Chapter 5 - Linear Equations in Two Variables
Practice set 5.1
(1) By using variables x and y form any five linear equations in two variables.
Solution:
The general form of a linear equation in two variables x and y is ax + by + c = 0, where a, b, c are real numbers and a ≠ 0, b ≠ 0.
Five linear equations in two variables are:
1. 4x + 3y – 10 = 0
2. 5x – 9y + 14 = 0
3. 3x + y – 5 = 0
4. 2x + 33y – 45 = 0
5. 3x – 2y + 8 = 0
(2) Write five solutions of the equation x + y = 7.
Solution:
(i) Let x = 4
Substituting x = 4 in equation x + y = 7, we get,
4 + y = 7
y = 7 – 4
y = 3
1st Solution is (x, y) = (4, 3)
(ii) Let x = 2
Substituting x = 2 in equation x + y = 7, we get,
2 + y = 7
y = 7 – 2
y = 5
2nd Solution is (x, y) = (2, 5)
(iii) Let x = -1
Substituting x = 4 in equation x + y = 7, we get,
-1 + y = 7
y = 7 – (-1)
y = 7 + 1
y = 8
3rd Solution is (x, y) = (-1, 8)
(iv) Let x = 7
Substituting x = 4 in equation x + y = 7, we get,
7 + y = 7
y = 7 – 7
y = 0
4th Solution is (x, y) = (7, 0)
(v) Let x = 10
Substituting x = 10 in equation x + y = 7, we get,
10 + y = 7
y = 7 – 10
y = -3
5th Solution is (x, y) = (10, -3)
(3) Solve the following sets of simultaneous equations.
(i) x + y = 4 ; 2x – 5y = 1
Solution:
1) By elimination method,
Let,
x + y = 4 … (i)
2x – 5y = 1 … (ii)
Multiplying both sides of equation (i) by 5,
5(x + y) = 5(4)
5x + 5y = 20 … (iii)
Adding equations (ii) and (iii),
2x – 5y = 1
+ 5x + 5y = 20
________________
7x – 0y = 21
∴ 7x = 21
∴ x = \(\large \frac{21}{7}\)
∴ x = 3
Substituting x = 3 in equation (i),
3 + y = 4
y = 4 – 3
∴ y = 1
2) By substitution method,
Let,
x + y = 4
∴ x = 4 – y …(i)
2x – 5y = 1 …(ii)
Substituting x = 4 – y in equation (ii),
2(4 – y) – 5y = 1
∴ 8 – 2y – 5y = 1
∴ 8 – 7y = 1
∴ -7y = 1 – 8
∴ -7y = -7
∴ y = \(\large \frac{-7}{-7}\)
∴ y = 1
Substituting y = 1 in equation (i),
x = 4 – 1
∴ x = 3
∴ (3,1) is the solution of the given equations.
(ii) 2x + y = 5; 3x – y = 5
Solution:
1) By elimination method,
Let,
2x + y = 5 … (i)
3x – y = 5 … (ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
________________
5x – 0 = 10
∴ 5x = 0
∴ x = \(\large \frac{10}{5}\)
∴ x = 2
Substituting x = 2 in equation (i),
2x + y = 5
y = 5 – 2(2)
y = 5 – 4
∴ y = 1
2) By substitution method,
Let,
2x + y = 5
∴ y = 5 – 2x … (i)
3x – y = 5 … (ii)
Substituting y = 5 – 2x in equation (ii),
3x – (5 – 2x) = 5
∴ 3x – 5 + 2x = 5
∴ 3x + 2x = 5 + 5
∴ 5x = 10
∴ x = \(\large \frac{10}{5}\)
∴ x = 2
Substituting x = 2 in equation (i),
y = 5 – 2(2)
∴ y = 5 – 4
∴ y = 1
∴ (2,1) is the solution of the given equations.
(iii) 3x – 5y = 16; x – 3y = 8
Solution:
1) By elimination method,
Let,
3x – 5y = 16 … (i)
x – 3y = 8 … (ii)
Multiplying both sides of equation (ii) by 3,
3(x – 3y) = 3(8)
3x – 9y = 24 … (iii)
Subtracting equations (i) and (iii),
3x – 5y = 16
3x – 9y = 24
(-) (+) (-) .
________________
0x + 4y = -8
∴ 4y = -8
∴ y = -8 ÷ 4
∴ y = -2
Substituting y = -2 in equation (i),
x – 3(-2) = 8
∴ x + 6 = 8
∴ x = 8 – 6
∴ x = 2
2) By substitution method,
Let,
3x – 5y = 16 …(i)
x – 3y = 8
∴x = 8 + 3y …..(ii)
Substituting x = 8 + 3y in equation (i),
3(8 + 3y) – 5y = 16
24 + 9y – 5y = 16
∴ 4y = 16 – 24
∴ 4y = -8
∴ y = -8 ÷ 4
y = -2
Substituting y = -2 in equation (ii),
x = 8 + 3(-2)
∴ x = 8 – 6
∴ x = 2
∴ (2,-2) is the solution of the given equations.
