Question

Solve the linear equation \[\dfrac{{(3t - 2)}}{4} - \dfrac{{(2t + 3)}}{3} = \dfrac{2}{3} - t\] linear equation in one variable.

Answer 1

Hint:Equations with linear expressions in one variable only are known as linear equations in one variable. Any term of an equation may be taken from one side to other with the change in its sign, this does not affect the equality of the statement and this process is called transposition. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS).

Complete step by step answer:
Given equation is as below,
\[\dfrac{{(3t - 2)}}{4} - \dfrac{{(2t + 3)}}{3} = \dfrac{2}{3} - t\]
Taking the LCM of denominators of LHS.
i.e. the LCM of 3 and 4 is 12, we get,
\[\dfrac{{3(3t - 2) - 4(2t + 3)}}{{12}} = \dfrac{2}{3} - t\]
Multiplying the numbers, we get,
\[\dfrac{{(9t - 6) - (8t + 12)}}{{12}} = \dfrac{2}{3} - t\]
Removing the brackets we get,
\[\dfrac{{9t - 6 - 8t - 12}}{{12}} = \dfrac{2}{3} - t\]
By transposing the above equation on the same side, we can write as,
\[\dfrac{{9t - 8t - 6 - 12}}{{12}} = \dfrac{2}{3} - t\]

Simplify the given above equation, we get,
\[\dfrac{{t - 18}}{{12}} = \dfrac{{2 - 3t}}{3}\]
\[ \Rightarrow \dfrac{{t - 18}}{{12}} = \dfrac{2}{3} - t\]
Reducing the denominators of both the sides, we get,
\[\dfrac{{t - 18}}{4} = \dfrac{{2 - 3t}}{1}\]
Simplify the above equation, we get,
\[ t - 18 = 4(2 - 3t)\]
Removing the brackets, we get,
\[t - 18 = 8 - 12t\]
Again by transposing the above equation, we can write as,
\[ t + 12t = 8 + 18 \\ \]
\[ \Rightarrow 13t = 26\]
Taking all numbers on one side and unknown variable on the other side, we get,
\[t = \dfrac{{26}}{{13}} \\
\Rightarrow t = 2 \\ \]
Let's see if the answer is correct or not.
First we will calculate the LHS part.
So, LHS =\[\dfrac{{(3t - 2)}}{4} - \dfrac{{(2t + 3)}}{3}\]
Substituting the value of t, we get,
\[LHS= \dfrac{{(3(2) - 2)}}{4} - \dfrac{{(2(2) + 3)}}{3} \\
\Rightarrow LHS= \dfrac{{(6 - 2)}}{4} - \dfrac{{(4 + 3)}}{3} \\ \]
Removing the brackets, we get,
\[LHS = \dfrac{4}{4} - \dfrac{7}{3} \\
\Rightarrow LHS = 1 - \dfrac{7}{3} \\ \]
Taking LCM of \[3\]and\[1\], which is, we get,
\[LHS = \dfrac{{3 - 7}}{3} \\
\Rightarrow LHS = \dfrac{{ - 4}}{3} \\ \]
Next, we will calculate the RHS part.
RHS = \[\dfrac{2}{3} - t\]
Substituting the value of t, we get,
\[RHS = \dfrac{2}{3} - 2\]
\[\Rightarrow RHS= \dfrac{{2 - 2(3)}}{3} \\
\Rightarrow RHS= \dfrac{{2 - 6}}{3} \\
\therefore RHS= \dfrac{{ - 4}}{3} \\ \]
Thus, LHS = RHS and so the answer is correct. Hence, it is verified.

Note:An algebraic equation is an equality involving variables i.e. equality sign (=). An algebraic equation is an equality involving variables, where the values of the expressions on the LHS and RHS are equal. We can verify the answer by putting the value of t in the above equation and check if LHS = RHS, then the answer is correct.