Chapter 5: Linear Inequalities

Start with $$4x + 3 < 6x + 7$$.

Subtract $$6x$$ from both sides:

$$4x + 3 - 6x < 6x + 7 - 6x$$

$$-2x + 3 < 7$$

Subtract $$3$$ from both sides:

$$-2x < 4$$

Divide both sides by $$-2$$. Since we divide by a negative number, the inequality sign is reversed:

$$x > -2$$

Hence the solution set is $$\{x \in \mathbb{R} : x > -2\} = (-2, \infty)$$.

Start with $$\dfrac{5 - 2x}{3} \leq \dfrac{x}{6} - 5$$.

Multiply both sides by $$6$$ (positive, so sign unchanged):

$$6 \cdot \dfrac{5 - 2x}{3} \leq 6 \cdot \dfrac{x}{6} - 6 \cdot 5$$

$$2(5 - 2x) \leq x - 30$$

$$10 - 4x \leq x - 30$$

Add $$4x$$ to both sides:

$$10 \leq 5x - 30$$

Add $$30$$ to both sides:

$$40 \leq 5x$$

Divide both sides by $$5$$:

$$8 \leq x$$, i.e. $$x \geq 8$$.

Hence the solution set is $$\{x \in \mathbb{R} : x \geq 8\} = [8, \infty)$$.

Start with $$7x + 3 < 5x + 9$$.

Subtract $$5x$$ from both sides:

$$2x + 3 < 9$$

Subtract $$3$$ from both sides:

$$2x < 6$$

Divide both sides by $$2$$ (positive, so sign unchanged):

$$x < 3$$

Hence the solution set is $$(-\infty, 3)$$.

Graph on the number line: Draw a horizontal number line marking integers near $$3$$. Place an open circle at $$x = 3$$ (to indicate $$x = 3$$ is not included) and shade the line to the left of $$3$$ (extending towards $$-\infty$$) with an arrow.

Start with $$\dfrac{3x - 4}{2} \geq \dfrac{x + 1}{4} - 1$$.

Multiply both sides by $$4$$ (positive, so sign unchanged):

$$4 \cdot \dfrac{3x - 4}{2} \geq 4 \cdot \dfrac{x + 1}{4} - 4$$

$$2(3x - 4) \geq (x + 1) - 4$$

$$6x - 8 \geq x - 3$$

Subtract $$x$$ from both sides:

$$5x - 8 \geq -3$$

Add $$8$$ to both sides:

$$5x \geq 5$$

Divide both sides by $$5$$:

$$x \geq 1$$

Hence the solution set is $$[1, \infty)$$.

Graph on the number line: Place a filled (closed) circle at $$x = 1$$ (since $$x = 1$$ is included) and shade the line to the right of $$1$$ with an arrow pointing towards $$+\infty$$.

Let $$x$$ be the marks the student obtains in the annual examination.

The average of the three examinations is $$\dfrac{62 + 48 + x}{3}$$.

For the average to be at least $$60$$, we need

$$\dfrac{62 + 48 + x}{3} \geq 60$$

$$\dfrac{110 + x}{3} \geq 60$$

Multiplying both sides by $$3$$ (positive):

$$110 + x \geq 180$$

Subtracting $$110$$ from both sides:

$$x \geq 70$$

Thus the student must obtain a minimum of $$70$$ marks in the annual examination.

Let $$x$$ be the smaller of the two consecutive odd natural numbers. Then the next consecutive odd number is $$x + 2$$.

Given conditions:

(a) Both numbers are larger than $$10$$, so the smaller one satisfies $$x > 10$$. Since $$x$$ is an odd natural number, this gives $$x \geq 11$$.

(b) Their sum is less than $$40$$:

$$x + (x + 2) < 40$$

$$2x + 2 < 40$$

$$2x < 38$$

$$x < 19$$

Combining (a) and (b): $$11 \leq x < 19$$, with $$x$$ odd. So $$x \in \{11, 13, 15, 17\}$$.

The corresponding pairs $$(x, x + 2)$$ are:

$$(11, 13),\ (13, 15),\ (15, 17),\ (17, 19).$$

Start with $$4x + 3 < 5x + 7$$.