(iv) 2y – x = 0; 10x + 15y = 105
Solution:
1) By elimination method,
Let,
2y – x = 0
-x + 2y = 0 … (i)
10x + 15y = 105 … (ii)
Multiplying both sides of equation (i) by 10,
10(-x + 2y) = 10(0)
-10x + 20y = 0 … (iii)
Adding equations (ii) and (iii),
10x + 15y = 105
+ (-10x) + 20y = 0
_____________________
0x + 35y = 105
∴ 35y = 105
∴ y = \(\large \frac{105}{35}\)
∴ y = 3
Substituting y = 3 in equation (i),
-x + 2(3) = 0
-x + 6 = 0
-x = -6
∴ x = 6
2) By substitution method,
Let,
2y – x = 0
∴ x = 2y …(i)
10x + 15y = 105 …(ii)
Substituting x = 2y in equation (ii),
10(2y) + 15y = 105
∴ 20y + 15y = 105
∴ 35y = 105
∴ y = \(\large \frac{105}{35}\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y
x = 2(3)
∴ x = 6
∴ (6, 3) is the solution of the given equations.
(v) 2x + 3y + 4 = 0; x – 5y = 11
Solution:
1) By elimination method,
Let,
2x + 3y + 4 = 0
2x + 3y = -4 … (i)
x – 5y = 11 … (ii)
Multiplying both sides of equation (ii) by 2,
2(x – 5y) = 2(11)
2x – 10y = 22 … (iii)
Subtracting equations (ii) and (iii),
2x + 3y = -4
2x – 10y = 22
(-) (+) (-)
________________
0x + 13y = -26
∴ 13y = -26
∴ y = \(\large \frac{-26}{13}\)
∴ y = -2
Substituting y = -2 in equation (i),
2x + 3y = -4
2x + 3(-2) = -4
∴ 2x – 6 = -4
∴ 2x = -4 + 6
∴ 2x = 2
∴ x = \(\large \frac{2}{2}\)
∴ x = 1
2) By substitution method,
Let,
2x + 3y + 4 = 0 …(i)
x – 5y = 11
∴x = 11 + 5y …(ii)
Substituting x = 11 + 5y in equation (i),
2(11 +5y) + 3y + 4 = 0
∴ 22 + 10y + 3y + 4 = 0
∴ 13y + 26 = 0
∴ 13y = -26
∴y = \(\large \frac{-26}{13}\)
∴ y = -2
Substituting y = -2 in equation (ii),
x = 11 + 5y
∴ x = 11 + 5(-2)
∴ x = 11 – 10
∴ x = 1
∴ (1, -2) is the solution of the given equations.
(vi) 2x – 7y = 7; 3x + y = 22
Solution:
1) By elimination method,
Let,
2x – 7y = 7 … (i)
3x + y = 22 … (ii)
Multiplying both sides of equation (ii) by 7,
7(3x + y) = 7(22)
21x + 7y = 154 … (iii)
Adding equations (i) and (iii),
2x – 7y = 7
+ 21x + 7y = 154
________________
23x – 0y = 161
∴ 23x = 161
∴ x = \(\large \frac{161}{23}\)
∴ x = 7
Substituting x = 7 in equation (i),
2x – 7y = 7
2(7) – 7y = 7
14 – 7y = 7
-7y = 7 – 14
∴ -7y = -7
∴ y = \(\large \frac{-7}{-7}\)
∴ y = 1
2) By substitution method,
Let,
2x – 7y = 7 …(i)
3x + y = 22
∴ y = 22 – 3x ……(ii)
Substituting y = 22 – 3x in equation (i),
2x – 7(22 – 3x) = 7
∴ 2x – 154 + 21x = 7
∴ 23x = 7 + 154
∴ 23x = 161
∴ x = \(\large \frac{161}{23}\)
∴ x = 7
Substituting x = 7 in equation (ii),
y = 22 – 3x
∴ y = 22 – 3(7)
∴ y = 22 – 21
∴ y = 1
∴ (7, 1) is the solution of the given equations.
Practice set 5.2
(1) In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.
Solution:
Let the number of ₹5 notes be ‘x’ and the number of ₹10 notes be ‘y’
Total amount of x notes of ₹ 5 = ₹ 5x
Total amount of y notes of ₹ 10 = ₹ 10y
∴ Total amount = 10y + 5x
According to the first condition,
Total amount of the notes together is ₹350.
∴ 5x + 10y = 350 …(i)
According to the second condition,
Number of ₹ 5 notes is less by 10 than twice the number of ₹ 10 notes.
∴ x = 2y – 10
∴ x – 2y = -10 …..(ii)
Multiplying equation (ii) by 5,
5x – 10y = -50 …(iii)
Adding equations (i) and (iii),
5x + 10y = 350
+ 5x – 10y = -50
10x = 300
∴ x = \(\large \frac{300}{10}\)
∴ x = 30
Substituting x = 30 in equation (ii),
x – 2y = -10
30 – 2y = -10
-2y = -10 -30
∴ -2y = -40
∴ y = \(\large \frac{-40}{-2}\)
∴ y = 20
Answer: There are 30 notes of ₹ 5 and 20 notes of ₹ 10 in the envelope.
(2) The denominator of a fraction is 1 less than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is 3 : 5. Find the fraction.
Solution:
Let the numerator of the fraction be ‘x’ and its denominator be ‘y’.
