Question

Solve for the given inequality $3-2x\ge x-32$ for $x\in I$.

Answer 1

Hint: Solve the inequality similar as we solve any equation with the sign of ‘equals’ to. So, separate variables on one side and constants on the other side. Add $2x$ to both sides and hence add 32 to both the sides to get a simple form of the equation. And use the relation $x\in I$ i.e. $x$ can be integer only. So, write down the set of integers using the calculated expression in the end.

Complete step-by-step solution -
Given inequality in the problem is
$3-2x\ge x-32$ for $x\in I$………………….(i)
Where $x\in I$ means $x$ is an integer.
Now, as the inequality (i) consists of variable and constants both, so we need to separate constant terms on one side and variable term on another side.
So, we have
$3-2x\ge x-32$
Adding 32 to both sides of the above equation, we can re-write the above expression as
$3-2x+32\ge x-32+32$
On simplifying the above equation, we get
$\begin{align}
  & 3+32-2x\ge x \\
 & \Rightarrow 35-2x\ge x \\
\end{align}$
Adding $2x$ to both the sides of the above equation, we get
$\begin{align}
  & 35-2x+2x\ge x+2x \\
 & 35\ge 3x \\
\end{align}$
Now, as the variable and constant are separated. So, we can divide the whole expression by 3 to get the value of $x$ as
$\dfrac{35}{3}\ge x$
We can re-write the above expression as
$x\le \dfrac{35}{3}..............\left( ii \right)$
Now, as $x\in I$ i.e. $x$ can only get integers. So, we can solve $\dfrac{35}{3}$ i.e. R.H.S. of the equation (ii). So, we can re-write the equation (ii) as
$x\le 11.667..............\left( iii \right)$
As $x$ is an integer less than 11.667, so the possible values of $x$ are given as
$x\in \{11,10,9,8,7.............\}\text{ or }x\in \{-\infty ......................,7,8,9,10,11\}$
Hence, the above set of integers are the possible values of $x$.

Note: Another approach would be that we can separate variables and constants in two different sides by transferring them to other side directly. We do not need to perform any addition of subtraction. It can be done as
$\begin{align}
  \Rightarrow & 3-2x\ge x-32 \\
   \Rightarrow & 3+32\ge x+2x \\
   \Rightarrow & 35\ge 3x \\
   \Rightarrow & 3x\le 35 \\
   \Rightarrow & x\le \dfrac{35}{3} \\
\end{align}$
So, it can be another approach to get the values of $x$ .
Need to take care of $x\in I$. One may go wrong and give the answer as $x\le \dfrac{35}{3}$, which is wrong. $x$ can take only integer values less than $\dfrac{35}{3}$. So, write down the whole set for this inequality and condition $x\in I$