Question

How do you solve $4t + 14 = \dfrac{{6t}}{5} + 7$ ?

Answer 1

Hint: The given equation is a linear equation in terms of a single variable $t$. We will write the equation in a way that the terms with the variable $t$ are written on one side of the equality sign and the rest terms on another side of the equality sign. If the same number is added to both the sides or the same number is subtracted from both the sides, the equation remains undisturbed.

Complete step-by-step solution:
We have the equation as
$4t + 14 = \dfrac{{6t}}{5} + 7$
We first take the terms with the variable $t$ to the Left Hand Side or the LHS.
Since subtracting the same number from both sides would not disturb the equation, we subtract $\dfrac{{6t}}{5}$ from both the LHS and the RHS. Thus, we get:
$
   \Rightarrow 4t + 14 - \dfrac{{6t}}{5} = \dfrac{{6t}}{5} + 7 - \dfrac{{6t}}{5} \\
   \Rightarrow 4t - \dfrac{{6t}}{5} + 14 = 7 \\
 $
This can be simplified into:
\[
   \Rightarrow \dfrac{{20t - 6t}}{5} + 14 = 7 \\
   \Rightarrow \dfrac{{14t}}{5} + 14 = 7 \\
 \]
Again, we subtract $14$ from both the LHS and the RHS. Thus, we get:
$
   \Rightarrow \dfrac{{14t}}{5} + 14 - 14 = 7 - 14 \\
   \Rightarrow \dfrac{{14t}}{5} = - 7 \\
 $
Since dividing both the LHS and the RHS by the same number would not disturb the equation, we divide both sides by $7$. Thus, we get:
$
   \Rightarrow \dfrac{{14t}}{5} \times \dfrac{1}{7} = \dfrac{{ - 7}}{7} \\
   \Rightarrow \dfrac{{2t}}{5} = - 1 \\
 $
Now dividing both sides by 2, we get:
$
   \Rightarrow \dfrac{{2t}}{5} \times \dfrac{1}{2} = \dfrac{{ - 1}}{2} \\
   \Rightarrow \dfrac{t}{5} = \dfrac{{ - 1}}{2} \\
 $
Now multiplying both sides by 5, we get:
$
   \Rightarrow \dfrac{t}{5} \times 5 = \dfrac{{ - 1}}{2} \times 5 \\
   \Rightarrow t = \dfrac{{ - 5}}{2} \\
 $
Hence, $t = \dfrac{{ - 5}}{2}$ is the solution to the equation $4t + 14 = \dfrac{{6t}}{5} + 7$.

Note: The equality sign is like a weighing balance which separates the Left Hand Side (or the LHS) and the Right Hand Side (or the RHS). When the same number is subtracted from both the LHS and the RHS or the same number is added to both the LHS and the RHS, the equation remains unchanged.
We can also check if the answer is correct or not by putting the value of $t$ in the equation given in the question. If LHS = RHS, the answer is said to be correct.