Question

Find the value of y:
$\dfrac{3y-2}{7}-\dfrac{5y-8}{4}=\dfrac{1}{14}$

Answer 1

Hint: For the type of question we need to find out the value of the variable in the equation we have. Here in our case it is ‘y’. We can find out by simplifying the equation, where we will get the linear form of equation in terms of y. Where we can easily find out the value of y through the method of linear equation in one variable.

Complete step by step answer:
So as we had a question;
$\dfrac{3y-2}{7}-\dfrac{5y-8}{4}=\dfrac{1}{14}$
As the equation had different denominators which needed to be subtracted. As we know that we can only add or subtract when denominators are the same. Otherwise we had to make them the same by taking the LCM.
So, by taking the LCM in LHS side i.e. of 7 and 4, It will be 28, so we will get equations as
$\begin{align}
  & \dfrac{4\left( 3y-2 \right)}{28}-\dfrac{7\left( 5y-8 \right)}{28}=\dfrac{1}{14} \\
 & \dfrac{12y-8}{28}-\dfrac{35y-56}{28}=\dfrac{1}{14} \\
 & \dfrac{12y-8-35y+56}{28}=\dfrac{1}{14} \\
 & \dfrac{-23y+48}{28}=\dfrac{1}{14} \\
\end{align}$
As on LHS and RHS denominator had 28 and 14 respectively which are divisible by 14. So we can further reduce equation to
$\dfrac{-23y+48}{2}=\dfrac{1}{1}$
By cross multiplying
$\begin{align}
  & -23y+48=2 \\
 & 48-2=23y \\
 & 46=23y \\
 & y=\dfrac{46}{23} \\
 & y=2 \\
\end{align}$
Hence y = 2.

Note: To solve such equations first make the denominator same by taking the LCM (Least Common Factor) as we cannot directly add or subtract the equation unless we had a denominator same.