Exercise 5.1: Solution of Linear Equations and Real-World Problems
Solve (1–8):
1. $\frac{ay}{b} – \frac{by}{a} = a^2 – b^2$
Solution: $\frac{ay}{b} – \frac{by}{a} = a^2 – b^2$
or, $y \left(\frac{a}{b} – \frac{b}{a}\right) = a^2 – b^2$
or, $y \left(\frac{a^2 – b^2}{ab}\right) = a^2 – b^2$
or, $y = (a^2 – b^2) \times \frac{ab}{a^2 – b^2}$
or, $y = ab$
Required solution: $y = ab$
2. $(z + 1)(z – 2) = (z – 4)(z + 2)$
Solution: $(z + 1)(z – 2) = (z – 4)(z + 2)$
or, $z(z – 2) + 1(z – 2) = z(z + 2) – 4(z + 2)$
or, $z^2 – 2z + z – 2 = z^2 + 2z – 4z – 8$
or, $z^2 – z – 2 = z^2 – 2z – 8$ [subtracting $z^2$ from both sides]
or, $-z – 2 = -2z – 8$
or, $-z + 2z = -8 + 2$
or, $z = -6$
Required solution: $z = -6$
3. $\frac{4}{2x + 1} + \frac{9}{3x + 2} = \frac{25}{5x + 4}$
Solution: $\frac{4}{2x + 1} + \frac{9}{3x + 2} = \frac{25}{5x + 4}$
or, $\frac{4}{2x + 1} + \frac{9}{3x + 2} = \frac{10 + 15}{5x + 4}$
or, $\frac{4}{2x + 1} + \frac{9}{3x + 2} = \frac{10}{5x + 4} + \frac{15}{5x + 4}$
or, $\frac{4}{2x + 1} – \frac{10}{5x + 4} = \frac{15}{5x + 4} – \frac{9}{3x + 2}$
or, $\frac{4(5x + 4) – 10(2x + 1)}{(2x + 1)(5x + 4)} = \frac{15(3x + 2) – 9(5x + 4)}{(5x + 4)(3x + 2)}$ [omitting $(5x + 4)$ from the denominators of both sides]
or, $\frac{20x + 16 – 20x – 10}{2x + 1} = \frac{45x + 30 – 45x – 36}{3x + 2}$
or, $\frac{6}{2x + 1} = \frac{-6}{3x + 2}$ [dividing both sides by 6]
or, $\frac{1}{2x + 1} = \frac{-1}{3x + 2}$
or, $1 \cdot (3x + 2) = -1 \cdot (2x + 1)$
or, $3x + 2 = -2x – 1$
or, $3x + 2x = -1 – 2$
or, $5x = -3$
or, $x = -\frac{3}{5}$
Required solution: $x = -\frac{3}{5}$
4. $\frac{1}{x + 1} + \frac{1}{x + 4} = \frac{1}{x + 2} + \frac{1}{x + 3}$
Solution: $\frac{1}{x + 1} + \frac{1}{x + 4} = \frac{1}{x + 2} + \frac{1}{x + 3}$
or, $\frac{(x + 4) + (x + 1)}{(x + 1)(x + 4)} = \frac{(x + 3) + (x + 2)}{(x + 2)(x + 3)}$
or, $\frac{2x + 5}{x^2 + 4x + x + 4} = \frac{2x + 5}{x^2 + 3x + 2x + 6}$
or, $\frac{2x + 5}{x^2 + 5x + 4} = \frac{2x + 5}{x^2 + 5x + 6}$
or, $\frac{2x + 5}{x^2 + 5x + 4} – \frac{2x + 5}{x^2 + 5x + 6} = 0$
or, $(2x + 5) \left(\frac{1}{x^2 + 5x + 4} – \frac{1}{x^2 + 5x + 6}\right) = 0$
Since $\frac{1}{x^2 + 5x + 4} – \frac{1}{x^2 + 5x + 6} \neq 0$,
therefore, $2x + 5 = 0$
or, $2x = -5$
or, $x = -\frac{5}{2}$
Required solution: $x = -\frac{5}{2}$
5. $\frac{a}{x – a} + \frac{b}{x – b} = \frac{a + b}{x – a – b}$
Solution: $\frac{a}{x – a} + \frac{b}{x – b} = \frac{a + b}{x – a – b}$
