In the previous section, we saw the rules for solving linear inequalities in one variable. We saw some solved examples also. In this section, we will see a few more solved examples.
Solved example 6.5
Solve 7x+3 < 5x+9
Solution:
• Given inequality is: 7x+3 < 5x+9
• This can be simplified as follows:
$\begin{array}{ll}
{}&7x+3 &{}<{}& {5x+9} &{} \\
\Rightarrow &7x+3-3&{}<{}& 5x+9-3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &7x&{}<{}& 5x+6 &{} \\
\Rightarrow &7x-5x&{}<{}& 5x+6-5x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 6 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{6}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 3 &{} \\
\end{array}$
• So all real numbers less than 3 are the solutions of this inequality.
• The solution set is: (-∞,3)
• The graph is shown in fig.6.4 below:
 |
| Fig.6.4 |
Solved example 6.6
Solve $\frac{3x-4}{2}\ge\frac{x+1}{4} -1$
Solution:
• Given inequality is: $\frac{3x-4}{2}\ge\frac{x+1}{4} -1$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{3x-4}{2}&{}\ge{}& \frac{x+1}{4} -1 &{} \\
\Rightarrow
&4 \left(\frac{3x-4}{2}\right)&{}\ge{}& 4\left(\frac{x+1}{4}
-1\right) &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &2 \left(3x-4 \right)&{}\ge{}& x+1-4 &{} \\
\Rightarrow &6x-8&{}\ge{}& x-3 &{} \\
\Rightarrow &6x-8+8&{}\ge{}& x-3+8 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &6x&{}\ge{}& x+5 &{} \\
\Rightarrow &6x-x&{}\ge{}& x+5-x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &5x&{}\ge{}& 5 &{} \\
\Rightarrow &x&{}\ge{}& 1 &{} \\
\end{array}$
• So all real numbers greater than or equal to 1 are the solutions of this inequality.
• The solution set is: [1,∞)
• The graph is shown in fig.6.5 below:
 |
| Fig.6.5 |
Solved example 6.7
The marks obtained by a student of class XI in first and second terminal examination are 62 and 48 respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Solution:
1. Let x be the marks in the annual examination. Then we can write:
$\frac{62+48+x}{3} \ge 60$
• Given inequality is: $\frac{3x-4}{2}\ge\frac{x+1}{4} -1$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{62+48+x}{3}&{}\ge{}&60 &{} \\
\Rightarrow &3 \left(\frac{62+48+x}{3}\right)&{}\ge{}& 3 × 60 &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &62+48+x&{}\ge{}& 180 &{} \\
\Rightarrow &110+x&{}\ge{}& 180 &{} \\
\Rightarrow &110+x-110&{}\ge{}& 180-110 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &x&{}\ge{}& 70 &{} \\
\end{array}$
• So the student must obtain at least 70 marks.
• If the maximum marks is 100, the solution set will be: [70,100]
• The graph is shown in fig.6.6 below:
 |
| Fig.6.6 |
Solved example 6.8
Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
Solution:
1. The consecutive odd numbers starting from 1 are: 1, 3, 5, 7, . . .
2. The consecutive odd numbers greater than 10 are: 11, 13, 15, 17, . . .
• We have to pick out pairs from this list.
• Some of the possible pairs are: (11,13), (13,15) etc.,
• The sum of the two members of any pair that we pick, must be less than 40
3. If x is the first member of a pair, the second member will be (x+2)
• So we can write: x+(x+2) < 40
4. Given that both members of the pairs are to be greater than 10.
• So the smaller one which is x, must be larger than 10. This will ensure that, the larger one which is (x+2) will also be greater than 10.
• We can write: x > 10
5. From step (3) we get: 2x +2 < 40
• This can be simplified as follows:
$\begin{array}{ll}
{}&2x+2 &{}<{}& 40 &{} \\
\Rightarrow &2x+2-2&{}<{}& 40-2 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 38 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{38}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 19 &{} \\
\end{array}$
6. So if the sum is to be less than 40, x must be less than 19
• Also from step (4), we have: x must be greater than 10
• The consecutive odd natural numbers satisfying both the above conditions are:
11, 13, 15 and 17
• So the possible pairs are:
(11, 11+2), (13, 13+2), (15, 15+2), (17, 17+2)
• That is: (11,13), (13,15), (15,17) and (17,19)
Solved example 6.9
Find all pairs of consecutive odd natural numbers, both of which are smaller than 18, such that their sum is more than 20.
Solution:
1. The consecutive odd natural numbers starting from 1 and smaller than 18 are:
1, 3, 5, 7, . . . , 17
2. Let x and x+2 form the pair.
• Given that both of them must be smaller than 18.
• So the larger one which is (x+2), must be smaller than 18. This will ensure that, the smaller one x will also be smaller than 18
• Thus we can write: x+2 < 18
• This can be simplified as follows:
$\begin{array}{ll}
{}&x+2 &{}<{}& 18 &{} \\
\Rightarrow &x+2-2&{}<{}& 18-2 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &x&{}<{}& 16 &{} \\
\end{array}$
• So first member of all pairs must be less than 16
3. Given that, the sum must be larger than 20.
So we can write: x+(x+2) > 20
• This can be simplified as follows:
$\begin{array}{ll}
{}&2x+2 &{}>{}& 20 &{} \\
\Rightarrow &2x+2-2&{}<>{}& 20-2 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}>{}& 18 &{} \\
\Rightarrow &\frac{2x}{2}&{}>{}& \frac{18}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow&x&{}>{}& 9 &{} \\
\end{array}$
4. So if the sum is to be greater than 40, x must be greater than 9
• Also from step (2), we have: x must be less than 16
• The consecutive odd natural numbers satisfying both the above conditions are:
11, 13, and 15
• So the possible pairs are:
(11, 11+2), (13, 13+2) and (15, 15+2)
• That is: (11,13), (13,15) and (15,17)
The link below gives some more solved examples:
• In the next section, we will see linear inequalities in two variables.
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