Subtract $$5x$$ from both sides:

$$4x + 3 - 5x < 7$$

$$-x + 3 < 7$$

Subtract $$3$$ from both sides:

$$-x < 4$$

Multiply both sides by $$-1$$ (negative, so the inequality reverses):

$$x > -4$$

Hence the solution set is $$(-4, \infty)$$.

Start with $$3x - 7 > 5x - 1$$.

Subtract $$5x$$ from both sides:

$$3x - 5x - 7 > -1$$

$$-2x - 7 > -1$$

Add $$7$$ to both sides:

$$-2x > 6$$

Divide both sides by $$-2$$ (negative, so the inequality reverses):

$$x < -3$$

Hence the solution set is $$(-\infty, -3)$$.

Expand both sides of $$3(x - 1) \leq 2(x - 3)$$:

$$3x - 3 \leq 2x - 6$$

Subtract $$2x$$ from both sides:

$$x - 3 \leq -6$$

Add $$3$$ to both sides:

$$x \leq -3$$

Hence the solution set is $$(-\infty, -3]$$.

Expand both sides of $$3(2 - x) \geq 2(1 - x)$$:

$$6 - 3x \geq 2 - 2x$$

Add $$3x$$ to both sides:

$$6 \geq 2 + x$$

Subtract $$2$$ from both sides:

$$4 \geq x$$, i.e. $$x \leq 4$$.

Hence the solution set is $$(-\infty, 4]$$.

Start with $$x + \dfrac{x}{2} + \dfrac{x}{3} < 11$$.

Take the LCM of the denominators on the left ($$\mathrm{LCM}(1, 2, 3) = 6$$):

$$\dfrac{6x + 3x + 2x}{6} < 11$$

$$\dfrac{11x}{6} < 11$$

Multiply both sides by $$6$$ (positive):

$$11x < 66$$

Divide both sides by $$11$$:

$$x < 6$$

Hence the solution set is $$(-\infty, 6)$$.

Start with $$\dfrac{x}{3} > \dfrac{x}{2} + 1$$.

Multiply both sides by $$6$$ (positive, so sign unchanged):

$$2x > 3x + 6$$

Subtract $$3x$$ from both sides:

$$-x > 6$$

Multiply both sides by $$-1$$ (negative, so the inequality reverses):

$$x < -6$$

Hence the solution set is $$(-\infty, -6)$$.

Start with $$\dfrac{3(x - 2)}{5} \leq \dfrac{5(2 - x)}{3}$$.

Multiply both sides by $$15$$ (positive, so the inequality sign is unchanged):

$$3 \cdot 3(x - 2) \leq 5 \cdot 5(2 - x)$$

$$9(x - 2) \leq 25(2 - x)$$

$$9x - 18 \leq 50 - 25x$$

Add $$25x$$ to both sides:

$$34x - 18 \leq 50$$

Add $$18$$ to both sides:

$$34x \leq 68$$

Divide both sides by $$34$$:

$$x \leq 2$$

Hence the solution set is $$(-\infty, 2]$$.

Start with $$\dfrac{1}{2}\left(\dfrac{3x}{5} + 4\right) \geq \dfrac{1}{3}(x - 6)$$.

Multiply both sides by $$30$$ (the LCM of $$2, 5, 3$$; positive, so the sign is unchanged):

$$30 \cdot \dfrac{1}{2}\left(\dfrac{3x}{5} + 4\right) \geq 30 \cdot \dfrac{1}{3}(x - 6)$$

$$15\left(\dfrac{3x}{5} + 4\right) \geq 10(x - 6)$$

$$\dfrac{15 \cdot 3x}{5} + 15 \cdot 4 \geq 10x - 60$$

$$9x + 60 \geq 10x - 60$$

Subtract $$9x$$ from both sides:

$$60 \geq x - 60$$

Add $$60$$ to both sides:

$$120 \geq x$$, i.e. $$x \leq 120$$.

Hence the solution set is $$(-\infty, 120]$$.

