NCERT
10th Mathematics
Exercise 3.5
Topic: 3 Pair of Linear Equations in Two Variables
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EXERCISE 3.5
Q1. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using the cross-multiplication method.
(i) x – 3y – 3 = 0; 3x – 9y – 2 = 0, (ii) 2x + y = 5; 3x + 2y = 8(iii) 3x – 5y = 20; 6x – 10y = 40, (iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0.
Explanation:
1) For the equations,
a) a1x + b1y + c1 = 0 ------------ equation 1b) a2x + b2y + c2 = 0 ------------ equation 2
In cross multiplication method, if a1b2 - a2b1 ≠ 0 we have,
c) x = [(b1c2 - b2c1)/(a1b2 - a2b1)] ------------ equation 3
2) So, we can say that,
e) If a1/a2 ≠ b1/b2,then we get a unique solution.
f) If a1/a2 = b1/b2 = c1/c2, then there are infinitely many solutions.g) If a1/a2 = b1/b2 ≠ c1/c2, then there is no solution.
3) From Equation 3 and Equation 4, we can say that,
h) x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1) ------------ equation 5.
4) Steps to solve these equations by cross multiplication method:
i) Write the given equations in the form of equation 1 and equation 2.
ii) Write your equations using the formula given in the equation 5.
iii) Simplify and get the values of x and y. (Here, we have a1b2 - a2b1 ≠ 0).
Solution:
(i) x – 3y – 3 = 0; 3x – 9y – 2 = 0
1) From equation x – 3y – 3 = 0, and equation a1x + b1y + c1 = 0, we have,
a1 = 1, b1= – 3, c1 = – 3.
2) From equation 3x – 9y – 2 = 0, and equation a2x + b2y + c2 = 0, we have,
a2 = 3, b2= – 9, c2 = – 2.
3) Here, we have
a) (a1/ a2) = 1/3
b) (b1/ b2) = (-3)/(-9) = 1/3c) (c1/ c2) = (-3)/(-2) = 3/2
4) From the above equations, we can say that,
a1/a2 = b1/b2 ≠ c1/c2, then there is no solution.
(ii) 2x + y = 5; 3x + 2y = 8
1) Our equations are 2x + y – 5 = 0; 3x + 2y – 8 = 0.
2) From equation 2x + y – 5 = 0, and equation a1x + b1y + c1 = 0, we have,
a1 = 2, b1= 1, c1 = – 5.
3) From equation 3x + 2y – 8 = 0, and equation a2x + b2y + c2 = 0, we have,
a2 = 3, b2= 2, c2 = – 8.
4) Here, we have
a) (a1/ a2) = 2/3
b) (b1/ b2) = 1/2c) (c1/ c2) = (-5)/(-8) = 5/8
5) From the above equations, we can say that,
a1/a2 ≠ b1/b2,then we get a unique solution.
6) We can find b1c2 - b2c1 as follows,
b1c2 - b2c1 = (1)(-8) - (2)(-5)b1c2 - b2c1 = -8 + 10b1c2 - b2c1 = 2 ------------ equation 1
7) We can find c1a2 - c2a1 as follows,
c1a2 - c2a1 = (-5)(3) - (-8)(2)c1a2 - c2a1 = -15 + 16c1a2 - c2a1 = 1 ------------ equation 2
8) We can find a1b2 - a2b1 as follows,
9) By cross multiplication method, we have,
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/2 = y/1 = 1/1, so
x/2 = 1, and y/(1) = 1, so
x = 2 and y = 1
10) So, here x = 2 and y = 1.
(iii) 3x – 5y = 20; 6x – 10y = 40
1) Our equations are 3x - 5y – 20 = 0; 6x - 10y – 40 = 0.
2) From equation 3x - 5y – 20 = 0, and equation a1x + b1y + c1 = 0, we have,
a1 = 3, b1= - 5, c1 = – 20.
3) From equation 6x - 10y – 40 = 0, and equation a2x + b2y + c2 = 0, we have,
a2 = 6, b2= - 10, c2 = – 40.
4) Here, we have
a) (a1/ a2) = 3/6 = 1/2
b) (b1/ b2) = (-5)/(-10) = 5/10 = 1/2c) (c1/ c2) = (-20)/(-40) = 1/2
5) From the above equations, we can say that,
a1/a2 = b1/b2 = c1/c2, then there are infinitely many solutions.
(iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0.
1) Our equations are x - 3y – 7 = 0; 3x - 3y – 15 = 0.
2) From equation x - 3y – 7 = 0, and equation a1x + b1y + c1 = 0, we have,
a1 = 1, b1= - 3, c1 = – 7.
3) From equation 3x - 3y – 15 = 0, and equation a2x + b2y + c2 = 0, we have,
a2 = 3, b2= - 3, c2 = – 15.