Then, the required fraction is \(\large \frac{x}{y}\)
According to the first condition,
The denominator is 1 less than twice its numerator.
∴ y = 2x – 1
∴ 2x – y = 1 …(i)
According to the second condition,
If 1 is added to the numerator and the denominator, the ratio of numerator to denominator is 3 : 5.
∴ \(\large \frac{(x\, +\, 1)}{(y\, +\, 1)}\) = \(\large \frac{3}{5}\)
∴ 5(x + 1) = 3(y + 1)
∴ 5x + 5 = 3y + 3
∴ 5x – 3y = 3 – 5
∴ 5x – 3y = -2 …(ii)
Multiplying equation (i) by 3,
6x – 3y = 3 …(iii)
Subtracting equation (ii) from (iii),
6x – 3y = 3
5x – 3y = -2
(-) (+) (+)
________________
x + 0y = 5
Substituting x = 5 in equation (i),
∴ 2x – y = 1
∴ 2(5) – y = 1
∴ 10 – y = 1
∴ – y = 1 – 10
∴ – y = – 9
∴ y = 9
The required fraction = \(\large \frac{x}{y}\) = \(\large \frac{5}{9}\)
Answer: The required fraction is \(\large \frac{5}{9}\).
(3) The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today’s ages.
Solution:
Let the present age of Priyanka be ‘x’ years and that of Deepika be ‘y’ years.
According to the first condition,
Priyanka’s age + Deepika’s age = 34 years
∴ x + y = 34 …(i)
According to the second condition,
Priyanka is elder to Deepika by 6 years.
∴ x = y + 6
∴ x – y = 6 …..(ii)
Adding equations (i) and (ii),
x + y = 34
+ x – y = 6
_____________
2x – 0y = 40
2x = 40
x = \(\large \frac{40}{2}\)
∴ x = 20
Substituting x = 20 in equation (i),
x + y = 34
∴ 20 + y = 34
∴ y = 34 – 20
∴ y = 14
Answer: The present age of Priyanka is 20 years and that of Deepika is 14 years.
(4) The total number of lions and peacocks in a certain zoo is 50. The total number of their legs are 140. Then find the number of lions and peacocks in the zoo.
Solution:
Let the number of lions in the zoo be ‘x’ and the number of peacocks be ‘y’.
According to the first condition,
The total number of lions and peacocks is 50.
∴ x + y = 50 …(i)
According to the second condition,
The total number of their legs is 140 (lions have 4 legs and peacocks have 2 legs).
∴ 4x + 2y = 140
Dividing both sides by 2,
2x + y = 70 …(ii)
Subtracting equation (i) from (ii),
2x + y = 70
x + y = 50
(-) (-) (-)
________________
x – 0y = 20
∴ x = 20
Substituting x = 20 in equation (i),
x + y = 50
∴ 20 + y = 50
∴ y = 50 – 20
∴ y = 30
Answer: The number of lions and peacocks in the zoo are 20 and 30 respectively.
(5) Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment.
Solution:
Let the original salary of Sanjay be ₹ ‘x’ and his yearly increment be ₹ ‘y’.
According to the first condition,
After 4 years his monthly salary was ₹ 4500.
∴ x + 4y = 4500 …..(i)
According to the second condition,
After 10 years his monthly salary became ₹ 5400.
∴ x + 10y = 5400 …(ii)
Subtracting equation (i) from (ii),
x + 10y = 5400
x + 4y = 4500
(-) (-) (-) .
________________
0x + 6y = 900
6y = 900
y = \(\large \frac{900}{6}\)
∴ y = 150
Substituting y = 150 in equation (i),
x + 4y = 4500
∴ x + 4(150) = 4500
∴ x + 600 = 4500
∴ x = 4500 – 600
∴ x = 3900
Answer: The original salary of Sanjay is ₹ 3900 and his yearly increment is ₹ 150.
(6) The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the total price of 2 chairs and 2 tables.
Solution:
Let the price of one chair be ₹ ‘x’ and that of one table be ₹ ‘y’.
According to the first condition,
The price of 3 chairs and 2 tables is ₹ 4500
∴ 3x + 2y = 4500 … (i)
According to the second condition,
The price of 5 chairs and 3 tables is ₹ 7000
∴ 5x + 3y = 7000 … (ii)
Multiplying equation (i) by 3,
3(3x + 2y) = 3(4500)
9x + 6y = 13500 … (iii)
Multiplying equation (ii) by 2,
2(5x + 3y) = 2(7000)
10x + 6y= 14000 … (iv)
Subtracting equation (iii) from (iv),
10x + 6y = 14000
9x + 6y = 13500
(-) (-) (-) .
________________
x + 0y = 500
∴ x = 500
Substituting x = 500 in equation (i),
3x + 2y = 4500
∴ 3(500) + 2y = 4500
∴ 1500 + 2y = 4500
∴ 2y = 4500 – 1500
∴ 2y = 3000
∴ y = \(\large \frac{3000}{2}\)
∴ y = 1500
∴ Price of 2 chairs and 2 tables = 2x + 2y
= 2(500) + 2(1500)
= 1000 + 3000
= ₹ 4000
Answer: The price of 2 chairs and 2 tables is ₹ 4000.
