Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Textbook Exercise Questions and Answers.
Haryana Board 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
Question 1.
Solve the following pairs of linear equations by reducing them to a pair of linear equations:
(i) \(\frac{1}{2 x}+\frac{1}{3 y}\) = 2
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)
(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = – 1
(iii) \(\frac{4}{x}\) + 3y = 14
\(\frac{3}{x}\) – 4y = 23
(iv) \(\frac{5}{x-1}+\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}-\frac{3}{y-2}\) = 1
(v) \(\frac{7 x-2 y}{x y}\) = 5
\(\frac{8 x+7 y}{x y}\) = 15
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
(vii) \(\frac{10}{x+y}+\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}-\frac{5}{x-y}\) = – 2
(viii) \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)
Solution:
(i) The given equations are:
\(\frac{1}{2 x}+\frac{1}{3 y}\) = 2 …………….(1)
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\) ………………(2)
Let \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b
∴ \(\frac{1}{2}\) a + \(\frac{1}{3}\) b = 2 ……………(3)
and \(\frac{1}{3}\) a + \(\frac{1}{2}\) b = \(\frac{13}{6}\) ……………(4)
Multiplying equation (3) by 4 and equation (4) by 6, we get
2a + \(\frac{4}{3}\)b = 8 ………….(5)
2a + 3b = 13 …………..(6)
Subtracting equation (6) from equation (5), we get
2a+ \(\frac{4}{3}\) b = 8
Substituting the value of b in equation (6), we get
2a + 3 × 3 = 13
⇒ 2a + 9 = 13
⇒ 2a = 13 – 9 = 4
⇒ a = \(\frac{4}{2}\) = 2
⇒ \(\frac{1}{x}\) = 2 [∵ a = \(\frac{1}{x}\)]
⇒ x = \(\frac{1}{2}\)
Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{3}\) is the required solution.
(ii) The given equations are:
\(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2 …………….(1)
\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}\) = – 1 ……………..(2)
Let \(\frac{1}{\sqrt{x}}\) = a, \(\frac{1}{\sqrt{y}}\) = b;
∴ 2a + 3b = 2 ……………..(3)
and 4a – 9b = – 1 ………………..(4)
Multiplying equation (3) by 3, we get
6a + 9b = 6 ……………(5)
Adding equation (5) and equation (4), we get
⇒ x = 4 [Squaring on both sides]
Substituting the value of a in equation (3), we get
2 × \(\frac{1}{2}\) + 3b = 2
⇒ 1 + 3b = 2
⇒ 3b = 2 – 1 = 1
⇒ b = \(\frac{1}{3}\)
⇒ \(\frac{1}{\sqrt{y}}=\frac{1}{3}\) [∵ b = \(\frac{1}{\sqrt{y}}\)]
⇒ √y = 3
⇒ y = 9 [Squaring on both sides]
Hence, x = 4, y= 9 is the required solution.
(iii) The given equations are :
\(\frac{4}{x}\) + 3y = 14 ……………….(1)
\(\frac{3}{x}\) – 4y = 23 ………………..(2)
Let \(\frac{1}{x}\) = a,
4a + 3y = 14 ………………(3)
3a – 4y = 23 ……………….(4)
Multiplying equation (3) by 4 and equation (4) by 3, we get
16a + 12y = 56 ……………….(5)
9a – 12y = 69 ……………….(6)
Adding equation (5) and equation (6), we get
⇒ a = 5
⇒ \(\frac{1}{x}\) = 5
⇒ x = \(\frac{1}{5}\)
Substituting the value of a in equation (3), we get
4 × 5 + 3y = 14
⇒ 20 + 3y = 14
⇒ 3y = 14 – 20 = – 6
⇒ y = \(-\frac{6}{3}\) = – 2
Hence, x = \(\frac{1}{5}\), y = – 2 is the required solution.
(iv) The given equations are:
\(\frac{5}{x-1}+\frac{1}{y-2}\) = 2 ……………..(1)
\(\frac{6}{x-1}-\frac{3}{y-2}\) = 1 ……………..(2)
Let \(\frac{1}{x-1}\) = p, \(\frac{1}{y-2}\) = q
∴ 5p + q = 2 ……………….(3)
and 6p – 3q = 1 ……………..(4)
Multiplying equation (3) by 3, we get
15p + 3q = 6 ………………..(5)
Adding equation (5) and equation (4), we get
x – 1 = 3
x = 3 + 1 = 4
Substituting the value of p in equation (3), we get
5 × \(\frac{1}{3}\) + q = 2
\(\frac{5}{3}\) + q = 2
q = 2 – \(\frac{5}{3}\)
q = \(\frac{4}{4}\)
\(\frac{1}{y-2}=\frac{1}{3}\) [∵ q = \(\frac{1}{y-2}\)]
y – 2 = 3
y = 3 + 2 = 5
Hence, x = 4, y = 5 is the required solution.
