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Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations :
(i)
Answer:
Multiply equation (1) and (2) by 6.
3a + 2b = 12 → (3)
2a + 3b = 13 → (4)
Multiply equation (1) by 3 and equation (4) by 2
4a + 6b = 26 → (6)
Subtract equation (5) and (6)
a = 2
Put a = 2 in equation (5)
9(2) + 6b = 36
6b = 36 – 18
6b – 18
b = \(\frac{18}{6}\) = 3
b = 3
x = \(\frac{1}{2}\) and y = \(\frac{1}{3}\)
(ii)
Answer:
2a + 3b = 2 → (1)
4a – 9b = -1 → (2)
Multiply equation (1) by 3.
6a + 9b = 6 → (3)
Adding equation (2) and (3)
a = \(\frac{5}{10}\)
a = \(\frac{1}{2}\)
Put a = \(\frac{1}{2}\) in equation (1).
2 × \(\frac{1}{2}\) + 3b = 2
1 + 3b = 2
3b = 2 – 1
3b = 1
b = \(\frac{1}{3}\)
Squaring both sides.
∴ x = 4 and y = 9.
Hence, the solution of the given pair of equations is x = 4 and y = 9.
(iii)
\(\frac{4}{x}\) + 3y = 14
\(\frac{3}{x}\) – 4y = 23
Answer:
\(\frac{4}{x}\) + 3y = 14
\(\frac{3}{x}\) – 4y = 23
put \(\frac{1}{x}\) = a
4a + 3y = 14 → (1)
3a – 4y = 23 → (2)
Multiply equation (1) by 4 and equation (2) by 3 and Add.
a = \(\frac{125}{25}\) = 5
Put a = 5 in equation (1) 4a + 3y = 14
4(5) + 3y = 14
20 + 3y = 14
3y = 14 – 20
y = \(=\frac{-6}{3}\) = – 2
y = – 2
a = \(\frac{1}{x}\) = 5
x = \(\frac{1}{5}\) and y = – 2
Hence the solution of the given pair of equations is x = \(\frac{1}{5}\) and y = – 2
(iv)
Answer:
Put \(\frac{1}{x-1}\) = and \(\frac{1}{y-2}\) = b
5a + b2 → (1)
6a – 3b= I → (2)
Multiply equation (1) by 3
15a + 3b = 6 → (3)
Adding equation (3) and (2).
a = \(\frac{7}{21}=\frac{1}{3}\)
a = \(\frac{1}{3}\)
put a = \(\frac{1}{3}\) in equation (2)
6 × \(\frac{1}{3}\) – 3b = 1
2 – 3b = 1
– 3b = 1 – 2
– 3b = – 1
b = \(\frac{-1}{-3}\)
b = \(\frac{1}{3}\)
∴ a = \(\frac{1}{x-1}\) = \(\frac{1}{3}\)
⇒ x – 1 = 3
x = 3 + 1 = 4
x = 4
and b = \(\frac{1}{y-2}=\frac{1}{3}\)
y – 2 = 3
y = 3 + 2 = 5
y = 5
Hence, the solution of the given pair of equation is x = 4, y = 5.
(v)
Answer:
Put \(\frac{1}{y}\) = a and \(\frac{1}{x}\) = b in equation (1) and (2)
7a – 2b = 5 → (3)
8a + 7b = 15 → (4)
Multiply equation (3) by 7 and equation (4) by 2.
49a – 14b = 35 → (5)
16a + 14b = 30 → (6)
Adding equation (5) and (6)
a = \(\frac{65}{65}\) = 1
a = 1
put a = 1 in equation (3)
7(a) – 2b = 5
– 2b = 5 – 7
– 2b = – 2
b = \(\frac{-2}{-2}\) = 1
b = 1
∴ a = \(\frac{1}{y}\) = 1 and b = \(\frac{1}{x}\) = 1
y = 1 and x = 1
Hence, the solution of pair of linear equation is x = 1, y = 1
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
Answer:
6x + 3y = 6xy
2x + 4y = 5xy
on dividing each equation by xy
Put a = \(\frac{1}{y}\) and b = \(\frac{1}{x}\) in equation (1) and (2).