4) Here, we have
a) (a1/ a2) = 1/3
b) (b1/ b2) = (-3)/(-3) = 1c) (c1/ c2) = (-7)/(-15) = 7/15
5) From the above equations, we can say that,
a1/a2 ≠ b1/b2,then we get a unique solution.
6) We can find b1c2 - b2c1 as follows,
b1c2 - b2c1 = (-3)(-15) - (-3)(-7)b1c2 - b2c1 = 45 - 21b1c2 - b2c1 = 24 ------------ equation 1
7) We can find c1a2 - c2a1 as follows,
c1a2 - c2a1 = (-7)(3) - (-15)(1)c1a2 - c2a1 = -21 + 15c1a2 - c2a1 = -6 ------------ equation 2
8) We can find a1b2 - a2b1 as follows,
9) By cross multiplication method, we have,
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/24 = y/(-6) = 1/6, so
x/24 = 1/6, and y/(-6) = 1/6, so
x = 24/6 and y = 6/(-6)
10) So, here x = 4 and y = -1.
Q2. (i) For which values of a and b do the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1
Explanation:
1) For the equations,
a) a1x + b1y + c1 = 0 ------------ equation 1b) a2x + b2y + c2 = 0 ------------ equation 2
2) So, we can say that,
c) If a1/a2 = b1/b2 = c1/c2, then there are infinitely many solutions.
d) If a1/a2 = b1/b2 ≠ c1/c2, then there is no solution.
Solution:
(i) For which values of a and b do the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2.
1) Our equations 2x + 3y - 7 = 0 and (a – b) x + (a + b) y - (3a + b – 2) = 0 have
infinite number of solutions, so we have,
a1/a2 = b1/b2 = c1/c2
2) From equation 2x + 3y – 7 = 0, and equation a1x + b1y + c1 = 0, we have,
a1 = 2, b1= 3, c1 = – 7.
3) From the equation,
(a – b) x + (a + b) y - (3a + b – 2) = 0, and a2x + b2y + c2 = 0, we have,
a2 = (a - b), b2= (a + b), c2 = –(3a + b - 2).
4) Here, we have
a) (a1/ a2) = (2)/(a - b)
b) (b1/ b2) = (3)/(a + b)c) (c1/ c2) = (-7)/(-(3a + b -2))
5) As our equations have an infinite number of solutions, so we have,
a1/a2 = b1/b2 = c1/c2 ,So we have,
(2)/(a - b) = (3)/(a + b) = (-7)/(-(3a + b - 2)) -------- equation 1
6) Equating the first two ratios of equation 1, we have
(2)/(a - b) = (3)/(a + b)
(2)(a + b) = (3)(a - b)
(2)(a + b) = (3)(a - b)
2a + 2b = 3a - 3b
2a - 3a = - 3b - 2b
- a = - 5b
a = 5b -------- equation 27) Equating the first and third ratios of equation 1, we have
(2)/(a - b) = (-7)/(-(3a + b - 2))
(2)/(a - b) = 7/(3a + b - 2)
(2)(3a + b - 2) = 7(a - b)
6a + 2b - 4 = 7a - 7b
7b + 2b - 7a + 6a = 4
9b - a = 4 -------- equation 3
8) Put the value of a = 5b from equation 2 in equation 3, we get,
9b - a = 4
9b - 5b = 4
4b = 4
b = 1-------------------------- equation 4.
9) Put the value of b = 1 from equation 4 in equation 2, we get,
a = 5b
a = 5(1)
a = 5 -------------------------- equation 5.
10) So, here a = 5, and b = 1.
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1
1) Our equations 3x + y - 1 = 0 and (2k – 1) x + (k – 1) y - (2k + 1) = 0 have
no solutions, so we have,
a1/a2 = b1/b2 ≠ c1/c2
2) From equation 3x + y - 1 = 0, and equation a1x + b1y + c1 = 0, we have,
a1 = 3, b1= 1, c1 = – 1.
3) From the equation,
(2k – 1) x + (k – 1) y - (2k + 1) = 0, and a2x + b2y + c2 = 0, we have,
a2 = (2k – 1), b2= (k – 1), c2 = – (2k + 1).
4) Here, we have
a) (a1/ a2) = (3)/(2k – 1)
b) (b1/ b2) = (1)/(k – 1)c) (c1/ c2) = (-1)/(-(2k + 1))
5) As our equations have no solutions, so we have,
a1/a2 = b1/b2 ≠ c1/c2 ,So we have,
(3)/(2k – 1) = (1)/(k – 1) = (-1)/(-(2k + 1)) -------- equation 1
6) Equating the first two ratios of equation 1, we have
(3)/(2k – 1) = (1)/(k – 1)
(3)(k – 1) = (1)(2k – 1)
3k - 3 = 2k - 1
3k - 2k = - 1 + 3
k = 2 -------- equation 2
7) So, here k = 2.