(v) The given equations are:
– 2p + 7q = 5 ………………….(3)
7p + 8q = 15 …………………..(4)
Multiplying equation (3) by 7 and equation (4) by 2, we get
– 14p + 49q = 35 ………………(5)
14p + 16q = 30 ………………….(6)
Adding equation (5) and equation (6), we get
⇒ y = 1
Substituting the value of q in equation (4), we get
7p + 8 × 1 = 15
⇒ 7p = 15 – 8 = 7
⇒ p = \(\frac{7}{7}\) = 1
⇒ \(\frac{1}{x}\) = 1 [∵ p = \(\frac{1}{x}\)]
⇒ x = 1
Hence, x = 1, y = 1 is the required solution.
(vi) The given equations are :
6x + 3y = 6xy
⇒ \(\frac{6 x}{x y}+\frac{3 y}{x y}=\frac{6 x y}{x y}\)
[Dividing both sides of the equation by xy]
\(\frac{6}{y}+\frac{3}{x}\) = 6
\(\frac{3}{x}+\frac{6}{y}\) = 6 ………………(1)
and 2x + 4y = 5xy
⇒ \(\frac{2 x}{x y}+\frac{4 y}{x y}=\frac{5 x y}{x y}\)
[Dividing both sides of the equation by xy]
⇒ \(\frac{2}{y}+\frac{4}{x}\) = 5
⇒ \(\frac{4}{x}+\frac{2}{y}\) = 5 …………..(2)
Let \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b
∴ 3a + 6b = 6
⇒ a + 2b = 2 ……………(3)
and 4a + 2b = 5 ……………….(4)
Subtracting equation (4) from (3), we get
⇒ x = 1
Substituting the value of a in equation (3), we get
⇒ 1 + 2b = 2
2b = 2 – 1 = 1
⇒ b = \(\frac{1}{2}\)
⇒ \(\frac{1}{y}=\frac{1}{2}\) [∵ b = \(\frac{1}{y}\)]
⇒ y = 2
Hence, x = 1, y = 2 is the required solution.
(vii) The given equations are :
\(\frac{10}{x+y}+\frac{2}{x-y}\) = 4 ……………….(1)
and \(\frac{15}{x+y}-\frac{5}{x-y}\) = – 2 ……………(2)
Let \(\frac{1}{x+y}\) = a, \(\frac{1}{x-y}\) = b
10a + 2b = 4
⇒ 5a + b = 2 ……………….(3)
and 15a – 5b = – 2 ……………..(4)
Adding equation (5) and equation (4), we get
⇒ x + y = 5
Substituting the value of a in equation (4), we get
15 × \(\frac{1}{5}\) – 5b = – 2
⇒ 3 – 5b = – 2
⇒ – 5b = – 2 – 3 = – 5
⇒ b = \(\frac{-5}{-5}\) = 1
⇒ \(\frac{1}{x-y}\) = 1
⇒ x – y = 1 …………………..(7)
Adding equation (6) and equation (7), we get
Substituting the value of x in equation (6), we get
3 + y = 5
⇒ y = 5 – 3 = 2
Hence, x = 3, y = 2 is the required solution.
(viii) The given equations are:
\(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) ………………..(1)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}\) …………………(2)
Let \(\frac{1}{3 x+y}\) = a, \(\frac{1}{3 x-y}\) = b
∴ a + b = \(\frac{3}{4}\) ……………(3)
and \(\frac{1}{2} a-\frac{1}{2} b=-\frac{1}{8}\) …………….(4)
Multiplying equation (4) by 2, we get
a – b = – \(\frac{2}{8}\)
or a – b = – \(\frac{1}{4}\) ………………(5)
Adding equation (3) and equation (5), we get
⇒ 3x + y = 4 ………………(6)
Substituting the value of a in equation (3), we get
⇒ \(\frac{1}{4}\) + b = \(\frac{3}{4}\)
⇒ b = \(\frac{3}{4}-\frac{1}{4}=\frac{2}{4}\)
⇒ b = \(\frac{1}{2}\)
⇒ \(\frac{1}{3 x-y}=\frac{1}{2}\) [∵ b = \(\frac{1}{3 x-y}\)]
⇒ 3x – y = 2 ……………….(7)
Adding equation (6) and equation (7), we get
Substituting the value of x in equation (6), we get
3 × 1 + y = 4
y = 4 – 3 = 1
y = 1
Hence, x = 1, y = 1 is the required solution.
Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions :
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
(i) Let speed of her rowing in still water be x km/hr.
and speed of current be y km/hr.
Speed of rowing downstream= (x + y) km/hr.
Speed of rowing upstream = (x – y) km/hr.
According to question,
\(\frac{20}{x+y}\) = 2 [∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
⇒ 2x + 2y = 20
⇒ x + y = 10 ……………..(1)
and \(\frac{4}{x-y}\) = 2
2x – 2y = 4
⇒ x – y = 2 ……………..(2)
Adding equation (1) and equation (2), we get
Substituting the value of x in equation (1), we
6 + y = 10
⇒ y = 10 – 6 = 4
Hence, the equations are : x + y = 10, x – y = 2 and x = 6, y = 4 is the required solution.
Therefore,speed of her rowing in still water = 6 km/hr and speed of current = 4 km/hr.
(ii) Let time taken by 1 woman to finish the embroidery work = x days
and time taken by 1 man to finish the embroidery work = y days
1 woman’s 1 day’s work = \(\frac{1}{x}\)
2 women’s 1 day’s work = \(\frac{2}{x}\)
and 1 man’s 1 day’s work = \(\frac{1}{y}\)
5 men’s 1 day’s work = \(\frac{5}{y}\)
Since 2 women and 5 men finish the work in 4 days.
∴ \(\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\)
⇒ \(\frac{8}{x}+\frac{20}{y}\) = 1 ……………..(1)
Again 3 women and 6 men finish the work in 3 days.
∴ \(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\)
⇒ \(\frac{9}{x}+\frac{18}{y}\) = 1 ……………..(2)
Let \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b,
∴ 8a + 20b = 1 ……………….(3)
and 9a + 186 = 1 ……………..(4)
Multiplying equation (3) by 9 and equation (4) by 8, we get
72a + 180b = 9 ………………..(5)
72 a + 144b = 8 ………………..(6)
Subtracting equation (6) from equation (5), we get
⇒ y = 36
Substituting the value of b in equation (3), we get
⇒ 8a + 20 × \(\frac{1}{36}\) = 1
⇒ 8a = 1 – \(\frac{20}{36}\)
⇒ 8a = \(\frac{36-20}{36}=\frac{16}{36}\)
⇒ a = \(\frac{16}{36 \times 8}=\frac{2}{36}\)
⇒ a = \(\frac{1}{18}\)
⇒ \(\frac{1}{x}=\frac{1}{18}\) [∵ a = \(\frac{1}{x}\)]
⇒ x = 18
Hence, the equations are :\(\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\), \(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\) and x = 18, y = 36 is the required solution.
Therefore, 1 woman can finish the embroidery work in 18 days and 1 man can finish the embroidery work in 36 days.
(iii) Let the speed of train be x km/hr and speed of bus be y km/hr.
According to question,
\(\frac{60}{x}+\frac{(300-60)}{y}\) = 4
[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
⇒ \(\frac{60}{x}+\frac{240}{y}\) = 4 …………….(1)
and \(\frac{100}{x}+\frac{(300-100)}{y}=4+\frac{10}{60}\)
\(\frac{100}{x}+\frac{200}{y}=4+\frac{1}{6}\)
\(\frac{100}{x}+\frac{200}{y}=\frac{\varepsilon}{6}\) ………………..(2)
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q
∴ 60p + 240q = 4 ……………..(3)
and 100p + 200q = \(\frac{25}{6}\) ………………(4)
Multiplying equation (3) by 5 and equation (4) by 3, we get
300p + 1200q = 20 ……………..(5)
300p + 600q = \(\frac{25}{2}\) …………….(6)
Subtracting equation (6) from equation (5), we get
y = 80
Substituting the value of q in equation (3), we get
60p + 240 × \(\frac{1}{80}\) = 4
⇒ 60p + 3 = 4
⇒ 60p = 4 – 3 = 1
⇒ p = \(\frac{1}{60}\)
⇒ \(\frac{1}{x}=\frac{1}{60}\) [∵ p = \(\frac{1}{x}\)]
⇒ x = 60
Hence, the equations are \(\frac{10}{x}+\frac{200}{y}=\frac{25}{6}\), \(\frac{100}{x}+\frac{200}{y}=\frac{25}{6}\)
and x = 60, y = 80 is the required solution.
Therefore,speed of the train = 60 km/hr
and speed of the bus = 80 km/hr