6a + 3b = 6 → (3)
2a + 4b = 5 → (4)
Multiply equation (4) by 3
6a + 12b = 15 → (5)
Subtract equation (3) and (5)
Put b = 1 in equation (3)
6a + 3(1) = 6
6a = 6 – 3 = 3
a = \(\frac{3}{6}=\frac{1}{2}\)
a = \(\frac{1}{2}\)
a = \(\frac{1}{y}=\frac{1}{2}\) and b = \(\frac{1}{x}\) = 1
∴ y = 2 and x = 1
Hence, the solution of the given pair of linear equations is x = 1 and y = 2.
(vii)
Answer:
15a – 5b = – 2 → (2)
Multiply equation (1) by 5.
25a + 5b = 10 → (3)
Adding equation (2) and (3)
x = 3
put x = 3 in equation (4)
3 + y = 5
y = 5 – 3
y = 2
Hence, the solution of the given pair of equation is x = 3, y = 2.
(viii)
Answer:
Adding equation (1) and (2).
3x – y = 2 → (4)
Adding (3) and (4)
x = 1
Put x = 1 in equation (3)
3(1) + y = 4
y = 4 – 3
y = 1
Hence, the solution of the given pair of the equation is x = 1, y = 1
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 the speed of the current.
Answer:
Let the speed of Ritu
in still water be x km/hr.
in current water be y km/hr
Speed of water while
upstream is (x + y) km/hr,
downstream is (x – y) km/hr,
x + y = 10 → (1)
case (ii) Ritu rows upstream.
x – y = 2 → (2)
Adding (1) and (2)
Put x = 6 in equation (1)
6 + y= 10
y = 10 – 6
y = 4
Speed of Ritu in still water = 6 km/hr and speed of current is 4 km/hr.
(ii) 2 women and 5 men can together finish a woman alone to finish the work, and also that time taken by 1 man alone.
Answer:
Let the time taken by 1 woman alone to finish the embroidery work be x days and the time is taken by 1 man alone to finish the embroidery work by y days.
1 woman’s 1 day work = \(\frac{1}{x}\)
1 man’s 1 day work = \(\frac{1}{y}\)
∴ 2 women’s 1 day’s work = \(\frac{2}{y}\)
5 men’s 1 day’s work = \(\frac{5}{y}\)
∴ 2 women and 5 men can together finish one embroidery work in 4 days.
Again, 3 women’s 1 day’s work = \(\frac{3}{x}\)
6 men’s 1 day’s work = \(\frac{6}{y}\)
∵3 women and 6 men can together finish one embroidery work in 3 days.
Put \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b
Then equation (1) and (2) becomes
8a + 20b = 1 → (3)
9a + 18 b = 1 → (4)
Multiply equation (3) by 9 and equation (4) by 8
72a + 180b = 9 → (5)
72a + 144b = 8 → (6)
Subtract equation (5) and (6).
b =\(\frac{1}{36}\)
Put b = \(\frac{1}{36}\) in equation (4)
9a + 18 × \(\frac{1}{36}\) = 1
9a + \(\frac{1}{2}\) = 1
9a = \(\frac{1}{1}\) – \(\frac{1}{2}\)
9a = \(\frac{1}{2}\)
a = \(\frac{1}{18}\)
∴ \(\frac{1}{x}\) = a = \(\frac{1}{18}\)
x = 18
\(\frac{1}{y}\) = b = \(\frac{1}{36}\)
y = 36
Hence, the time taken by one woman to finish embroidery work is 18 days, and the time taken by one man to finish embroidery work is 36 days.
(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 the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Answer:
Let the speed of the bus be x km/hr and the speed of the train be y km/hr.
Put \(\frac{1}{x}\) = a = \(\frac{1}{y}\) = b in equation (1) and (2)
60a + 240b =4 → (3)
100a + 200b = \(\frac{25}{6}\)→ (4)
Multiply equation (3) by 5 and equation (4) by 6.
300a + 1200b = 20 → (5)
600a + 1200b = 25 → (6)
Subtract equation (5) and (6)
a = \(\frac{-5}{-300}\) = \(\frac{1}{60}\)
a = \(\frac{1}{60}\)
Put a = \(\frac{1}{60}\) in equation (3)
60 × \(\frac{1}{60}\) + 240 b = 4
240 b = 4 – 1 = 3
b = \(\frac{3}{240}\)
b = \(\frac{1}{80}\)
\(\frac{1}{x}\) = a = \(\frac{1}{60}\)
x = \(\frac{1}{60}\)
and \(\frac{1}{y}\) = b = \(\frac{1}{80}\)
y = 80
Hence the speed of bus is 60 km/hr and speed of train is 80 km/hr.