In the previous section, we saw the disadvantages of using trial and error method for solving inequalities. In this section, we will see a systematic method.
For developing a systematic method, we must first learn some basic properties of inequalities. We can use those basic properties as rules for solving inequalities. This can be explained in 7 steps:
1. Let us recall the two rules that we used while solving linear equations:
First rule:
• Any number can be added to any side of the equation. The same number should be added to the other side also.
• Any number can be subtracted from any side of the equation. The same number should be subtracted from the other side also.
Second rule:
• Any side of the equation can be multiplied by a non-zero number. The other side also should be multiplied by the same number.
• Any side of the equation can be divided by a non-zero number. The other side also should be divided by the same number.
2. In the same way, three rules can be developed for inequalities.
• If we add the same number to both sides of the inequality, there will be no change for the sign.
♦ For example, consider the inequality 2 < 8.
♦ Let us add 7 on both sides. We get 9 < 15.
♦ The ‘<’ sign has not changed.
• If we subtract the same number from both sides of the inequality, there will be no change for the sign.
♦ For example, consider the inequality 12 < 21.
♦ Let us subtract 4 from both sides. We get 8 < 17.
♦ The ‘<’ sign has not changed.
3. So we can write the first rule for solving inequalities.
Rule 1 for solving inequalities:
• Any number can be added to any side of the inequality. The same number should be added to the other side also. Then the sign will not change.
• Any number can be subtracted from any side of the inequality. The same number should be subtracted from the other side also. Then the sign will not change.
4. If we multiply both sides of an inequality by the same +ve number, there will be no change for the sign.
♦ For example, consider the inequality 2 < 8.
♦ Let us multiply both sides by 3. We get 6 < 24.
♦ The ‘<’ sign has not changed.
• If we divide both sides of an inequality by the same +ve number, there will be no change for the sign.
♦ For example, consider the inequality 9 < 12.
♦ Let us divide both sides by 4. We get 2.25 < 3.
♦ The ‘<’ sign has not changed.
5. So we can write the second rule for solving inequalities.
Rule 2 for solving inequalities:
• Any side of the inequality can be multiplied by a +ve number. The other side also should be multiplied by the same number. The sign will not change.
• Any side of the inequality can be divided by a +ve number. The other side also should be divided by the same number. The sign will not change.
6. If we multiply both sides of an inequality by the same -ve number, the sign will be reversed.
♦ For example, consider the inequality 2 < 8.
♦ Let us multiply both sides by -3. We get -6 > -24.
♦ The ‘<’ sign has become ‘>’.
• If we divide both sides of an inequality by the same -ve number, the sign will be reversed.
♦ For example, consider the inequality 9 < 12.
♦ Let us divide both sides by -4. We get -2.25 > -3.
♦ The ‘<’ sign has become ‘>'.
7. So we can write the third rule for solving inequalities.
Rule 3 for solving inequalities:
• Any side of the inequality can be multiplied by a -ve number. The other side also should be multiplied by the same number. The sign will be reversed.
• Any side of the inequality can be divided by a -ve number. The other side also should be divided by the same number. The sign will be reversed.
• Reversal of sign means:
♦ < will become >
♦ > will become <
♦ ≤ will become ≥
♦ ≥ will become ≤
Now we will see some solved examples
Solved example 6.1
Solve 30x < 200 when (i) x is a natural number (ii) x is an integer
Solution:
• Given inequality is: 30x < 200
• Applying Rule 2, we get:
$\frac{30x}{30}~<~\frac{200}{30}$
$\Rightarrow x~<~\frac{20}{3}$
$\Rightarrow~x~<~6.667$
• So whichever value we select for x, must be less than 6.667
Part (i):
• The set of natural number is {1, 2, 3, 4, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {1, 2, 3, 4, 5, 6}
Part (ii):
• The set of integers is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}
Why do we specifically say natural numbers, integers etc.,?
The answer can be written in 2 steps:
1. In some problems, x may represent number of pens, number of books, number of cars etc., In such cases, x must be a natural number (or whole number).
2. In some problems, x may represent the floor level of a building.
♦ The floor immediately below the ground level is indicated by -1
♦ The floor below that is indicated by -2 and so on . . .
• In such cases, x must be an integer.
Solved example 6.2
Solve 5x-3 < 3x+1 when (i) x is an integer (ii) x is a real number
Solution:
• Given inequality is: 5x-3 < 3x+1
• This can be simplified as follows:
$\begin{array}{ll}
{}&5x-3 &{}<{}& {3x+1} &{} \\
\Rightarrow &5x-3+3&{}<{}& 3x+1+3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &5x&{}<{}& 3x+4 &{} \\
\Rightarrow &5x-3x&{}<{}& 3x+4-3x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 4 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{4}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 2 &{} \\
\end{array}$
Part (i):
• The set of integers is {. . . , -4, -3, -2, -1, 0, 1, 2, 3, . . .}
• We must select the appropriate values from this set. The solution set will contain those appropriate values.