Expand both sides of $$2(2x + 3) - 10 < 6(x - 2)$$:

$$4x + 6 - 10 < 6x - 12$$

$$4x - 4 < 6x - 12$$

Subtract $$4x$$ from both sides:

$$-4 < 2x - 12$$

Add $$12$$ to both sides:

$$8 < 2x$$

Divide both sides by $$2$$ (positive):

$$4 < x$$, i.e. $$x > 4$$.

Hence the solution set is $$(4, \infty)$$.

Simplify both sides of $$37 - (3x + 5) \geq 9x - 8(x - 3)$$.

Left side: $$37 - 3x - 5 = 32 - 3x$$.

Right side: $$9x - 8x + 24 = x + 24$$.

So the inequality becomes

$$32 - 3x \geq x + 24$$

Subtract $$x$$ from both sides:

$$32 - 4x \geq 24$$

Subtract $$32$$ from both sides:

$$-4x \geq -8$$

Divide both sides by $$-4$$ (negative, so the inequality reverses):

$$x \leq 2$$

Hence the solution set is $$(-\infty, 2]$$.

Start with $$\dfrac{x}{4} < \dfrac{5x - 2}{3} - \dfrac{7x - 3}{5}$$.

Multiply both sides by $$60$$ (LCM of $$4, 3, 5$$; positive, so the sign is unchanged):

$$15x < 20(5x - 2) - 12(7x - 3)$$

$$15x < (100x - 40) - (84x - 36)$$

$$15x < 100x - 40 - 84x + 36$$

$$15x < 16x - 4$$

Subtract $$15x$$ from both sides:

$$0 < x - 4$$

So $$x > 4$$.

Hence the solution set is $$(4, \infty)$$.

Start with $$\dfrac{2x - 1}{3} \geq \dfrac{3x - 2}{4} - \dfrac{2 - x}{5}$$.

Multiply both sides by $$60$$ (LCM of $$3, 4, 5$$; positive, so the sign is unchanged):

$$20(2x - 1) \geq 15(3x - 2) - 12(2 - x)$$

$$40x - 20 \geq (45x - 30) - (24 - 12x)$$

$$40x - 20 \geq 45x - 30 - 24 + 12x$$

$$40x - 20 \geq 57x - 54$$

Subtract $$40x$$ from both sides:

$$-20 \geq 17x - 54$$

Add $$54$$ to both sides:

$$34 \geq 17x$$

Divide both sides by $$17$$ (positive):

$$2 \geq x$$, i.e. $$x \leq 2$$.

Hence the solution set is $$(-\infty, 2]$$.

Start with $$3x - 2 < 2x + 1$$.

Subtract $$2x$$ from both sides:

$$x - 2 < 1$$

Add $$2$$ to both sides:

$$x < 3$$

Hence the solution set is $$(-\infty, 3)$$.

Graph on the number line: Draw a number line and place an open circle at $$x = 3$$ (since $$x = 3$$ is excluded). Shade the portion of the line to the left of $$3$$, with an arrow pointing towards $$-\infty$$.

Start with $$5x - 3 \geq 3x - 5$$.

Subtract $$3x$$ from both sides:

$$2x - 3 \geq -5$$

Add $$3$$ to both sides:

$$2x \geq -2$$

Divide both sides by $$2$$ (positive):

$$x \geq -1$$

Hence the solution set is $$[-1, \infty)$$.

Graph on the number line: Place a filled (closed) circle at $$x = -1$$ (since $$x = -1$$ is included). Shade the portion of the line to the right of $$-1$$, with an arrow pointing towards $$+\infty$$.

Expand both sides of $$3(1 - x) < 2(x + 4)$$:

$$3 - 3x < 2x + 8$$

Subtract $$2x$$ from both sides:

$$3 - 5x < 8$$

Subtract $$3$$ from both sides:

$$-5x < 5$$

Divide both sides by $$-5$$ (negative, so the inequality reverses):

$$x > -1$$

Hence the solution set is $$(-1, \infty)$$.

Graph on the number line: Place an open circle at $$x = -1$$ (since $$x = -1$$ is excluded). Shade the portion of the line to the right of $$-1$$, with an arrow pointing towards $$+\infty$$.