(7) The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.
∴ Original number = 10y + x
And, number obtained by interchanging digits = 10x + y
According to the first condition,
The sum of the digits in a two-digit number is 9
x + y = 9 … (i)
According to the second condition,
The number obtained by interchanging the digits exceeds the original number by 27
∴ 10x + y = 10y + x + 27
∴ 10x – x + y – 10y = 27
∴ 9x – 9y = 27
Dividing both sides by 9,
x – y = 3 …(ii)
Adding equations (i) and (ii),
x + y = 9
+ x – y = 3
___________
2x – 0y = 12
2x = 12
∴ x = \(\large \frac{12}{2}\)
∴ x = 6
Substituting x = 6 in equation (i),
x + y = 9
∴ 6 + y = 9
∴ y = 9 – 6
∴ y = 3
∴ Original number = 10y + x
= 10(3)+ 6
= 30 + 6
= 36
Answer: The two digit number is 36.
(8*) In ∆ABC, the measure of ∠A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle.
Solution:
Let the measure of ∠B be ‘x°’ and that of ∠C be ‘y°’.
According to the first condition,
m∠A = m∠B + m∠C
∴ m∠A = x° + y°
In ∆ABC, m∠A + m∠B + m∠C = 180°
…(Sum of the measures of the angles of a triangle is 180°)
∴ x + y + x + y = 180 ,
∴ 2x + 2y = 180
Dividing both sides by 2,
x + y = 90 …(i)
According to the second condition,
The ratio of the measures of ∠B and ∠C is 4 : 5.
∴ \(\large \frac{x}{y}\) = \(\large \frac{4}{5}\)
∴ 5x = 4y
∴ 5x – 4y = 0 … (ii)
Multiplying equation (i) by 4,
4(x + y) = 4(90)
∴ 4x + 4y = 360 …(iii)
Adding equations (ii) and (iii),
5x – 4y = 0
+ 4x + 4y = 360
_______________
9x – 0y = 360
9x = 360
∴ x = \(\large \frac{360}{9}\)
∴ x = 40
Substituting x = 40 in equation (i),
x + y = 90
∴ 40 + y = 90
∴ y = 90 – 40
∴ y = 50
∴ m∠A = x° + y° = 40° + 50° = 90°
and m∠B = x° = 40°
and m∠C = y° = 50°
Answer: The measures of ∠A, ∠B and ∠C are 90°, 40°, and 50° respectively.
(9*) Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to \(\large \frac{1}{3}\) of the larger part. Then find the length of the larger part.
Solution:
Let the length of the smaller part of the rope be ‘x’ cm and that of the larger part be ‘y’ cm.
According to the first condition,
Total length of the rope is 560 cm.
∴ x + y = 560 …(i)
Twice the length of the smaller part = 2x
\(\large \frac{1}{3}\) rd length of the larger part = \(\large \frac{1}{3}\) y
According to the second condition,
Twice the length of the smaller part is equal to ⅓ of the larger part,
2x = \(\large \frac{1}{3}\) y
∴ 6x = y
∴ 6x – y = 0 … (ii)
Adding equations (i) and (ii),
x + y = 560
+ 6x – y = 0
_______________
7x – 0y = 560
7x = 560
∴ x = \(\large \frac{560}{7}\)
∴ x = 80
Substituting x = 80 in equation (ii),
6x – y = 0
∴ 6(80) – y = 0
∴ 480 – y = 0
∴ y = 480
Answer: The length of the larger part of the rope is 480 cm.
(10) In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong ?
Solution:
Let us suppose that Yashwant got ‘x’ questions right and ‘y’ questions wrong.
According to the first condition,
Total number of questions in the examination are 60.
∴ x + y = 60 … (i)
Yashwant got 2 marks for each correct answer and 1 mark was deducted for each wrong answer.
∴ He got 2x – y marks.
According to the second condition,
He got 90 marks,
∴ 2x – y = 90 … (ii)
Adding equations (i) and (ii),
x + y = 60
+ 2x – y = 90
_______________
3x – 0y = 150
3x = 150
x = \(\large \frac{150}{3}\)
∴ x = 50
Substituting x = 50 in equation (i),
x + y = 60
∴ 50 + y = 60
∴ y = 60 – 50
∴ y = 10
Answer: Yashwant got 10 questions wrong.
Problem set 5
(1) Choose the correct alternative answers for the following questions.
(i) If 3x + 5y = 9 and 5x + 3y = 7 then what is the value of x + y ?
