In the previous section, we completed a discussion on solving a system of linear inequalities in two variables. In this section we will see some miscellaneous examples.
Solved example 6.17
Solve –8 ≤ 5x – 3 < 7
Solution:
1. This is a double inequality. We can rearrange it as two inequalities:
(i) –8 ≤ 5x – 3
(ii) 5x – 3 < 7
2. The first inequality is: –8 ≤ 5x – 3
• This can be simplified as follows:
$\begin{array}{ll}
{}&-8 &{}\le{}& {5x-3} &{} \\
\Rightarrow &-8-5x&{}\le{}& 5x-3-5x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &-8-5x&{}\le{}& -3 &{} \\
\Rightarrow &-8-5x+8&{}\le{}& -3+8 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &-5x&{}\le{}& 5 &{} \\
\Rightarrow &\frac{-5x}{-5}&{}\ge{}& \frac{5}{-5} &\color {green}{\text{(Rule 3)}} \\
\Rightarrow &x&{}\ge{}& -1 &{} \\
\end{array}$
3. The second inequality is: 5x – 3 < 7
• This can be simplified as follows:
$\begin{array}{ll}
{}&5x-3 &{}<{}& {7} &{} \\
\Rightarrow &5x-3+3&{}<{}& 7+3 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &5x&{}<{}& 10 &{} \\
\Rightarrow &\frac{5x}{5}&{}<{}& \frac{10}{5} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 2 &{} \\
\end{array}$
4. Now we can assemble the results:
♦ From (2), we see that, x must be greater than or equal to -1
♦ From (3), we see that, x must be less than 2
• These two results can be combined as: -1 ≤ x < 2
• In interval form, this can be written as: [-1,2)
• The graphical representation is shown in fig.6.26 below:
 |
| Fig.6.26 |
Solved example 6.18
Solve $-5 \le \frac{5-3x}{2}\le 8$
Solution:
1. This is a double inequality. We can rearrange it as two inequalities:
(i) $-5 \le \frac{5-3x}{2}$
(ii) $\frac{5-3x}{2}\le 8$
2. The first inequality is: $-5 \le \frac{5-3x}{2}$
• This can be simplified as follows:
$\begin{array}{ll}
{}&-5 &{}\le{}& {\frac{5-3x}{2}} &{} \\
\Rightarrow &-5 × 2&{}\le{}& {\frac{5-3x}{2}} × 2 &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &-10&{}\le{}& 5-3x &{} \\
\Rightarrow &-10+3x&{}\le{}& 5-3x+3x &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &-10+3x&{}\le{}& 5 &{} \\
\Rightarrow &-10+3x+10&{}\le{}& 5+10 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &3x&{}\le{}& 15 &{} \\
\Rightarrow &\frac{3x}{3}&{}\le{}& \frac{15}{3} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}\le{}& 5 &{} \\
\end{array}$
3. The second inequality is: $\frac{5-3x}{2}\le 8$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\frac{5-3x}{2} &{}\le{}& {8} &{} \\
\Rightarrow &\frac{5-3x}{2} × 2&{}\le{}& 8 × 2 &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &5-3x&{}\le{}& 16 &{} \\
\Rightarrow &5-3x-5&{}\le{}& 16-5 &{\color {green}{\text{(Rule 1)}}} \\
\Rightarrow &-3x&{}\le{}& 11 &{} \\
\Rightarrow &\frac{-3x}{-3}&{}\ge{}& \frac{11}{-3} &\color {green}{\text{(Rule 3)}} \\
\Rightarrow &x&{}\ge{}& -\frac{11}{3} &{} \\
\end{array}$
4. Now we can assemble the results:
♦ From (2), we see that, x must be less than or equal to 5
♦ From (3), we see that, x must be greater than or equal to $-\frac{11}{3}$
• These two results can be combined as: $-\frac{11}{3} \le x \le 5$
• In interval form, this can be written as: $\left[-\frac{11}{3}, 5 \right]$
• The graphical representation is shown in fig.6.27 below:
 |
| Fig.6.27 |
Solved example 6.19
Solve the system of inequalities:
3x – 7 < 5 + x
11 – 5x ≤ 1
and represent the solutions on the number line.
Solution:
1. The first inequality is: 3x – 7 < 5 + x
• This can be simplified as follows:
$\begin{array}{ll}
{}&3x-7 &{}<{}& 5+x &{} \\
\Rightarrow &3x-7+(7-x)&{}<{}& 5+x+(7-x) &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &2x&{}<{}& 12 &{} \\
\Rightarrow &\frac{2x}{2}&{}<{}& \frac{12}{2} &\color {green}{\text{(Rule 2)}} \\
\Rightarrow &x&{}<{}& 6 &{} \\
\end{array}$
2. The second inequality is: 11 – 5x ≤ 1
• This can be simplified as follows:
$\begin{array}{ll}
{}&11-5x &{}\le{}& {1} &{} \\
\Rightarrow &11-5x-11&{}\le{}& 1-11 &\color {green}{\text{(Rule 1)}} \\
\Rightarrow &-5x&{}\le{}& -10 &{} \\
\Rightarrow &\frac{-5x}{-5}&{}\ge{}& \frac{-10}{-5} &\color {green}{\text{(Rule 3)}} \\
\Rightarrow &x&{}\ge{}& 2 &{} \\
\end{array}$
3. Now we can assemble the results:
♦ From (1), we see that, x must be less than 6
♦ From (2), we see that, x must be greater than or equal to 2
• These two results can be combined as: 2 ≤ x < 6
• In interval form, this can be written as: [2,6)
• The graphical representation is shown in fig.6.28 below:
 |
| Fig.6.28 |
Solved example 6.20
In an experiment, a solution of hydrochloric acid is to be kept between 30° and 35° Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by $C=\frac{5}{9}(F-32)$, where C and F represent temperature in degree Celsius and degree Fahrenheit, respectively.