Start with $$\dfrac{x}{2} \geq \dfrac{5x - 2}{3} - \dfrac{7x - 3}{5}$$.

Multiply both sides by $$30$$ (LCM of $$2, 3, 5$$; positive, so the sign is unchanged):

$$15x \geq 10(5x - 2) - 6(7x - 3)$$

$$15x \geq 50x - 20 - 42x + 18$$

$$15x \geq 8x - 2$$

Subtract $$8x$$ from both sides:

$$7x \geq -2$$

Divide both sides by $$7$$ (positive):

$$x \geq -\dfrac{2}{7}$$

Hence the solution set is $$\left[-\dfrac{2}{7}, \infty\right)$$.

Graph on the number line: Place a filled circle at $$x = -\dfrac{2}{7}$$ (since $$x = -\tfrac{2}{7}$$ is included). Shade the portion of the line to the right of $$-\dfrac{2}{7}$$, with an arrow pointing towards $$+\infty$$.

Let $$x$$ be the marks Ravi obtains in the third unit test.

The average of his three tests is $$\dfrac{70 + 75 + x}{3}$$.

For the average to be at least $$60$$, we require

$$\dfrac{70 + 75 + x}{3} \geq 60$$

$$\dfrac{145 + x}{3} \geq 60$$

Multiply both sides by $$3$$ (positive):

$$145 + x \geq 180$$

Subtract $$145$$ from both sides:

$$x \geq 35$$

Hence Ravi must obtain at least $$35$$ marks in the third test.

Let $$x$$ be Sunita's marks in the fifth examination.

Her average across the five exams is $$\dfrac{87 + 92 + 94 + 95 + x}{5}$$.

For Grade 'A', this average must be at least $$90$$:

$$\dfrac{87 + 92 + 94 + 95 + x}{5} \geq 90$$

Sum the constants: $$87 + 92 + 94 + 95 = 368$$, so

$$\dfrac{368 + x}{5} \geq 90$$

Multiply both sides by $$5$$ (positive):

$$368 + x \geq 450$$

Subtract $$368$$ from both sides:

$$x \geq 82$$

Hence Sunita must obtain at least $$82$$ marks in the fifth examination. (Since each exam is out of $$100$$, this is achievable.)

Let $$x$$ be the smaller of the two consecutive odd positive integers. Then the larger is $$x + 2$$.

Given:

(a) Both numbers are smaller than $$10$$. Since the larger one is $$x + 2$$, we need $$x + 2 < 10$$, i.e. $$x < 8$$.

(b) Their sum is more than $$11$$:

$$x + (x + 2) > 11$$

$$2x + 2 > 11$$

$$2x > 9$$

$$x > \dfrac{9}{2} = 4.5$$

So we need $$4.5 < x < 8$$ with $$x$$ an odd positive integer. The odd integers in this range are $$x = 5$$ and $$x = 7$$.

The corresponding pairs $$(x, x + 2)$$ are $$(5, 7)$$ and $$(7, 9)$$.

Let $$x$$ be the smaller of the two consecutive even positive integers. Then the larger is $$x + 2$$.

Given:

(a) Both numbers are larger than $$5$$. Since $$x$$ is the smaller one, we need $$x > 5$$. As $$x$$ is an even positive integer, this gives $$x \geq 6$$.

(b) Their sum is less than $$23$$:

$$x + (x + 2) < 23$$

$$2x + 2 < 23$$

$$2x < 21$$

$$x < \dfrac{21}{2} = 10.5$$

So $$6 \leq x \leq 10$$ with $$x$$ even. Hence $$x \in \{6, 8, 10\}$$.

The corresponding pairs $$(x, x + 2)$$ are $$(6, 8)$$, $$(8, 10)$$ and $$(10, 12)$$.

Let the length (in cm) of the shortest side of the triangle be $$x$$.

Then:

Longest side $$= 3x$$ cm.

Third side $$= 3x - 2$$ cm (it is $$2$$ cm shorter than the longest side).

Perimeter $$= x + 3x + (3x - 2) = 7x - 2$$ cm.