(A) 2
(B) 16
(C) 9
(D) 7
Solution:
Let,
3x + 5y = 9 … (i)
5x + 3y = 7 … (ii)
Multiplying equation (i) by 3,
3(3x + 5y) = 3(9)
9x + 15y = 27 … (iii)
Multiplying equation (ii) by 5,
5(5x + 3y) = 5(7)
25x + 15y = 35 … (iv)
Subtracting equation (iii) from (iv)
25x + 15y = 35
9x + 15y = 27
(-) (-) (-)
________________
16x + 0y = 8
16x = 8
x = \(\large \frac{8}{16}\)
∴ x = \(\large \frac{1}{2}\)
Substituting x = 1/2 in equation (i),
3x + 5y = 9
∴ 3(\(\large \frac{1}{2}\)) + 5y = 9
∴ \(\large \frac{3}{2}\) + 5y = 9
5y = 9 – (\(\large \frac{3}{2}\))
5y = (\(\large \frac{18}{2}\)) – (\(\large \frac{3}{2}\))
…(equalizing denominators)
5y = \(\large \frac{15}{2}\)
y = \(\large \frac{15}{2\, ×\, 5}\)
y = \(\large \frac{15}{10}\)
∴ y = \(\large \frac{3}{2}\)
Now,
x + y = \(\large \frac{1}{2}\) + \(\large \frac{3}{2}\) = \(\large \frac{4}{2}\) = 2
Option (A)
(ii) ‘When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.’ What is the mathematical form of the statement ?
(A) x – y = 8
(B) x + y = 8
(C) x + y = 23
(D) 2x + y = 21
Solution:
Let the length of the rectangle be ‘x’ and its breadth be ‘y’
According to the given condition,
Length becomes (x – 5)
Breadth becomes (y – 5)
Perimeter of the rectangle = 2l + 2b = 26
2(x – 5) + 2 (y – 5) = 26
∴ 2x – 10 + 2y – 10 = 26
∴ 2x + 2y – 20 = 26
∴ 2x + 2y = 26 + 20
∴ 2x + 2y = 46
Dividing the equation by 2, we get,
∴ x + y = 23
Option (C)
(iii) Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay’s age ?
(A) 20
(B) 15
(C) 10
(D) 5
Solution:
Let Ajay’s age be x and Vijay’s age be y.
According to the first condition,
Ajay is younger than Vijay by 5 years.
∴ x = y – 5
∴ x – y = -5 … (i)
According to the second condition,
Sum of their ages is 25 years.
∴ x + y = 25 …(ii)
Adding equations (i) and (ii), we get,
x – y = -5
+ x + y = 25
____________
2x – 0y = 20
2x = 20
∴ x = \(\large \frac{20}{2}\)
∴ x = 10
∴ Ajay’s age is 10 years.
Option (C)
(2) Solve the following simultaneous equations.
(i) 2x + y = 5 ; 3x – y = 5
Solution:
Let,
2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
______________
5x – 0y = 10
5x = 10
∴ x = \(\large \frac{10}{5}\)
∴ x = 2
Substituting x = 2 in equation (i),
2x + y = 5
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4
∴ y = 1
∴ (2, 1) is the solution of the given equations.
(ii) x – 2y = -1 ; 2x- y = 7
Solution:
Let,
x – 2y = -1
∴ x = –1 + 2y… (i)
∴ 2x – y = 7 … (ii)
Substituting x = 2y – 1 in equation (ii),
2x – y = 7
2(–1 + 2y) – y = 7
∴ –2 + 4y – y = 7
∴ 3y = 7 + 2
∴ 3y = 9
∴ y = \(\large \frac{9}{3}\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 1
∴ x = 2(3) – 1
∴ x = 6 – 1
∴ x = 5
∴ (5, 3) is the solution of the given equations.
(iii) x + y = 11 ; 2x – 3y = 7
Solution:
Let,
x + y = 11
∴ x = 11 – y …(i)
2x – 3y = 7 …….(ii)
Substituting x = 11 – y in equation (ii),
2(11 – y) – 3y = 7
∴ 22 – 2y – 3y = 1
∴ 22 – 5y = 7
∴ -5y = 7 – 22
∴ -5y = -15
∴ y = \(\large \frac{-15}{-5}\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 11 – y
∴ x = 11 – 3
∴ x = 8
∴ (8, 3) is the solution of the given equations.
(iv) 2x + y = -2 ; 3x – y = 7
Solution:
Let,
2x + y = -2 … (i)
3x – y = 7 … (ii)
Adding equations (i) and (ii),
2x + y = -2
+ 3x – y = 7
____________
5x – 0y = 5
5x = 5
∴ x = \(\large \frac{5}{5}\)
∴ x = 1
Substituting x = 1 in equation (i),
2x + y = -2
∴ 2(1) + y = -2
2 + y = -2
∴ y = –2 – 2
∴ y = -4
∴ (1, -4) is the solution of the given equations
(v) 2x – y = 5 ; 3x + 2y = 11
Solution:
Let,
2x – y = 5
∴ -y = 5 – 2x
∴ y = 2x – 5 …(i)
3x + 2y = 11 …(ii)
Substituting y = 2x – 5 in equation (ii),
3x + 2y = 11
3x + 2(2x – 5) = 11
∴ 3x + 4x – 10= 11
∴ 7x = 11 + 10
∴ 7x = 21
∴ x = \(\large \frac{21}{7}\)
∴ x = 3
Substituting x = 3 in equation (i),
y = 2x – 5
∴ y = 2(3) – 5
∴ y = 6 – 5
∴ y = 1
∴(3, 1) is the solution of the given equations.