Q3. Solve the following pair of linear equations by the substitution and cross-multiplication methods :
8x + 5y = 9; 3x + 2y = 4
Explanation:
Substitution method:
2) Simply get the value of x or y with the easy steps (say).
3) Put the obtained value of x in the other equation and get the value of y.
Cross-multiplication methods:
1) For the equations,
a) a1x + b1y + c1 = 0 ------------ equation 1b) a2x + b2y + c2 = 0 ------------ equation 2
In cross multiplication method, if a1b2 - a2b1 ≠ 0 we have,
c) x = [(b1c2 - b2c1)/(a1b2 - a2b1)] ------------ equation 3
2) So, we can say that,
e) If a1/a2 ≠ b1/b2,then we get a unique solution.
f) If a1/a2 = b1/b2 = c1/c2, then there are infinitely many solutions.g) If a1/a2 = b1/b2 ≠ c1/c2, then there is no solution.
3) From Equation 3 and Equation 4, we can say that,
h) x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1) ------------ equation 5.
4) Steps to solve these equations by cross multiplication method:
i) Write the given equations in the form of equation 1 and equation 2.
ii) Write your equations using the formula given in the equation 5.
iii) Simplify and get the values of x and y. (Here, we have a1b2 - a2b1 ≠ 0).
Solution:
8x + 5y = 9; 3x + 2y = 4
a) Substitution method:
1) 8x + 5y = 9 ------------ equation 1
2) 3x + 2y = 4 ------------ equation 2
3) Simplify equation 2, we get
3x + 2y = 4
3x = - 2y + 4
x = (- 2y + 4)/3 ------------ equation 3
4) Substitute the value of x = (- 2y + 4)/3 from equation 3 in equation 1, we get
8x + 5y = 9
8(- 2y + 4)/3 + 5y = 9
[8(- 2y + 4) + 15y]/3 = 9
[-16y + 32 + 15y] = 9(3)
- y + 32 = 27
y = 32 - 27
y = 5 ------------ equation 4
5) Put the value of y = 5 from equation 4 in equation 3, we get
x = (- 2y + 4)/3
x = (- 2(5) + 4)/3
x = (- 10 + 4)/3
x = (- 6)/3x = - 2 ------------ equation 5
6) So, x = - 2 and y = 5.
8x + 5y = 9; 3x + 2y = 4
b) Cross-multiplication methods:
7) Our equations are 8x + 5y – 9 = 0; 3x + 2y – 4 = 0.
8) From equation 8x + 5y – 9 = 0, and equation a1x + b1y + c1 = 0, we have,
a1 = 8, b1= 5, c1 = – 9.
9) From equation 3x + 2y – 4 = 0, and equation a2x + b2y + c2 = 0, we have,
a2 = 3, b2= 2, c2 = – 4.
10) Here, we have
a) (a1/ a2) = 8/3
b) (b1/ b2) = 5/2c) (c1/ c2) = (-9)/(-4) = 9/4
11) From the above equations, we can say that,
a1/a2 ≠ b1/b2,then we get a unique solution.
12) We can find b1c2 - b2c1 as follows,
b1c2 - b2c1 = (5)(-4) - (2)(-9)b1c2 - b2c1 = - 20 + 18b1c2 - b2c1 = - 2 ------------ equation 6
13) We can find c1a2 - c2a1 as follows,
c1a2 - c2a1 = (-9)(3) - (-4)(8)c1a2 - c2a1 = -27 + 32c1a2 - c2a1 = 5 ------------ equation 7
14) We can find a1b2 - a2b1 as follows,
15) By cross multiplication method, we have,
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
x/(-2) = y/5 = 1/1, so
x/(-2) = 1, and y/(5) = 1, so
x = - 2 and y = 5
16) So, here x = - 2 and y = 5.
Q4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Explanation:
1) Here, let x and y be two variables.
2) Apply the given conditions and frame the equations.
3) We will get two equations from the above two conditions, then solve these
equations to get the values of x and y.
Solution:
(i) A part of monthly hostel charges is fixed and the remaining depends on
the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
1) Let the fixed charges be Rs x.
2) Let the charges for food per day be Rs y.
3) Student A paid Rs 1000 for 20 days, so we have,
x + 20 y = 1000 ------------ equation 1
4) Student A paid Rs 1180 for 26 days, so we have,
x + 26 y = 1180 ------------ equation 2
5) Subtract equation 1 from equation 2, and we get,
x + 26 y = 1180
x + 20 y = 1000
(-) (-) (-)
----------------------------
6 y = 180
y = 30 ---------------equation 3
5) Put the value of y = 30 from equation 3 in equation 1, we get,
x + 20 y = 1000
x + 20 (30) = 1000
x + 600 = 1000
x = 1000 - 600
x = 400 -------------------------- equation 4.
