Solve the following equation and check your results: \[5t - 3 = 3t - 5\]

Hint: Here, we will first add or subtract the above equation to get all the variable terms on one side and the constants on the other. Then we will simplify the obtained equation to find the required value. Now we will substitute the obtained value of \[t\] in the left hand side and right hand side of the equation separately to check our result if both the sides are equal.

Complete step-by-step answer:
We are given that the equation is
\[5t - 3 = 3t - 5{\text{ ......eq(1)}}\]
Subtracting the above equation by \[3t\] on both sides, we get
\[
   \Rightarrow 5t - 3 - 3t = 3t - 5 - 3t \\
   \Rightarrow 2t - 3 = - 5 \\
 \]
Adding the above equation by \[3\] on both sides, we get
\[
   \Rightarrow 2t - 3 + 3 = - 5 + 3 \\
   \Rightarrow 2t = - 2 \\
 \]
Dividing the above equation by \[2\] on both sides, we get
\[
   \Rightarrow \dfrac{{2t}}{2} = \dfrac{{ - 2}}{2} \\
   \Rightarrow t = - 1 \\
 \]
We will now check the above value of \[t\] in the equation (1) is the solution of this equation.
We know that a solution is a value we can put in place of a variable that makes the equation true, that is, the left hand side is equal to the right hand side in the equation.
Substituting the value of \[t\] in the left hand side of the equation (1), we get
\[
   \Rightarrow 5\left( { - 1} \right) - 3 \\
   \Rightarrow - 5 - 3 \\
   \Rightarrow - 8 \\
 \]
Replacing \[ - 1\] for \[t\] in the right hand side of the equation (1), we get
\[
   \Rightarrow 3\left( { - 1} \right) - 5 \\
   \Rightarrow - 3 - 5 \\
   \Rightarrow - 8 \\
 \]
Therefore, LHS is equal to RHS.
Hence, proved.

Note: While solving these types of questions, we have to find some value of the variable like \[t\] from the given equation. You can directly solve such a simple equation to get the answer but the more complicated equation will have more work involved. Students forget to check the answer, which is an incomplete solution for this problem.