Solve the inequalities in Exercises 1 to 6

Q1. \(\quad 2 \leq 3x – 4 \leq 5\)

Solution

$$\begin{aligned}2&\leq 3x-4\leq 5\\ 2&\leq 3x-4\\ 3x-4&\geq 2\\ 3x&\geq 2+4\\ x&\geq \dfrac{6}{3}\\ x&\geq 2\\\\ 3x-4&\leq 5\\ 3x&\leq 9\\ x&\leq \dfrac{9}{3}\\ x&\leq 3\\\\ x&\in \left[ 2,3\right] \end{aligned}$$

Q2. \(\quad 6 \leq – 3 (2x – 4) \lt 12\)

Solution

$$\begin{aligned}6\leq -3\left( 2x-4\right) &\lt 12\\ -3\left( 2x-4\right) &\geq 6\\ -6x+12&\geq 6\\ -6x&\geq -6\\ -x&\geq -1\\ x&\leq 1\\\\ -3\left( 2x-4\right) &\lt12\\ -6x+12 &\lt12\\ -6x &\lt0\\ x &\gt0\\\\ x&\in \left[ 0,1\right] \end{aligned}$$

Q3. \(\quad -3\leq 4-\dfrac{7x}{2}\leq 18\)

Solution

$$\begin{aligned}-3\leq 4-\dfrac{7x}{2}&\leq 18\\ 4-\dfrac{7x}{2}&\geq -3\\ 8-7x&\geq -6\\ -7x&\geq -6-8\\ x&\leq \dfrac{14}{7}\\ x&\leq 2\\\\ 4-\dfrac{7x}{2}&\leq 18\\ 8-7x&\leq 36\\ -7x&\leq 36-8\\ -7x&\leq 28\\ x&\geq \dfrac{-28}{7}\\ x&\geq -4\\\\ x&\in \left[ -4,2\right] \end{aligned}$$

Q4. \(\quad -15 <\dfrac{3\left( x-2\right) }{5}\leq 0\)

Solution

$$\begin{aligned}-15 <\dfrac{3\left( x-2\right) }{5}&\leq 0\\ \dfrac{3\left( x-2\right) }{5} &\gt-15\\ 3x-6 &\gt-75\\ 3x &\gt-75+6\\ x &\gt-\dfrac{69}{3}\\ x &\gt-23\\\\ \dfrac{3\left( x-2\right) }{5}&\leq 0\\ 3x-6&\leq 0\\ 3x&\leq 6\\ x&\leq \dfrac{6}{3}\\ x&\leq 2\\\\ x&\in \left[ -23,2\right] \end{aligned}$$

Q5. \(\quad -12 \lt 4-\dfrac{3x}{-5}\leq 2\)

Solution

$$\begin{aligned}-12 \lt 4-\dfrac{3x}{-5}&\leq 2\\ 4-\dfrac{3x}{-5} &\gt-12\\ 20+3x &\gt-60\\ 3x &\gt-60-20\\ 3x &\gt-80\\ x &\gt\dfrac{-80}{3}\\\\ 4-\dfrac{3x}{-5}&\leq 2\\ 20+3x&\leq 10\\ 3x&\leq 10-20\\ x&\leq \dfrac{-10}{3}\\ x&\leq -\dfrac{10}{3}\\\\ x&\in \left[ -\dfrac{80}{3},-\dfrac{10}{3}\right] \end{aligned}$$

Q6. \(\quad 7\leq \dfrac{3x+11}{2}\leq 11\)

Solution

$$\begin{aligned}7\leq \dfrac{3x+11}{2}&\leq 11\\ \dfrac{3x+11}{2}&\geq 7\\ 3x+11&\geq 14\\ 3x&\geq 14-11\\ x&\geq 1\\\\ \dfrac{3x+11}{2}&\leq 11\\ 3x+11&\leq 22\\ 3x&\leq 22-11\\ x&\leq \dfrac{11}{3}\\\\ x&\in \left[ 1,\dfrac{11}{3}\right] \end{aligned}$$

Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.

