Question

Solve the inequality for real $x$:
$\dfrac{x}{3} > \dfrac{x}{2} + 1$.

Answer 1

Firstly, we will combine the like terms, that is, terms $\dfrac{x}{3}$ and $\dfrac{x}{2}$ by taking the term $\;\dfrac{x}{2}$ on the left side of the inequality $\dfrac{x}{3} > \dfrac{x}{2} + 1$ .
Then, we will simplify the left-hand side of the inequality and will take all the terms except $x$ on the right-hand side of the inequality.
The resultant inequality will be our solution to the given inequality.

Complete step by step solution:
We are given the inequality:
$\dfrac{x}{3} > \dfrac{x}{2} + 1$.
We need to solve the given inequality for real $x$.
Combining like terms $\dfrac{x}{3}$ and $\;\dfrac{x}{2}$, by taking the term $\;\dfrac{x}{2}$ on the left side of the inequality, we get:
$\dfrac{x}{3} - \dfrac{x}{2} > 1$.
Now, we know that that $lcm\left( {2,3} \right) = 6$,
because $2 = 2 \times 1$ and $3 = 3 \times 1$.
Therefore, the above inequality can be written as:
$\dfrac{{2x - 3x}}{6} > 1$.
On simplifying the numerator of the term present on the left-hand side of the above inequality, we get:
$\dfrac{{ - x}}{6} > 1$.
Taking $6$ from the left-hand side to the right-hand side of the above inequality, we get:
$- x > 1 \times 6$.
On multiplying $6$ with $1$, we get:
$- x > 6$.
Multiplying both the sides of the inequality with $- 1$, we get:
$- 1 \times ( - x) < \- 1 \times 6$.
Observe that the inequality sign reverses if we multiply both sides of the inequality by a negative number.
Since we have the following two properties:
$- a \times - b = ab$ and
$- a \times b = ab$,
So, we can use them to simplify the above inequality as:
$x < \- 6$.
Therefore, the solutions of the inequality $\dfrac{x}{3} > \dfrac{x}{2} + 1$ are all the real numbers $x$ that are less than $- 6$.

Note:
We can solve any linear inequality by performing inverse operations to isolate the variable on one side of the inequality.
Always, remember to reverse the inequality while multiplying or dividing the inequality with a negative number.