The perimeter is at least $$61$$ cm:

$$7x - 2 \geq 61$$

Add $$2$$ to both sides:

$$7x \geq 63$$

Divide both sides by $$7$$ (positive):

$$x \geq 9$$

Hence the shortest side must be at least $$9$$ cm. The minimum length of the shortest side is $$9$$ cm.

Let $$x$$ cm be the length of the shortest piece. Then the three pieces have lengths $$x$$, $$x + 3$$ and $$2x$$ cm (as stated in the hint).

Condition 1: The three pieces together cannot exceed the original board length of $$91$$ cm:

$$x + (x + 3) + 2x \leq 91$$

$$4x + 3 \leq 91$$

$$4x \leq 88$$

$$x \leq 22$$

Condition 2: The third piece is at least $$5$$ cm longer than the second:

$$2x \geq (x + 3) + 5$$

$$2x \geq x + 8$$

$$x \geq 8$$

Combining the two conditions:

$$8 \leq x \leq 22$$

Hence the length of the shortest board may be anywhere from $$8$$ cm to $$22$$ cm inclusive.

We have the double inequality $$-8 \leq 5x - 3 < 7$$.

Add $$3$$ to each of the three members:

$$-8 + 3 \leq 5x - 3 + 3 < 7 + 3$$

$$-5 \leq 5x < 10$$

Divide each member by $$5$$ (positive, so the inequalities are preserved):

$$-1 \leq x < 2$$

Hence the solution set is $$[-1, 2)$$.

We have $$-5 \leq \dfrac{5 - 3x}{2} \leq 8$$.

Multiply every member by $$2$$ (positive, so the inequalities are preserved):

$$-10 \leq 5 - 3x \leq 16$$

Subtract $$5$$ from each member:

$$-15 \leq -3x \leq 11$$

Divide every member by $$-3$$ (negative — both inequality signs reverse):

$$\dfrac{-15}{-3} \geq x \geq \dfrac{11}{-3}$$

$$5 \geq x \geq -\dfrac{11}{3}$$

Rewriting in the usual order:

$$-\dfrac{11}{3} \leq x \leq 5$$

Hence the solution set is $$\left[-\dfrac{11}{3}, 5\right]$$.

Solve each inequality individually, then intersect the solution sets.

Inequality 1: $$3x - 7 < 5 + x$$.

Subtract $$x$$ from both sides: $$2x - 7 < 5$$.

Add $$7$$ to both sides: $$2x < 12$$.

Divide by $$2$$: $$x < 6$$. So the solution set is $$(-\infty, 6)$$.

Inequality 2: $$11 - 5x \leq 1$$.

Subtract $$11$$ from both sides: $$-5x \leq -10$$.

Divide both sides by $$-5$$ (negative — the sign reverses): $$x \geq 2$$. So the solution set is $$[2, \infty)$$.

Intersection: $$(-\infty, 6) \cap [2, \infty) = [2, 6)$$.

Hence the solution set of the system is $$\{x \in \mathbb{R} : 2 \leq x < 6\} = [2, 6)$$.

Graph on the number line: Place a filled circle at $$x = 2$$ (included) and an open circle at $$x = 6$$ (excluded). Shade the segment between them.

The temperature in Celsius must satisfy $$30 < C < 35$$.

Substitute $$C = \dfrac{5}{9}(F - 32)$$:

$$30 < \dfrac{5}{9}(F - 32) < 35$$

Multiply every member by $$\dfrac{9}{5}$$ (positive, so the inequalities are preserved):

$$30 \cdot \dfrac{9}{5} < F - 32 < 35 \cdot \dfrac{9}{5}$$

$$54 < F - 32 < 63$$

Add $$32$$ to each member:

$$86 < F < 95$$

Hence the temperature must be kept between $$86^\circ$$ F and $$95^\circ$$ F.

Let $$x$$ litres of $$30\%$$ acid solution be added.

Total volume of resulting mixture $$= (600 + x)$$ litres.