or, $\frac{a}{x – a} + \frac{b}{x – b} = \frac{a}{x – a – b} + \frac{b}{x – a – b}$
or, $\frac{a}{x – a} – \frac{a}{x – a – b} = \frac{b}{x – a – b} – \frac{b}{x – b}$
or, $a \left(\frac{1}{x – a} – \frac{1}{x – a – b}\right) = b \left(\frac{1}{x – a – b} – \frac{1}{x – b}\right)$
or, $a \left[\frac{(x – a – b) – (x – a)}{(x – a)(x – a – b)}\right] = b \left[\frac{(x – b) – (x – a – b)}{(x – a – b)(x – b)}\right]$
or, $a \left[\frac{x – a – b – x + a}{(x – a)(x – a – b)}\right] = b \left[\frac{x – b – x + a + b}{(x – a – b)(x – b)}\right]$
or, $a \left[\frac{-b}{(x – a)(x – a – b)}\right] = b \left[\frac{a}{(x – a – b)(x – b)}\right]$
or, $\frac{-ab}{(x – a)(x – a – b)} = \frac{ab}{(x – a – b)(x – b)}$ [multiplying both sides by $(x – a – b)$ and dividing by $ab$]
or, $\frac{-1}{x – a} = \frac{1}{x – b}$
or, $-1 \cdot (x – b) = 1 \cdot (x – a)$
or, $-x + b = x – a$
or, $-x – x = -a – b$
or, $-2x = -(a + b)$
or, $2x = a + b$
or, $x = \frac{a + b}{2}$
Required solution: $x = \frac{a + b}{2}$
6. $\frac{x – a}{b} + \frac{x – b}{a} + \frac{x – 3a – 3b}{a + b} = 0$
Solution: $\frac{x – a}{b} + \frac{x – b}{a} + \frac{x – 3a – 3b}{a + b} = 0$
or, $\left(\frac{x – a}{b} – 1\right) + \left(\frac{x – b}{a} – 1\right) + \left(\frac{x – 3a – 3b}{a + b} + 2\right) = 0$
or, $\frac{x – a – b}{b} + \frac{x – b – a}{a} + \frac{x – 3a – 3b + 2(a + b)}{a + b} = 0$
or, $\frac{x – a – b}{b} + \frac{x – a – b}{a} + \frac{x – 3a – 3b + 2a + 2b}{a + b} = 0$
or, $\frac{x – a – b}{b} + \frac{x – a – b}{a} + \frac{x – a – b}{a + b} = 0$
or, $(x – a – b) \left(\frac{1}{b} + \frac{1}{a} + \frac{1}{a + b}\right) = 0$
Since $\left(\frac{1}{b} + \frac{1}{a} + \frac{1}{a + b}\right) \neq 0$,
therefore, $x – a – b = 0$
or, $x = a + b$
Required solution: $x = a + b$
7. $\frac{x – a}{a^2 – b^2} = \frac{x – b}{b^2 – a^2}$
Solution: $\frac{x – a}{a^2 – b^2} = \frac{x – b}{b^2 – a^2}$
or, $\frac{x – a}{a^2 – b^2} = \frac{x – b}{-(a^2 – b^2)}$ [multiplying both sides by $(a^2 – b^2)$]
or, $\frac{x – a}{1} = \frac{x – b}{-1}$
or, $x – a = -(x – b)$
or, $x – a = -x + b$
or, $x + x = a + b$
or, $2x = a + b$
or, $x = \frac{a + b}{2}$
Required solution: $x = \frac{a + b}{2}$
8. $(3 + \sqrt{3})z + 2 = 5 + 3\sqrt{3}$
Solution: $(3 + \sqrt{3})z + 2 = 5 + 3\sqrt{3}$
or, $(3 + \sqrt{3})z = 5 + 3\sqrt{3} – 2$
or, $(3 + \sqrt{3})z = 3 + 3\sqrt{3}$
or, $(3 + \sqrt{3})z = 3(1 + \sqrt{3})$
or, $\sqrt{3}(\sqrt{3} + 1)z = 3(\sqrt{3} + 1)$ [dividing both sides by $(\sqrt{3} + 1)$]