Q7. \(\quad 5x + 1 \gt – 24,\; 5x – 1 \lt 24\)

Solution

$$\begin{aligned}5x+1 &\gt-24\\ 5x&\gt -25\\ x&\gt -5\\\\ 5x-1 &\lt 24\\ 5x &\lt25\\ x &\lt5\\\\ x&\in \left( -5,5\right) \end{aligned}$$
Fig. 5.2-misc-7-xi

Q8. \(\quad 2 (x – 1) \lt x + 5,\; 3 (x + 2) \gt 2 – x\)

Solution

$$\begin{aligned}2\left( x-1\right) &\lt x+5\\ 2x-2 &\lt x+5\\ 2x-x &\lt5+2\\ x &\lt7\\\\ 3\left( x+2\right) &\gt2-x\\ 3x+6 &\gt2-x\\ 3x+x &\gt2-6\\ 4x &\gt-4\\ x &\gt-1\\\\ x&\in \left( -1,7\right) \end{aligned}$$
Fig. 5.2-misc-8-xi

Q9. \(\quad 3x – 7 \gt 2 (x – 6) ,\; 6 – x \gt 11 – 2x\)

Solution

$$\begin{aligned}3x-7 &\gt 2\left( x-6\right) \\ 3x-7 &\gt 2x-12\\ 3x-2x &\gt -12+7\\ x &\gt -5\\\\ 6-x &\gt 11-2x\\ -x+2x &\gt 11-6\\ x &\gt 5\\\\ x&\in \left( 5,\infty \right) \end{aligned}$$
Fig. 5.2-misc-9-xi

Q10. \(\quad 5 (2x – 7) – 3 (2x + 3) \leq 0 ,\; 2x + 19 \leq 6x + 47\)

Solution

$$\begin{aligned}5\left( 2x-7\right) -3\left( 2x+3\right) &\leq 0\\ 10x-35-6x-9&\leq 0\\ 4x&\leq 44\\ x&\leq 11\\\\ 2x+19&\leq 6x+47\\ 2x-6x&\leq 47-19\\ -4x&\leq 28\\ x&\geq -\dfrac{28}{4}\\ x&\geq -7\\\\ x&\in \left[ -7,11\right] \end{aligned}$$
Fig. 5.2-misc-10-xi

Q11. A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by \[F = \dfrac{9}{5}C+32\]

Solution

The solution is to be kept between 68°F and 77°F. Using the given conversion formula \(F = \dfrac{9}{5}C + 32\), we convert this temperature range from Fahrenheit to Celsius.

\[ \begin{aligned} 77 &\ge \dfrac{9}{5}C + 32 \\ 385 &\ge 9C + 160 \\ 9C &\le 385 - 160 \\ C &\le \dfrac{225}{9} \\ C &\le 25 \\\\ 68 &\le \dfrac{9}{5}C + 32 \\ 340 &\le 9C + 160 \\ 9C &\ge 340 - 160 \\ C &\ge \dfrac{180}{9} \\ C &\ge 20 \\\\ C &\in [20, 25] \end{aligned} \]

Therefore, the required range of temperature in degree Celsius is between 20°C and 25°C.

Q12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Solution

Let the number of litres of the 2% boric acid solution added to the given solution be \(x\). The amount of pure boric acid present in 640 litres of the 8% solution is \(0.08 \times 640\) litres, and the amount of pure boric acid present in \(x\) litres of the 2% solution is \(0.02x\) litres.

After mixing, the total volume of the solution becomes \(640 + x\) litres, and the total amount of boric acid becomes \(0.08 \times 640 + 0.02x\) litres. Since the resulting mixture is to be more than 4% but less than 6% boric acid, the concentration of the mixture must satisfy the following inequality.

\[ \begin{aligned} 0.04 &\lt \frac{0.08 \times 640 + 0.02x}{640 + x} \lt 0.06 \\ 0.04(640 + x) &\lt 51.2 + 0.02x < 0.06(640 + x) \\ 25.6 + 0.04x &\lt 51.2 + 0.02x \\ 0.02x &\lt 25.6 \\ x &\lt 1280 \\ 51.2 + 0.02x &\lt 38.4 + 0.06x \\ 12.8 &\lt 0.04x \\ x &\gt 320 \end{aligned} \]

Combining both conditions, the value of \(x\) must satisfy \(320 < x < 1280\).