Total acid content $$=$$ acid from $$12\%$$ solution $$+$$ acid from $$30\%$$ solution

$$= \dfrac{12}{100} \times 600 + \dfrac{30}{100} \times x = 72 + 0.3x \text{ litres}.$$

The mixture must be more than $$15\%$$ but less than $$18\%$$ acid, so:

$$15\% < \dfrac{72 + 0.3x}{600 + x} \times 100 < 18\%$$

$$\dfrac{15}{100} < \dfrac{72 + 0.3x}{600 + x} < \dfrac{18}{100}$$

Multiply throughout by $$100(600 + x)$$. Since $$x > 0$$, $$(600 + x) > 0$$, so the inequality signs are preserved:

$$15(600 + x) < 100(72 + 0.3x) < 18(600 + x)$$

$$15(600 + x) < 7200 + 30x < 18(600 + x)$$

Lower part: $$15(600 + x) < 7200 + 30x$$

$$9000 + 15x < 7200 + 30x$$

$$1800 < 15x$$

$$x > 120$$

Upper part: $$7200 + 30x < 18(600 + x)$$

$$7200 + 30x < 10800 + 18x$$

$$12x < 3600$$

$$x < 300$$

Combining: $$120 < x < 300$$.

Hence between $$120$$ litres and $$300$$ litres (exclusive of both endpoints) of $$30\%$$ acid solution must be added.

We have $$2 \leq 3x - 4 \leq 5$$.

Add $$4$$ to each member:

$$2 + 4 \leq 3x - 4 + 4 \leq 5 + 4$$

$$6 \leq 3x \leq 9$$

Divide every member by $$3$$ (positive, so the inequalities are preserved):

$$2 \leq x \leq 3$$

Hence the solution set is $$[2, 3]$$.

Start with $$6 \leq -3(2x - 4) < 12$$.

Expand the middle expression: $$-3(2x - 4) = -6x + 12$$. So:

$$6 \leq -6x + 12 < 12$$

Subtract $$12$$ from every member:

$$-6 \leq -6x < 0$$

Divide every member by $$-6$$ (negative — both inequality signs reverse):

$$1 \geq x > 0$$

Rewriting in usual order:

$$0 < x \leq 1$$

Hence the solution set is $$(0, 1]$$.

Start with $$-3 \leq 4 - \dfrac{7x}{2} \leq 18$$.

Subtract $$4$$ from each member:

$$-3 - 4 \leq -\dfrac{7x}{2} \leq 18 - 4$$

$$-7 \leq -\dfrac{7x}{2} \leq 14$$

Multiply every member by $$-\dfrac{2}{7}$$ (negative — both inequality signs reverse):

$$-7 \cdot \left(-\dfrac{2}{7}\right) \geq x \geq 14 \cdot \left(-\dfrac{2}{7}\right)$$

$$2 \geq x \geq -4$$

Rewriting in usual order:

$$-4 \leq x \leq 2$$

Hence the solution set is $$[-4, 2]$$.

Start with $$-15 < \dfrac{3(x - 2)}{5} \leq 0$$.

Multiply every member by $$5$$ (positive, so the inequalities are preserved):

$$-75 < 3(x - 2) \leq 0$$

Divide every member by $$3$$ (positive):

$$-25 < x - 2 \leq 0$$

Add $$2$$ to each member:

$$-23 < x \leq 2$$

Hence the solution set is $$(-23, 2]$$.

Simplify the middle term first: $$-\dfrac{3x}{-5} = \dfrac{3x}{5}$$, so the inequality is

$$-12 < 4 + \dfrac{3x}{5} \leq 2$$

Subtract $$4$$ from each member:

$$-16 < \dfrac{3x}{5} \leq -2$$

Multiply every member by $$5$$ (positive):

$$-80 < 3x \leq -10$$

Divide every member by $$3$$ (positive):

$$-\dfrac{80}{3} < x \leq -\dfrac{10}{3}$$

Hence the solution set is $$\left(-\dfrac{80}{3}, -\dfrac{10}{3}\right]$$.

Start with $$7 \leq \dfrac{3x + 11}{2} \leq 11$$.

Multiply every member by $$2$$ (positive, so the inequalities are preserved):

$$14 \leq 3x + 11 \leq 22$$

Subtract $$11$$ from each member:

$$3 \leq 3x \leq 11$$

Divide every member by $$3$$ (positive):

$$1 \leq x \leq \dfrac{11}{3}$$

Hence the solution set is $$\left[1, \dfrac{11}{3}\right]$$.