or, $\sqrt{3}z = 3$
or, $z = \frac{3}{\sqrt{3}}$
or, $z = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}}$
or, $z = \sqrt{3}$
Required solution: $z = \sqrt{3}$
Find the solution set (9–14):
9. $2x + \sqrt{2} = 3x – 4 – 3\sqrt{2}$
Solution: $2x + \sqrt{2} = 3x – 4 – 3\sqrt{2}$
or, $2x – 3x = -4 – 3\sqrt{2} – \sqrt{2}$
or, $-x = -4 – 4\sqrt{2}$ [multiplying both sides by $-1$]
or, $x = 4 + 4\sqrt{2}$
or, $x = 4(1 + \sqrt{2})$
Required solution set: $S = \{4(1 + \sqrt{2})\}$
10. $\frac{z – 2}{z – 1} = 2 – \frac{1}{z – 1}$
Solution: $\frac{z – 2}{z – 1} = 2 – \frac{1}{z – 1}$
or, $\frac{z – 2}{z – 1} + \frac{1}{z – 1} = 2$
or, $\frac{z – 2 + 1}{z – 1} = 2$
or, $\frac{z – 1}{z – 1} = 2$
Here, if $z = 1$, the denominator $(z – 1)$ of the fractions in the given equation becomes $0$, which is undefined.
Also, simplifying the left-hand side gives $1 = 2$, which is impossible.
Therefore, the equation has no real solution.
Required solution set: $S = \emptyset$ (or $\{\}$)
11. $\frac{1}{x} + \frac{1}{x + 1} = \frac{2}{x – 1}$
Solution: $\frac{1}{x} + \frac{1}{x + 1} = \frac{2}{x – 1}$
or, $\frac{(x + 1) + x}{x(x + 1)} = \frac{2}{x – 1}$
or, $\frac{2x + 1}{x^2 + x} = \frac{2}{x – 1}$ [cross-multiplying]
or, $(2x + 1)(x – 1) = 2(x^2 + x)$
or, $2x^2 – 2x + x – 1 = 2x^2 + 2x$
or, $2x^2 – x – 1 = 2x^2 + 2x$ [subtracting $2x^2$ from both sides]
or, $-x – 1 = 2x$
or, $-x – 2x = 1$
or, $-3x = 1$
or, $x = -\frac{1}{3}$
Required solution set: $S = \left\{-\frac{1}{3}\right\}$
12. $\frac{m}{m – x} + \frac{n}{n – x} = \frac{m + n}{m + n – x}$
Solution: $\frac{m}{m – x} + \frac{n}{n – x} = \frac{m + n}{m + n – x}$
or, $\frac{m}{m – x} + \frac{n}{n – x} = \frac{m}{m + n – x} + \frac{n}{m + n – x}$
or, $\frac{m}{m – x} – \frac{m}{m + n – x} = \frac{n}{m + n – x} – \frac{n}{n – x}$
or, $m \left(\frac{1}{m – x} – \frac{1}{m + n – x}\right) = n \left(\frac{1}{m + n – x} – \frac{1}{n – x}\right)$
or, $m \left[\frac{(m + n – x) – (m – x)}{(m – x)(m + n – x)}\right] = n \left[\frac{(n – x) – (m + n – x)}{(m + n – x)(n – x)}\right]$
or, $m \left[\frac{m + n – x – m + x}{(m – x)(m + n – x)}\right] = n \left[\frac{n – x – m – n + x}{(m + n – x)(n – x)}\right]$
or, $m \cdot \frac{n}{(m – x)(m + n – x)} = n \cdot \frac{-m}{(m + n – x)(n – x)}$ [multiplying both sides by $(m + n – x)$ and dividing by $mn$]
or, $\frac{1}{m – x} = \frac{-1}{n – x}$
or, $1 \cdot (n – x) = -1 \cdot (m – x)$
