# Ex.2.5 Q5 Linear Equations in One Variable Solution - NCERT Maths Class 8

## Question

Solve the linear equation\begin{align}\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t\end{align}

Video Solution
Linear Equations
Ex 2.5 | Question 5

## Text Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

Steps:

\begin{align}\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t\end{align}

LCM of the denominators, $$3$$ and $$4,$$ is $$12.$$

Multiplying both sides by $$12,$$ we obtain

\begin{align}3\left( {3t \! - \! 2} \right) \! - \! 4\left( {2t \! + \! 3} \right) &= \! 8 \! - \! 12t \\9t \! - \! 6 \! - \! 8t \! - \! 12 &= \! 8 \! - \! 12t\, \\ \quad\quad\quad\text{(Opening }& \text{the brackets)}\\9t \! - \! 8t \! + \! 12t &= \! 8 \! + \! 6 \! + \! 12 \\13t &= \! 26 \\t &= \! \frac{{26}}{{13}} \\t &= \! 2 \\\end{align}

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