• So the solution set is {. . . , -4, -3, -2, -1, 0, 1}
Part (ii):
• We know that the set of real numbers contain all types of numbers.
♦ They include natural numbers, whole numbers, integers
♦ They also include rational numbers and irrational numbers.
♦ So there will be numbers like -√3, -√2, √3, √2, -π, π etc.,
• Since they do not occur at regular intervals, we pick out the "relevant portion of the number line" and write it as an interval.
• In our present case, the interval will be: (-∞,2)
• That means, the relevant portion on the number line in our case, begins from -∞ at the extreme left end. It extends upto +2.
• But the exact +2 should not be included. Numbers like 1.999, 1.9999 etc., are allowed. This is indicated by the ')' on the right side of 2.
• This interval can be shown graphically as in fig.6.1 below:
 |
| Fig.6.1 |
• The yellow line represents the number line. The red line is the graph.
• The arrow at the left end of the red line indicates that, the left boundary of the interval is at -∞.
• The circle at the right end of the red line indicates that, the right boundary of the interval is at 2.
♦ Since 2 is not included in the interval, it is an ordinary circle.
♦ If 2 is also included, we give a filled circle.
• All points on the red line is a solution of the inequality. We can write:
When x is a real number, the solution set is: (-∞,2)
Why do we specifically say integers, real numbers etc.,?
The answer can be written in 2 steps:
1.
In solved example 6.1 above, we saw the situations where integers are specified. So we need not discuss about it again.
2. In some problems, x may represent lengths or distances. Such quantities are not always available as integers.
• We may encounter lengths like 3.5 cm, 121.667 meter etc.,
• Some times distances towards the left are considered -ve and those towards the right are considered +ve
♦ Then we may encounter lengths like -3.5 cm, -121.667 meter etc.,
• Also it is possible to encounter lengths like -2√3 meter, 5√2 meter, -3π cm etc.,
• In some problems, x may represent temperature.
• In some problems, x may represent volume.
• In all such cases, x must be a real number.
We have seen the significance of specifying whether x is natural number, integer or real number. For the rest of our discussion in this chapter, the variables x, y etc., will be considered as representing real numbers.
Solved example 6.3
Solve 4x+3 < 6x+7
Solution:
• Given inequality is: 4x+3 < 6x+7
• This can be simplified as follows:
$\begin{array}{ll}
{}&4x+3 &{}<{}& {6x+7} &{} \\
\Rightarrow &4x+3-3&{}<{}& 6x+7-3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &4x&{}<{}& 6x+4 &{} \\
\Rightarrow &4x-6x&{}<{}& 6x+4-6x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &-2x&{}<{}& 4 &{} \\
\Rightarrow &\frac{-2x}{-2}&{}<{}& \frac{4}{-2} &\color {green}{\text{(Rule 3)}} \\
\Rightarrow &x&{}>{}& -2 &{} \\
\end{array}$
• So all real numbers greater than -2 are the solutions of this inequality.
• The solution set is: (-2,∞)
• The graph is shown in fig.6.2 below:
 |
| Fig.6.2 |
Solved example 6.4
Solve $\frac{5-2x}{3}\le\frac{x}{6} -5$
Solution:
• Given inequality is: $\frac{5-2x}{3}\le\frac{x}{6} -5$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{5-2x}{3}&{}\le{}& \frac{x}{6} -5 &{} \\
\Rightarrow
&6 \left(\frac{5-2x}{3}\right)&{}\le{}& 6\left(\frac{x}{6}
-5\right) &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &2 \left(5-2x \right)&{}\le{}& x-30 &{} \\
\Rightarrow &10-4x&{}\le{}& x-30 &{} \\
\Rightarrow &10-4x+4x&{}\le{}& x-30+4x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &10&{}\le{}& 5x-30 &{} \\
\Rightarrow &10+30&{}\le{}& 5x-30+30 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &40&{}\le{}& 5x &{} \\
\Rightarrow &\frac{40}{5}&{}\le{}& \frac{5x}{5} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &8&{}\le{}& x &{} \\
\Rightarrow &x&{}\ge{}& 8 &{} \\
\end{array}$
• So all real numbers greater than or equal to 8 are the solutions of this inequality.
• The solution set is: [8,∞)
• The graph is shown in fig.6.3 below:
 |
| Fig.6.3 |
• In the next section, we will see a few more solved examples.
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