(vi) x – 2y = -2 ; x + 2y = 10
Solution:
Let,
x – 2y = -2
∴ x = 2y – 2 …(i)
x + 2y = 10 …(ii)
Substituting x = 2y – 2 in equation (ii),
x + 2y = 10
2y – 2 + 2y = 10
∴ 4y = 10 + 2
∴ 4y = 12
∴ y = \(\large \frac{12}{4}\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 2
∴ x = 2(3) – 2
∴ x = 6 – 2
∴ x = 4
∴ (4, 3) is the solution of the given equations.
(3) By equating coefficients of variables, solve the following equations.
(i) 3x – 4y = 7; 5x + 2y = 3
Solution:
Let,
3x – 4y = 7 … (i)
5x + 2y = 3 … (ii)
Multiplying equation (ii) by 2,
10x + 4y = 6 … (iii)
Adding equations (i) and (iii),
3x – 4y = 7
+ 10x + 4y = 6
_______________
13x – 0y = 13
13x = 13
x = \(\large \frac{13}{13}\)
∴ x = 1
Substituting x = 1 in equation (i),
3x – 4y = 7
∴ 3(1) – 4y = 7
∴ 3 – 4y = 7
∴ -4y = 7 – 3
∴ -4y = 4
∴ y = \(\large \frac{4}{-4}\)
∴ y = -1
∴ (1, -1) is the solution of the given equations.
(ii) 5x + 7y = 17 ; 3x – 2y = 4
Solution:
Let,
5x + 7y = 17 … (i)
3x – 2y = 4 … (ii)
Multiplying equation (i) by 2,
2(5x + 7y) = 2(17)
∴ 10x + 14y = 34 … (iii)
Multiplying equation (ii) by 7,
7(3x – 2y) = 7(4)
∴ 21x – 14y = 28 … (iv)
Adding equations (iii) and (iv),
10x + 14y = 34
+ 21x – 14y = 28
__________________
31x – 0y = 62
31x = 62
∴ x = \(\large \frac{62}{31}\)
∴ x = 2
Substituting x = 2 in equation (ii),
3x – 2y = 4
∴ 3(2) – 2y = 4
∴ 6 – 2y = 4
∴ -2y = 4 – 6
∴ -2y = -2
∴ y = \(\large \frac{-2}{-2}\)
∴ y = 1
∴ (2, 1) is the solution of the given equations.
(iii) x – 2y = -10; 3x – 5y = -12
Solution:
Let,
x – 2y = -10 … (i)
3x – 5y = -12 … (ii)
Multiplying equation (i) by 3,
3x – 6y = -30 … (iii)
Subtracting equation (ii) from (iii),
3x – 6y = -30
3x – 5y = -12
(-) (+) (+) .
________________
0x – 1y = -18
-y = -18
∴ y = 18
Substituting y = 18 in equation (i),
x – 2y = -10
∴ x – 2(18) = -10
∴ x – 36 = -10
∴ x = -10 + 36
∴ x = 26
∴ (26, 18) is the solution of the given equations.
(iv) 4x + y = 34 ; x + 4y = 16
Solution:
Let,
4x + y = 34 …(i)
x + 4y = 16 …… (ii)
Multiplying equation (i) by 4,
4(4x + y) = 4(34)
16x + 4y = 136 … (iii)
Subtracting equation (ii) from (iii),
16x + 4y = 136
x + 4y = 16
(-) (-) (-) .
________________
15x – 0y = 120
15x = 120
∴ x = \(\large \frac{120}{15}\)
∴ x = 8
Substituting x = 8 in equation (i),
4x + y = 34
∴ 4(8) + y = 34
∴ 32 + y = 34
∴ y = 34 – 32
∴ y = 2
∴ (8, 2) is the solution of the given equations.
(4) Solve the following simultaneous equations.
(i) \(\large \frac {x}{3}\) + \(\large \frac {y}{4}\) = 4 ; \(\large \frac {x}{2}\) – \(\large \frac {y}{4}\) = 1
Solution:
\(\large \frac {x}{3}\) + \(\large \frac {y}{4}\) = 4
Multiplying both sides by 12,
12 × \(\large \frac {x}{3}\) + 12 × \(\large \frac {y}{4}\) = 12 × 4
∴ 4x + 3y = 48 …(i)
\(\large \frac {x}{2}\) – \(\large \frac {y}{4}\) = 1
Multiplying both sides by 8,
8 × \(\large \frac {x}{2}\) – 8 × \(\large \frac {y}{4}\) = 8 × 1
∴ 4x – 2y = 8 …(ii)
Subtracting equation (ii) from (i),
4x + 3y = 48
4x – 2y = 8
(-) (+) (-)
________________
0x + 5y = 40
∴ 5y = 40
∴ y = \(\large \frac {40}{5}\)
∴ y = 8
Substituting y = 8 in equation (ii),
4x – 2y = 8
∴ 4x – 2(8) = 8
∴ 4x – 16 = 8
∴ 4x = 8+ 16
∴ 4x = 24
∴ x = \(\large \frac {24}{4}\)
∴ x = 6
∴ (6, 8) is the solution of the given equations.