Solve each inequality separately and intersect the solution sets.

Inequality 1: $$5x + 1 > -24$$.

Subtract $$1$$: $$5x > -25$$. Divide by $$5$$ (positive): $$x > -5$$.

Solution set: $$(-5, \infty)$$.

Inequality 2: $$5x - 1 < 24$$.

Add $$1$$: $$5x < 25$$. Divide by $$5$$ (positive): $$x < 5$$.

Solution set: $$(-\infty, 5)$$.

Intersection: $$(-5, \infty) \cap (-\infty, 5) = (-5, 5)$$.

Hence the solution set is $$\{x \in \mathbb{R} : -5 < x < 5\} = (-5, 5)$$.

Graph on the number line: Place open circles at $$x = -5$$ and $$x = 5$$ and shade the segment between them.

Inequality 1: $$2(x - 1) < x + 5$$.

Expand the left side: $$2x - 2 < x + 5$$.

Subtract $$x$$ from both sides: $$x - 2 < 5$$.

Add $$2$$ to both sides: $$x < 7$$. Solution set: $$(-\infty, 7)$$.

Inequality 2: $$3(x + 2) > 2 - x$$.

Expand the left side: $$3x + 6 > 2 - x$$.

Add $$x$$ to both sides: $$4x + 6 > 2$$.

Subtract $$6$$ from both sides: $$4x > -4$$.

Divide by $$4$$ (positive, so the inequality direction is preserved): $$x > -1$$. Solution set: $$(-1, \infty)$$.

Intersection of the two solution sets: $$(-\infty, 7) \cap (-1, \infty) = (-1, 7)$$.

Hence the solution set is $$\{x \in \mathbb{R} : -1 < x < 7\} = (-1, 7)$$.

Graph on the number line: Place open circles at $$x = -1$$ and $$x = 7$$, and shade the segment between them.

Inequality 1: $$3x - 7 > 2(x - 6)$$.

$$3x - 7 > 2x - 12$$

$$x > -5$$. Solution set: $$(-5, \infty)$$.

Inequality 2: $$6 - x > 11 - 2x$$.

Add $$2x$$ to both sides: $$6 + x > 11$$.

Subtract $$6$$: $$x > 5$$. Solution set: $$(5, \infty)$$.

Intersection: $$(-5, \infty) \cap (5, \infty) = (5, \infty)$$.

Hence the solution set is $$\{x \in \mathbb{R} : x > 5\} = (5, \infty)$$.

Graph on the number line: Place an open circle at $$x = 5$$ and shade the part of the line to the right of $$5$$, with an arrow pointing towards $$+\infty$$.

Inequality 1: $$5(2x - 7) - 3(2x + 3) \leq 0$$.

$$10x - 35 - 6x - 9 \leq 0$$

$$4x - 44 \leq 0$$

$$4x \leq 44$$

$$x \leq 11$$. Solution set: $$(-\infty, 11]$$.

Inequality 2: $$2x + 19 \leq 6x + 47$$.

Subtract $$6x$$: $$-4x + 19 \leq 47$$.

Subtract $$19$$: $$-4x \leq 28$$.

Divide by $$-4$$ (negative — sign reverses): $$x \geq -7$$. Solution set: $$[-7, \infty)$$.

Intersection: $$(-\infty, 11] \cap [-7, \infty) = [-7, 11]$$.

Hence the solution set is $$\{x \in \mathbb{R} : -7 \leq x \leq 11\} = [-7, 11]$$.

Graph on the number line: Place filled circles at $$x = -7$$ and $$x = 11$$, and shade the segment between them.

The temperature must satisfy $$68 < F < 77$$.

Substitute $$F = \dfrac{9}{5}C + 32$$:

$$68 < \dfrac{9}{5}C + 32 < 77$$

Subtract $$32$$ from each member:

$$36 < \dfrac{9}{5}C < 45$$

Multiply every member by $$\dfrac{5}{9}$$ (positive, so the inequalities are preserved):

$$36 \cdot \dfrac{5}{9} < C < 45 \cdot \dfrac{5}{9}$$

$$20 < C < 25$$

Hence the temperature in Celsius must lie strictly between $$20^\circ$$ C and $$25^\circ$$ C.