or, $n – x = -m + x$
or, $-x – x = -m – n$
or, $-2x = -(m + n)$
or, $2x = m + n$
or, $x = \frac{m + n}{2}$
Required solution set: $S = \left\{\frac{m + n}{2}\right\}$
13. $\frac{1}{x + 2} + \frac{1}{x + 5} = \frac{1}{x + 3} + \frac{1}{x + 4}$
Solution: $\frac{1}{x + 2} + \frac{1}{x + 5} = \frac{1}{x + 3} + \frac{1}{x + 4}$
or, $\frac{(x + 5) + (x + 2)}{(x + 2)(x + 5)} = \frac{(x + 4) + (x + 3)}{(x + 3)(x + 4)}$
or, $\frac{2x + 7}{x^2 + 5x + 2x + 10} = \frac{2x + 7}{x^2 + 4x + 3x + 12}$
or, $\frac{2x + 7}{x^2 + 7x + 10} = \frac{2x + 7}{x^2 + 7x + 12}$
or, $\frac{2x + 7}{x^2 + 7x + 10} – \frac{2x + 7}{x^2 + 7x + 12} = 0$
or, $(2x + 7) \left(\frac{1}{x^2 + 7x + 10} – \frac{1}{x^2 + 7x + 12}\right) = 0$
Since $\left(\frac{1}{x^2 + 7x + 10} – \frac{1}{x^2 + 7x + 12}\right) \neq 0$,
therefore, $2x + 7 = 0$
or, $2x = -7$
or, $x = -\frac{7}{2}$
Required solution set: $S = \left\{-\frac{7}{2}\right\}$
14. $\frac{2t – 6}{9} + \frac{15 – 2t}{12 – 5t} = \frac{4t – 15}{18}$
Solution: $\frac{2t – 6}{9} + \frac{15 – 2t}{12 – 5t} = \frac{4t – 15}{18}$
or, $\frac{15 – 2t}{12 – 5t} = \frac{4t – 15}{18} – \frac{2t – 6}{9}$
or, $\frac{15 – 2t}{12 – 5t} = \frac{(4t – 15) – 2(2t – 6)}{18}$
or, $\frac{15 – 2t}{12 – 5t} = \frac{4t – 15 – 4t + 12}{18}$
or, $\frac{15 – 2t}{12 – 5t} = \frac{-3}{18}$
or, $\frac{15 – 2t}{12 – 5t} = -\frac{1}{6}$ [cross-multiplying]
or, $6(15 – 2t) = -1(12 – 5t)$
or, $90 – 12t = -12 + 5t$
or, $-12t – 5t = -12 – 90$
or, $-17t = -102$
or, $t = \frac{-102}{-17}$
or, $t = 6$
Required solution set: $S = \{6\}$
Solve by forming equations (15–24):
15. A number is $\frac{2}{5}$ times of another number. If the sum of the numbers is 98, find the two numbers.
Solution: Let, the other number $= x$
Therefore, one number $= \frac{2}{5} \times x = \frac{2x}{5}$
According to the question,
$x + \frac{2x}{5} = 98$
or, $\frac{5x + 2x}{5} = 98$
or, $\frac{7x}{5} = 98$
or, $7x = 98 \times 5$
or, $7x = 490$
or, $x = \frac{490}{7}$
or, $x = 70$
So, the other number is $70$
and one number $= \frac{2 \times 70}{5} = 2 \times 14 = 28$
Answer: The two numbers are $28$ and $70$.
16. Difference of numerator and denominator of a proper fraction is 1; if 2 is subtracted from numerator and 2 is added to denominator of the fraction, it will be equal to $\frac{1}{6}$. Find the fraction.
Solution: Since, the fraction is a proper fraction, the denominator will be greater than the numerator.