(ii) \(\large \frac {x}{3}\) + 5y = 13 ; 2x + \(\large \frac {y}{2}\) = 19
Solution:
\(\large \frac {x}{3}\) + 5y = 13
Multiplying both sides by 3,
3 × \(\large \frac {x}{3}\) + 3 × 5y = 3 × 13
x + 15y = 39 …(i)
2x + \(\large \frac {y}{2}\) = 19
Multiplying both sides by 2,
2 × 2x + 2 × \(\large \frac {y}{2}\) = 2 × 19
4x + y = 38 …(ii)
Multiplying equation (i) by 4,
4x + 60y = 156 …(iii)
Subtracting equation (ii) from (iii),
4x + 60y = 156
4x + y = 38
(-) (-) (-)
________________
0x + 59y = 118
∴ 59y = 118
∴ y = \(\large \frac {118}{59}\)
∴ y = 2
Substituting y = 2 in equation (i),
x + 15y = 39
∴ x + 15(2) = 39
∴ x + 30 = 39
∴ x = 39 – 30
∴ x = 9
∴ (9,2) is the solution of the given equations.
(iii) \(\large \frac {2}{x}\) + \(\large \frac {3}{y}\) = 13 ; \(\large \frac {5}{x}\) – \(\large \frac {4}{y}\) = -2
Solution:
Let \(\large \frac {1}{x}\) = a and \(\large \frac {1}{y}\) = b
The equations become,
2a + 3b = 13 …(i)
5a – 4b = -2 …(ii)
Multiplying equation (i) by 4,
8a + 12b = 52 …(iii)
Multiplying equation (ii) by 3,
15a – 12b = -6 …(iv)
Adding equations (iii) and (iv),
8a + 12b = 52
15a – 12b = – 6
________________
23a + 0b = 46
∴ 23a = 46
∴ a = \(\large \frac {46}{23}\)
∴ a = 2
Substituting a = 2 in equation (i),
2a + 3b = 13
∴ 2(2) + 3b = 13
∴ 4 + 3b = 13
∴ 3b = 13 – 4
∴ 3b = 9
∴ b = \(\large \frac {9}{3}\)
∴ b = 3
Resubstituting,
\(\large \frac {1}{x}\) = a
∴ \(\large \frac {1}{x}\) = 2
∴ 1 = 2 × x
∴ x = \(\large \frac {1}{2}\)
And,
\(\large \frac {1}{y}\) = b
∴ \(\large \frac {1}{y}\) = 3
∴ 1 = 3 × y
∴ y = \(\large \frac {1}{3}\)
∴ (\(\large \frac {1}{2}\), \(\large \frac {1}{3}\)) is the solution of the given equations.
(5*) A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Solution:
Let the digit in the unit’s place be ‘x’ and the digit in ten’s place be ‘y’.
| STORY TYPE | Digit in tens place | Digit in units place | Number | Sum of the digits |
|---|---|---|---|---|
For original number | y | x | 10y + x | y + x |
Number obtained by interchanging the digits | x | y | 10x + y | x + y |
According to the first condition,
The two digit number is 3 more than 4 times the sum of its digits.
∴ 10y + x = 4(x + y) + 3
∴ 10y + x = 4x + 4y + 3
∴ x – 4x + 10y – 4y = 3
∴ -3x + 6y = 3
Dividing both sides by -3,
x – 2y = -1 …(i)
According to the second condition,
If 18 is added to the number, the sum is equal to the number obtained by interchanging the digits.
10y + x + 18 = 10x + y
∴ x – 10x + 10y – y = -18
∴ -9x + 9y = -18
Dividing both sides by – 9,
x – y = 2 …(ii)
Subtracting equation (ii) from (i),
x – 2y = -1
x – y = 2
(-) (+) (-)
________________
0x + -y = -3
∴ -y = -3
∴ y = 3
Substituting y = 3 in equation (ii),
x – y = 2
∴ x – 3 = 2
∴ x = 2 + 3
∴ x = 5
∴ Original number = 10y + x
= 10(3) + 5
= 30 + 5
= 35
∴ The required number is 35.
(6) The total cost of 8 books and 5 pens is 420 rupees and the total cost of 5 books and 8 pens is 321 rupees. Find the cost of 1 book and 2 pens.
Solution:
Let the cost of one book be ₹ x and the cost of one pen be ₹ y.
According to the first condition,
The total cost of 8 books and 5 pens is ₹ 420.
∴ 8x + 5y = 420 …(i)
According to the second condition,
The total cost of 5 books and 8 pens is ₹ 321.
∴ 5x + 8y = 321 …(ii)
Multiplying equation (i) by 5,
40x + 25y = 2100 …(iii)
Multiplying equation (ii) by 8,
40x + 64y = 2568 …(iv)
Subtracting equation (iii) from (iv),
40x + 64y = 2568
40x + 25y = 2100
(-) (-) (-)
________________
0x + 39y = 468
∴ 39y = 468
∴ y = \(\large \frac {468}{39}\)
∴ y = 12
Substituting y = 12 in equation (i),
8x + 5y = 420
∴ 8x + 5(12) = 420
∴ 8x + 60 = 420
∴ 8x = 420 – 60
∴ 8x = 360
∴ x = \(\large \frac {360}{8}\)
∴ x = 45
Cost of 1 book and 2 pens = x + 2y
= 45 + 2(12)
= 45 + 24
= ₹69
∴ The cost of 1 book and 2 pens is ₹69
(7*) The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves rupees 200, find the income of each.