Let $$x$$ litres of the $$2\%$$ boric acid solution be added.

Total volume of resulting mixture $$= (640 + x)$$ litres.

Total boric acid in the mixture $$= 8\%$$ of $$640 + 2\%$$ of $$x$$

$$= \dfrac{8}{100} \times 640 + \dfrac{2}{100} \times x = 51.2 + 0.02x \text{ litres}.$$

The mixture must be more than $$4\%$$ but less than $$6\%$$ boric acid:

$$4\% < \dfrac{51.2 + 0.02x}{640 + x} \times 100 < 6\%$$

i.e. $$\dfrac{4}{100} < \dfrac{51.2 + 0.02x}{640 + x} < \dfrac{6}{100}$$.

Since $$x > 0$$, $$(640 + x) > 0$$, so multiplying through by $$100(640 + x)$$ preserves the inequalities:

$$4(640 + x) < 100(51.2 + 0.02x) < 6(640 + x)$$

$$4(640 + x) < 5120 + 2x < 6(640 + x)$$

Lower part: $$4(640 + x) < 5120 + 2x$$

$$2560 + 4x < 5120 + 2x$$

$$2x < 2560$$

$$x < 1280$$

Upper part: $$5120 + 2x < 6(640 + x)$$

$$5120 + 2x < 3840 + 6x$$

$$1280 < 4x$$

$$x > 320$$

Combining: $$320 < x < 1280$$.

Hence between $$320$$ and $$1280$$ litres (exclusive) of the $$2\%$$ solution must be added.

Let $$x$$ litres of water be added.

Total volume of mixture $$= (1125 + x)$$ litres.

Amount of acid is unchanged when water (no acid) is added:

Acid $$= \dfrac{45}{100} \times 1125 = 506.25$$ litres.

The resulting mixture must have an acid concentration of more than $$25\%$$ and less than $$30\%$$:

$$25\% < \dfrac{506.25}{1125 + x} \times 100 < 30\%$$

$$\dfrac{25}{100} < \dfrac{506.25}{1125 + x} < \dfrac{30}{100}$$

Since $$x \geq 0$$, $$(1125 + x) > 0$$, so multiplying through by $$100(1125 + x)$$ preserves the inequalities:

$$25(1125 + x) < 100 \cdot 506.25 < 30(1125 + x)$$

$$25(1125 + x) < 50625 < 30(1125 + x)$$

Lower part: $$25(1125 + x) < 50625$$

$$28125 + 25x < 50625$$

$$25x < 22500$$

$$x < 900$$

Upper part: $$50625 < 30(1125 + x)$$

$$50625 < 33750 + 30x$$

$$16875 < 30x$$

$$x > 562.5$$

Combining: $$562.5 < x < 900$$.

Hence the volume of water to be added must lie strictly between $$562.5$$ litres and $$900$$ litres.

For the given group, the chronological age $$\mathrm{CA} = 12$$ years.

Using the formula $$\mathrm{IQ} = \dfrac{\mathrm{MA}}{\mathrm{CA}} \times 100 = \dfrac{\mathrm{MA}}{12} \times 100 = \dfrac{100 \, \mathrm{MA}}{12}$$.

The given condition is

$$80 \leq \mathrm{IQ} \leq 140$$

Substitute the expression for IQ:

$$80 \leq \dfrac{100 \, \mathrm{MA}}{12} \leq 140$$

Multiply every member by $$\dfrac{12}{100}$$ (positive, so the inequalities are preserved):

$$80 \cdot \dfrac{12}{100} \leq \mathrm{MA} \leq 140 \cdot \dfrac{12}{100}$$

$$\dfrac{960}{100} \leq \mathrm{MA} \leq \dfrac{1680}{100}$$

$$9.6 \leq \mathrm{MA} \leq 16.8$$

Hence the mental age of the children in this group lies between $$9.6$$ and $$16.8$$ years (inclusive).