Let the numerator of the fraction $= x$
Therefore, the denominator of the fraction $= x + 1$
So, the fraction $= \frac{x}{x + 1}$
According to the question,
$\frac{x – 2}{(x + 1) + 2} = \frac{1}{6}$
or, $\frac{x – 2}{x + 3} = \frac{1}{6}$ [cross-multiplying]
or, $6(x – 2) = 1(x + 3)$
or, $6x – 12 = x + 3$
or, $6x – x = 3 + 12$
or, $5x = 15$
or, $x = \frac{15}{5}$
or, $x = 3$
Therefore, the required fraction $= \frac{3}{3 + 1} = \frac{3}{4}$
Answer: The fraction is $\frac{3}{4}$.
17. Sum of the digits of a number consisting of two digits is 9; if the number obtained by interchanging the places of the digits is less by 45 than the given number. What is the number?
Solution: Let, the unit’s digit $= x$
Since, the sum of the digits is $9$,
the ten’s digit $= 9 – x$
Therefore, the original number $= 10(9 – x) + x$
$= 90 – 10x + x$
$= 90 – 9x$
If the digits interchange their places, the interchanged number $= 10x + (9 – x) = 10x + 9 – x = 9x + 9$
According to the question,
$(90 – 9x) – (9x + 9) = 45$
or, $90 – 9x – 9x – 9 = 45$
or, $81 – 18x = 45$
or, $-18x = 45 – 81$
or, $-18x = -36$
or, $x = \frac{-36}{-18}$
or, $x = 2$
Therefore, the required number $= 90 – 9(2) = 90 – 18 = 72$
Answer: The number is $72$.
18. The digit of the tens place of a number of two digits is twice the digit of unit place. Show that, the number is seven times the sum of the digits.
Solution: Let, the unit’s digit $= x$
Therefore, the ten’s digit $= 2x$
So, the number $= 10(2x) + x = 20x + x = 21x$
and the sum of the digits $= 2x + x = 3x$
Now,
$\frac{\text{Number}}{\text{Sum of digits}} = \frac{21x}{3x} = 7$
That is,
$\text{Number} = 7 \times \text{Sum of digits}$
Therefore, the number is seven times the sum of its digits. (Shown)
19. A petty merchant, by investing Tk. 5600 got the 5% profit on some of the money and 4% profit of the rest of the money. If the total profit is 256, then on how much money did he get 5% profit?
Solution: Let, the amount invested at $5\%$ profit $= x$ Tk.
Therefore, the remaining amount invested at $4\%$ profit $= (5600 – x)$ Tk.
$5\%$ profit on $x$ Tk $= x \times \frac{5}{100} = \frac{5x}{100}$ Tk.
and
$4\%$ profit on $(5600 – x)$ Tk $= (5600 – x) \times \frac{4}{100} = \frac{4(5600 – x)}{100}$ Tk.
According to the question,
$\frac{5x}{100} + \frac{4(5600 – x)}{100} = 256$
or, $\frac{5x + 22400 – 4x}{100} = 256$
or, $\frac{x + 22400}{100} = 256$
or, $x + 22400 = 256 \times 100$
or, $x + 22400 = 25600$
or, $x = 25600 – 22400$
or, $x = 3200$
Answer: He made a $5\%$ profit on $3200$ Tk.
20. In a girls school if 6 students sit on each bench, 2 benches remain empty. But if 5 students sit in each bench, 6 students have to remain standing. What is the number of benches in the classroom.
Solution: Let, the total number of benches in that classroom $= x$
According to the first condition,
If $2$ benches remain empty, students sit on $(x – 2)$ benches.
If $6$ students sit on each bench, the total number of students $= 6(x – 2)$
According to the second condition,
If $5$ students sit on each bench, $5x$ students sit on $x$ benches, and $6$ students remain standing.
Therefore, the total number of students $= 5x + 6$
Since the total number of students is equal in both cases,
According to the question,
$6(x – 2) = 5x + 6$
or, $6x – 12 = 5x + 6$
or, $6x – 5x = 6 + 12$
or, $x = 18$
Answer: The number of benches in that classroom is $18$.
21. Number of passengers in a launch is 47. The fare per head for the cabin is twice that for the deck. The fare per head for the deck is Tk. 30. If the total fare collected is Tk. 1680, what is the number of passengers in the cabin?