Therefore, the quantity of the 2% boric acid solution that must be added lies between 320 litres and 1280 litres.

Q13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Solution

Let the number of litres of water added to the given solution be \(x\). The amount of pure acid present in 1125 litres of the 45% solution is \(0.45 \times 1125\) litres.

After adding water, the total volume of the solution becomes \(1125 + x\) litres, while the amount of acid remains unchanged. Since the resulting mixture must contain more than 25% but less than 30% acid, the concentration of the mixture must satisfy the following inequality.

\[ \begin{aligned} 0.25 &\lt \frac{0.45 \times 1125}{1125 + x} \lt 0.30 \\ 0.25(1125 + x) &\lt 506.25 < 0.30(1125 + x) \\ 281.25 + 0.25x &\lt 506.25 \\ 0.25x &\lt 225 \\ x &\lt 900 \\ 506.25 &\lt 337.5 + 0.30x \\ 168.75 &\lt 0.30x \\ x &\gt 562.5 \end{aligned} \]

Combining both conditions, the value of \(x\) must satisfy \(562.5 < x < 900\).

Therefore, the quantity of water that must be added lies between 562.5 litres and 900 litres.

Q14. IQ of a person is given by the formula \[IQ=\dfrac{MA}{CA}\times 100\] where MA is mental age and CA is chronological age. If \(80 \leq IQ \leq 140\) for a group of 12 years old children, find the range of their mental age.

Solution

The IQ of a person is given by the formula \(IQ = \dfrac{MA}{CA} \times 100\). For the given group, the chronological age is \(CA = 12\) years. Let the mental age be \(MA\) years.

Since the IQ of the children lies between 80 and 140, the given condition can be written as the following double inequality.

\[ \begin{aligned} 80 &\le \dfrac{MA}{12} \times 100 \le 140 \\ 0.8 &\le \dfrac{MA}{12} \le 1.4 \\ 9.6 &\le MA \le 16.8 \end{aligned} \]

Therefore, the range of the mental age of the children lies between 9.6 years and 16.8 years.

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    Frequently Asked Questions

    A linear inequality is an inequality of the form \(ax + b < 0\), \(ax + b \le 0\), \(ax + b > 0\), or \(ax + b \ge 0\), where \(a\) and \(b\) are real numbers and \(a \ne 0\).

    A linear equation uses an equality sign \(=\) and has a unique solution, while a linear inequality uses \(<, \le, >, \ge\) and has a range of solutions.

    The symbols are less than \((<)\), less than or equal to \((\le)\), greater than \((>)\), and greater than or equal to \((\ge)\).

    The solution set is the collection of all real numbers that satisfy the given inequality.

    It is solved by isolating the variable using standard algebraic operations while maintaining the inequality sign.

    The inequality sign is reversed when both sides are multiplied or divided by a negative number.

    For \(2x - 5 < 3\), we get \(2x < 8\) and hence \(x < 4\).

    It is a graphical method where solutions are shown as points or intervals on the number line.

    Strict inequalities \((<, >)\) are represented using open circles to exclude the boundary point.

    Inclusive inequalities \((\le, \ge)\) are represented using closed circles to include the boundary point.

    Compound linear inequalities involve two inequalities connected by “and” or “or”.

    “And” means the intersection of solution sets, where both inequalities must be satisfied simultaneously.

    “Or” means the union of solution sets, where at least one inequality must be satisfied.

    For \(1 < x < 5\), the solution is all real numbers between 1 and 5.

    For \(x < -2\) or \(x > 3\), the solution includes numbers less than \(-2\) and greater than \(3\).

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