following equation for the value of variable ‘p’ : $5\left( p-3 \right)=3\left( p-2 \right)$.
Answer
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Hint:We will first expand the brackets on both sides and then separate the variables having the term with variable ‘p’ on one side and the constant terms on the other side. We will then solve the obtained equation to get the value of variable ‘p’.
Complete step-by-step answer:
It is given in the question that we have to solve the equation $5\left( p-3 \right)=3\left( p-2 \right)$.
We will first expand the brackets on both the side of the equation individually and the solve both the sides individually, we get - $5\left( p-3 \right)=3\left( p-2 \right)$, opening brackets, we get
$5p-15=3p-6$.
Now, we will separate the variable term, that is, the term containing ‘p’ on one side and the constant terms on the other side.
On transposing -15 from LHS to RHS, we get
$5p=3p-6+15$, solving the constant operation, we get, $5p=3p+9$. Now, taking the term $3p$ to LHS and changing its sign, we get,
$5p-3p=9$, now, solving the LHS we get,
$2p=9$ or further
$p=\dfrac{9}{2}$ which can be written as $p=4.5$.
Thus, the value of p is $\dfrac{9}{2}$ or \[4.5\].
Note: It is observed that most of the students make mistakes in taking signs while transposing the terms from one side to another. They may take $5p=3p-6-15$ in the first step after transposing -15 from LHS to RHS which is not correct at all. It is important to know that whenever we transpose any term from one side to another the sign will change.
Complete step-by-step answer:
It is given in the question that we have to solve the equation $5\left( p-3 \right)=3\left( p-2 \right)$.
We will first expand the brackets on both the side of the equation individually and the solve both the sides individually, we get - $5\left( p-3 \right)=3\left( p-2 \right)$, opening brackets, we get
$5p-15=3p-6$.
Now, we will separate the variable term, that is, the term containing ‘p’ on one side and the constant terms on the other side.
On transposing -15 from LHS to RHS, we get
$5p=3p-6+15$, solving the constant operation, we get, $5p=3p+9$. Now, taking the term $3p$ to LHS and changing its sign, we get,
$5p-3p=9$, now, solving the LHS we get,
$2p=9$ or further
$p=\dfrac{9}{2}$ which can be written as $p=4.5$.
Thus, the value of p is $\dfrac{9}{2}$ or \[4.5\].
Note: It is observed that most of the students make mistakes in taking signs while transposing the terms from one side to another. They may take $5p=3p-6-15$ in the first step after transposing -15 from LHS to RHS which is not correct at all. It is important to know that whenever we transpose any term from one side to another the sign will change.
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