Solution:
1. Let the temperature in degree Celsius be TC
Then we can write: 30 < TC <35
2. The conversion formula is: $C=\frac{5}{9}(F-32)$
• This can be rearranged as:
$\begin{array}{ll}
{}&C &{}={}& \frac{5}{9}(F-32) &{} \\
\Rightarrow &\frac{9C}{5}&{}={}& F-32 &{} \\
\Rightarrow &F&{}={}& \frac{9C}{5}+32 &{} \\
\end{array}$
3. Now we can convert from C to F:
• When the Celsius thermometer shows a temperature of 30° Celsius, the Fahrenheit thermometer will show a temperature of $\frac{9 × 30}{5}+32=86$
• When the Celsius thermometer shows a temperature of 35° Celsius, the
Fahrenheit thermometer will show a temperature of $\frac{9 ×
35}{5}+32=95$
4. So based on step (1), we can write:
If the temperature in degree Fahrenheit is TF, then the required range will be:
86 < TF < 95
Solved example 6.21
A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18%?
Solution:
1. "a 12% solution of acid" means:
If we take 100 litres of that solution, 12 litres will be acid and the remaining 88 litres will be water.
• It is a method for specifying the acid content of a solution.
2. So out of the 600 litres, (6 × 12) = 72 litres will be acid.
3. Now consider the 30% solution.
♦ If we take 100 litres of that solution, there will be 30 litres of acid in it.
♦ So if we take 1 litre of that solution, there will be 0.3 litres of acid in it.
♦ So if we take V litres of that solution, there will be 0.3V litres of acid in it.
4. Let us add V litres of the 30% solution to the original 600 litres of 12 % solution.
• Then the total volume = (600+V) litres
♦ This total volume has an original 72 litres of acid.
♦ Additionally, now it has 0.3V litres of acid from the newly added V litres
• So the total volume of acid in the newly prepared solution will be (72+0.3V)
5. Now we can write the quantities:
(600+V) litres of the newly prepared solution has (72+0.3V) litres of acid
6. So the acid content of the newly prepared solution = $\left(\frac{72+0.3V}{600+V}\right) × 100$
7. The acid content of the newly prepared solution must be more than 15% but less than 18%.
• So we can write: $15 < \left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right] < 18$
8. This is a double inequality. We can rearrange it as two inequalities:
(i) $15 < \left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right]$
(ii) $\left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right] < 18$
9. The first inequality is: $15 < \left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right]$
• This can be simplified as follows:
$\begin{array}{ll}
{}&15 &{}<{}& \left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right] &{} \\
\Rightarrow
&15 × (600+V)&{}<{}&
\left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right] × (600+V)
&\color {green}{\text{(Rule 2)}} \\
\Rightarrow &15 × (600+V)&{}<{}& (72+0.3V) × 100
&{} \\
\Rightarrow &15 × (600+V) × \frac{1}{100}&{}<{}& (72+0.3V) × 100 × \frac{1}{100}
&{\color {green}{\text{(Rule 2)}}} \\
\Rightarrow &0.15 × (600+V)&{}<{}& 72+0.3V
&{} \\
\Rightarrow &90+0.15V&{}<{}& 72+0.3V
&{} \\
\Rightarrow &90+0.15V-90 - 0.3V&{}<{}& 72+0.3V - 90 -0.3V
&{\color {green}{\text{(Rule 1)}}} \\
\Rightarrow &-0.15V&{}<{}& -18
&{} \\
\Rightarrow &-0.15V × -1 × \frac{1}{0.15}&{}>{}& -18 × -1 × \frac{1}{0.15}
&{\color {green}{\text{(Rule 3)}}} \\
\Rightarrow &V&{}>{}& 18 × \frac{1}{0.15}
&{} \\
\Rightarrow &V&{}>{}& 120
&{} \\
\end{array}$
10. The second inequality is: $\left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right] < 18$
• This can be simplified as follows:
$\begin{array}{ll}
{}&\left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right] &{}<{}& {18} &{} \\
\Rightarrow
&\left[\left(\frac{72+0.3V}{600+V}\right) × 100 \right] ×
\frac{600+V}{100}&{}<{}& 18 × \frac{600+V}{100} &\color
{green}{\text{(Rule 2)}} \\
\Rightarrow &72+0.3V&{}<{}& 0.18 × (600+V) &{} \\
\Rightarrow &72+0.3V&{}<{}& 108+0.18V &{} \\
\Rightarrow &72+0.3V-72-0.18V&{}<{}& 108+0.18V-72-0.18V &{\color {green}{\text{(Rule 1)}}} \\
\Rightarrow &0.12V&{}<{}& 36 &{} \\
\Rightarrow &0.12V × \frac{1}{0.12}&{}<{}& 36 × \frac{1}{0.12} &{\color {green}{\text{(Rule 2)}}} \\
\Rightarrow &V&{}<{}& 300 &{} \\
\end{array}$
11. Now we can assemble the results:
♦ From (9), we see that, V must be greater than 120 litres.
♦ From (10), we see that, V must be less than 300 litres.
• These two results can be combined as: 120 < V < 300
12. If the additional volume V taken from the 30% solution is greater than 120 litres but less than 300 litres, then the acid strength of the resulting solution will be greater than 15% but less than 18%
The link below gives some more miscellaneous examples:
• In the next chapter, we will see permutations and combinations.
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