Question

How do you solve \[\dfrac{4}{3}s - 3 < s + \dfrac{2}{3} - \dfrac{1}{3}s\] ?

Answer 1

Hint: In this question, we are given that \[\dfrac{4}{3}s - 3\] is smaller than $ s + \dfrac{2}{3} - \dfrac{1}{3}s $ , it contains one unknown quantity (s) so it is an algebraic expression containing one unknown variable quantity. We know that we need “n” number of equations to find the value of “n” unknown variables. In the given algebraic expression; we have 1 unknown quantity and exactly one equation. So we can easily find the value of s by rearranging the equation such that the terms containing “s” are present on one side of the equation and the other side contains all other terms. Then we can find the value of “s” by applying the given arithmetic operations.

Complete step-by-step answer:
We are given that \[\dfrac{4}{3}s - 3 < s + \dfrac{2}{3} - \dfrac{1}{3}s\]
To find the value of s, we will take $ s\,and\,\dfrac{1}{3}s $ to the left-hand side and 3 to the right-hand side –
 \[
  \dfrac{4}{3}s + \dfrac{1}{3}s - s < \dfrac{2}{3} + 3 \\
   \Rightarrow \dfrac{{5 - 3}}{3}s < \dfrac{{11}}{3} \\
   \Rightarrow \dfrac{2}{3}s < \dfrac{{11}}{3} \;
 \]
Now, we will take $ \dfrac{2}{3} $ to the right-hand side –
 $
  s < \dfrac{{11}}{3} \times \dfrac{3}{2} \\
   \Rightarrow s < \dfrac{{11}}{2}\,or\,s < 5.5 \;
  $
Hence, when \[\dfrac{4}{3}s - 3 < s + \dfrac{2}{3} - \dfrac{1}{3}s\] , we get $ s < \dfrac{{11}}{2} $ .
So, the correct answer is “ $ s < \dfrac{{11}}{2} $ ”.

Note: In the expression $ s < \dfrac{{11}}{3} \times \dfrac{3}{2} $ , we see that 3 is present in both the numerator and the denominator so we cancel it out and get the value of “s” in simplified form, now the numerator and the denominator are prime numbers so it cannot be simplified further. In the given algebraic expression, “s” is raised to the power 1, that is, the alphabet representing the unknown quantity has a non-negative integer as power. So the given expression is a polynomial equation and is known as a linear equation as the degree of s is 1.