Solution:
Let the income of the first person be ₹ x and that of the second person be ₹ y.
According to the first condition,
Ratio of their incomes is 9 : 7.
∴ \(\large \frac {x}{y}\) = \(\large \frac {9}{7}\)
∴ 7x = 9y
∴ 7x – 9y = 0 …….(i)
Now,
Each person saves ₹ 200
∴ Expense = Income – Savings
Expenses of first person = x – 200
Expenses of second person = y – 200
According to the second condition,
The ratio of their expenses is 4 : 3
∴ \(\large \frac {x\,–\,200}{y\,–\,200}\) = 43
∴ 3(x – 200) = 4(y – 200)
∴ 3x – 600 = 4y – 800
∴ 3x – 4y = -800 + 600
∴ 3x – 4y = -200 …(ii)
Multiplying equation (i) by 4,
28x – 36y = 0 …(iii)
Multiplying equation (ii) by 9,
27x – 36y = -1800 …(iv)
Subtracting equation (iii) from (iv),
27x – 36y = -1800
28x – 36y = 0
(-) (+) (-)
________________
-x + 0y = -1800
∴ -x = -1800
∴ x = 1800
Substituting x = 1800 in equation (i),
7x – 9y = 0
∴ 7(1800) – 9y = 0
∴ 9y = 7 x 1800
∴ y = \(\large \frac {7\, ×\, 1800}{9}\)
y = 7 x 200
∴ y = 1400
∴ The income of first person is ₹ 1800 and that of second person is ₹ 1400.
(8*) If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of the rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Solution:
Let the length of the rectangle be ‘x’ units and the breadth of the rectangle be ‘y’ units.
Area of the rectangle = xy sq. units
According to the first condition,
Length of the rectangle is reduced by 5 units
∴ Length = x – 5
Breadth of the rectangle is increased by 3 units
∴ Breadth = y + 3
Area of the rectangle is reduced by 9 square units
∴ Area of the rectangle = xy – 9
∴ (x – 5)(y + 3) = xy – 9
∴ xy + 3x – 5y – 15 = xy – 9
∴ 3x – 5y = -9 + 15
∴ 3x – 5y = 6 …(i)
According to the second condition,
Length of the rectangle is reduced by 3 units
∴ Length = x – 3
Breadth of the rectangle is increased by 2 units
∴ Breadth = y + 2
Area of the rectangle is increased by 67 square units
∴ Area of the rectangle = xy + 67
(x – 3)(y + 2) = xy + 67
∴ xy + 2x – 3y – 6 = xy + 67
∴ 2x – 3y = 67 + 6
∴ 2x – 3y = 73 …(ii)
Multiplying equation (i) by 3,
9x – 15y = 18 …(iii)
Multiplying equation (ii) by 5,
10x – 15y = 365 …(iv)
Subtracting equation (iii) from (iv),
10x – 15y = 365
9x – 15y = 18
(-) (+) (-)
________________________
x + 0y = 347
∴ x = 347
Substituting x = 347 in equation (ii),
2x – 3y = 73
∴ 2(347) – 3y = 73
∴ 694 – 73 = 3y
∴ 621 = 3y
∴ y = \(\large \frac {621}{3}\)
∴ y = 207
∴ The length and breadth of rectangle are 347 units and 207 units respectively.
(9*) The distance between two places A and B on the road is 70 kilometers. A car starts from A and the other from B. If they travel in the same direction, they will meet after 7 hours. If they travel towards each other they will meet after 1 hour, then find their speeds.
Solution:
Let the speed of the car starting from A (first car) be ‘x’ km/hr and that starting from B (second car) be ‘y’ km/hr. (x > y)
According to the first condition,
Distance covered by the first car in 7 hours = 7x km
Distance covered by the second car in 7 hours = 7y km
If the cars are travelling in the same direction,
7x – 7y = 70
Dividing both sides by 7,
x – y = 10 …(i)
According to the second condition,
Distance covered by the first car in 1 hour = x km
Distance covered by the second car in 1 hour = y km
If the cars are travelling in the opposite direction
x + y = 70 …(ii)
Adding equations (i) and (ii),
x – y = 10
+ x + y = 70
________________
2x + 0y = 80
∴ 2x = 80
∴ x = \(\large \frac {80}{2}\)
∴ x = 40
Substituting x = 40 in equation (ii),
x + y = 70
∴ 40 + y = 70
∴ y = 70 – 40
∴ y = 30
∴ The speed of the cars starting from places A and B are 40 km/hr and 30 km/hr respectively.
(10*) The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.
| STORY TYPE | Digit in tens place | Digit in units place | Number | Sum of the digits |
|---|---|---|---|---|
For original number | y | x | 10y + x | y + x |
Number obtained by interchanging the digits | x | y | 10x + y | x + y |
According to the given condition,
The sum of a two digit number and the number obtained by interchanging its digits is 99.
∴ 10y + x + 10x +y = 99
∴ 11x + 11y = 99
Dividing both sides by 11,
x + y = 9
Since only 1 equation can be formed from the given information, hence there exist many solutions.
If y = 1, then x = 8
If y = 2, then x = 7
If y = 3, then x = 6 and so on.
∴ The number can be 18, 27, 36, … etc.