Question

Solve the inequality \[\dfrac{1}{2}\left( {\dfrac{{3x}}{5} + 4} \right) \geqslant \dfrac{1}{3}\left( {x - 6} \right)\].

Answer 1

Hint: The given inequality is a simple inequality having one variable only. To solve the equation of one variable, we need only one equation to solve for that variable. At first, we will simplify the given inequality to a possible extent by using operations of inequality. We will also use the Least common multiple to add both the terms of the bracket part on the left hand side of the inequality. We will use any of addition, subtraction, multiplication and division possible to reach the desired answer.

Complete step-by-step solution:
Multiplying both sides by\[6\], we obtain,
\[ 3\left( {\dfrac{{3x}}{5} + 4} \right) \geqslant 2\left( {x - 6} \right) \\
 \Rightarrow 3\left( {\dfrac{{3x + 20}}{5}} \right) \geqslant 2\left( {x - 6} \right) \]
Multiplying both sides again by\[5\], we obtain,
\[3\left( {3x + 20} \right) \geqslant 10\left( {x - 6} \right)\]
Use distributive property of inequality,
\[9x + 60 \geqslant 10x - 60\]
Subtract \[9x\] from both sides, we get,
\[60 \geqslant x - 60\]
Now, adding \[60\] both sides, we get,
\[120 \geqslant x\] \[\]
Or, It is same as:
\[x \leqslant 120\]
So the final answer: This statement means \[x\] is less than or equal to \[120\].

Note: The major difference between an inequality and an equation is the result that we get after solving it. On solving an equation in one variable, we get a point on a number line. On the other hand, on solving an inequality with one variable, we get a range of points or a region on a number line.
Like, in the above solution, we get a range of points for \[x\], \[\left( { - \infty ,\left. {120} \right]} \right.\].
Rules for solving the inequalities: (1). Eliminate fraction and decimals. (2). Flip the inequality symbol if multiplying or dividing by a negative number. (3). Isolate terms containing variables. (4). Combine like terms. (5). Divide by numerical coefficient of variable (only and only if coefficient is greater than 1).