Solution: Let, the number of cabin passengers $= x$
Therefore, the number of deck passengers $= (47 – x)$
Given that,
Deck fare per head $= 30$ Tk
Therefore, cabin fare per head $= 30 \times 2 = 60$ Tk
According to the question,
$60x + 30(47 – x) = 1680$
or, $60x + 1410 – 30x = 1680$
or, $30x + 1410 = 1680$
or, $30x = 1680 – 1410$
or, $30x = 270$
or, $x = \frac{270}{30}$
or, $x = 9$
Answer: The number of cabin passengers is $9$.
22. 120 coins of twenty five paisa and fifty paisa together is Tk. 35. What is the number of coins of each kind?
Solution: We know, $1$ Tk $= 100$ paisa
So, $35$ Tk $= 35 \times 100 = 3500$ paisa
Let, the number of 25-paisa coins $= x$
Therefore, the number of 50-paisa coins $= (120 – x)$
According to the question,
$25x + 50(120 – x) = 3500$
or, $25x + 6000 – 50x = 3500$
or, $-25x + 6000 = 3500$
or, $-25x = 3500 – 6000$
or, $-25x = -2500$
or, $x = \frac{-2500}{-25}$
or, $x = 100$
Therefore, the number of 25-paisa coins $= 100$
and the number of 50-paisa coins $= 120 – 100 = 20$
Answer: There are $100$ coins of 25-paisa and $20$ coins of 50-paisa.
23. A car passed over some distance at the speed of 60 km per hour and passed over the rest of the distance at the speed of 40 km per hour. The car passed over the total distance of 240 km in 5 hours. How far did the car pass over at the speed of 60 km per hour?
Solution: Let, the distance traveled at $60$ km/h $= x$ km
Therefore, the distance traveled at $40$ km/h $= (240 – x)$ km
We know, $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$
So, time taken at $60$ km/h $= \frac{x}{60}$ hours
and time taken at $40$ km/h $= \frac{240 – x}{40}$ hours
According to the question,
$\frac{x}{60} + \frac{240 – x}{40} = 5$
or, $\frac{2x + 3(240 – x)}{120} = 5$
or, $\frac{2x + 720 – 3x}{120} = 5$
or, $\frac{720 – x}{120} = 5$
or, $720 – x = 5 \times 120$
or, $720 – x = 600$
or, $-x = 600 – 720$
or, $-x = -120$
or, $x = 120$
Answer: The car traveled $120$ km at $60$ km/h.
24. Distance between Dhaka New Market and Gabtoli is 12 km. From new market, Sajal started for Gabtoli by rickshaw at a speed of 6 km/h, and Kajal from the same place started for Gabtoli on foot at the speed of 4 km/h. After reaching Gabtoli Sajal rest for 30 minutes, and then started for New Market at the same speed. At which distance from New Market will they meet?
Solution: Given that,
Distance from New Market to Gabtoli $= 12$ km
Speed of Sajal’s rickshaw $= 6$ km/h
So, time taken by Sajal to reach Gabtoli from New Market $= \frac{12}{6} = 2$ hours
Sajal rests after reaching Gabtoli $= 30$ minutes $= \frac{30}{60} = 0.5$ hours
Total elapsed time for Sajal (including travel and rest) $= 2 + 0.5 = 2.5$ hours
In this $2.5$ hours, distance covered by Kajal on foot $= \text{Speed} \times \text{Time} = 4 \times 2.5 = 10$ km
Therefore, when Sajal finishes resting and starts heading back toward New Market, Kajal is located $10$ km away from New Market.
Then, Kajal’s distance from Gabtoli was $= 12 – 10 = 2$ km.
Now, Sajal and Kajal are coming toward each other from opposite directions.
Their relative speed $= (6 + 4) = 10$ km/h
Time required to cover the $2$ km distance between them $= \frac{2}{10} = 0.2$ hours
In this $0.2$ hours, distance covered by Kajal on foot $= 4 \times 0.2 = 0.8$ km
Therefore, the distance of their meeting point from New Market $= 10 + 0.8 = 10.8$ km
Answer: They will meet at a distance of $10.8$ km from New Market.