6) So, the fixed charges are Rs 400 and the cost of food per day is Rs 30.
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it
becomes 1/4 when 8 is added to its denominator. Find the fraction.
1) Let the numerator be x and the denominator be y.
(x - 1)/y = 1/3
3(x - 1) = y
3x - 3 = y
3x - y = 3 ------------ equation 1
2) According to the second condition, we have,
(x)/(y + 8) = 1/4
4x = y + 8
y = 4x - 8 ------------ equation 2
3) Put the value of y = 4x - 8 from equation 2 in equation 1, we get,
3x - y = 3
3x - (4x - 8) = 3
3x - 4x + 8 = 3
3x - 4x = 3 - 8
- x = - 5
x = 5 ---------------equation 3
4) Put the value of x = 5 from equation 3 in equation 2, we get,
y = 4x - 8
y = 4(5) - 8
y = 20 - 8
y = 12 -------------------------- equation 4.
5) So, here, the numerator is 5 and the denominator is 12. So the fraction is 5/12.
(iii) Yash scored 40 marks on a test, getting 3 marks for each right answer and
losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
1) Let the right answers be x and the wrong answers be y.
2) So the total questions will be (x + y).
3) According to the first condition, we have,
3x - y = 40 ------------ equation 1
4) According to the second condition, we have,
4x - 2y = 50
2x - y = 25
y = 2x - 25 ------------ equation 2
5) Put the value of y = 2x - 25 from equation 2 in equation 1, we get,
3x - y = 40
3x - (2x - 25) = 40
3x - 2x + 25 = 40
3x - 2x = 40 - 25
x = 15 -------------------------- equation 3.
6) Put the value of x = 15 from equation 3 in equation 2, we get,
y = 2x - 25
y = 2(15) - 25
y = 30 - 25
y = 5 -------------------------- equation 4.
6) The right answers are 15 and the wrong answers are 5, so the total number of questions is 20.
(iv) Places A and B are 100 km apart on a highway. One car starts from A and
another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
1) Let the speed of the car starting from place A be x km/h.
2) Let the speed of the car starting from place B be y km/h.
3) When a car travels in the same direction, and meets in 5 hrs, (speed = distance/time),
a) Car started from place A covered 5x km.
b) Car started from place B covered 5y km.
4) According to the first condition, as the places are 100 km apart, we have,
5x - 5y = 100
x - y = 20
y = x - 20 ------------ equation 1
5) When car travels towards each other, and meet in 1 hrs, (speed = distance/time),
a) Car started from place A covered x km.
b) Car started from place B covered y km.
6) According to the second condition, we have,
x + y = 100 ------------ equation 2
7) Put the value of y = x - 20 from equation 1 in equation 2, we get,
x + y = 100
x + (x - 20) = 100
2x = 100 + 20
2x = 120
x = 60 -------------------------- equation 3.
8) Put the value of x = 60 from equation 3 in equation 1, we get,
y = x - 20
y = 60 - 20
y = 40 -------------------------- equation 4.
9) The speed of the car starting from place A is 60 km/h and the speed of the car
starting from place B be 40 km/h.
(v) The area of a rectangle gets reduced by 9 square units if its length is
reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
1) Let the length be x units, and the breadth be y units.
2) So the area of the rectangle will be xy square units.
3) According to the first condition,
a) New area = xy - 9
b) New length = x - 5
c) New breadth = y + 3
(x - 5)(y + 3) = xy - 9
xy + 3x - 5y - 15 = xy - 9
3x - 5y - 15 = - 93x - 5y = 15 - 9
3x - 5y = 6
5y = 3x - 6
y = (3x - 6)/5 -------------------------- equation 1.
4) According to the second condition,
a) New area = xy + 67
b) New length = x + 3
c) New breadth = y + 2
(x + 3)(y + 2) = xy + 67
xy + 2x + 3y + 6 = xy + 67
2x + 3y + 6 = 67
2x + 3y = 67 - 6
2x + 3y = 61 -------------------------- equation 2.
5) Put the value of y = (3x - 6)/5 from equation 1 in equation 2, we get,
2x + 3y = 61
2x + [3(3x - 6)/5] = 61
[10x + 3(3x - 6)]/5 = 61
[10x + 3(3x - 6)] = 61(5)
[10x + 9x - 18] = 305
19x - 18 = 305
19x = 305 + 18
19x = 323
x = 323/19x = 17 -------------------------- equation 3.
6) Put the value of x = 17 from equation 3 in equation 1, we get,
y = (3x - 6)/5
y = (3(17) - 6)/5
y = (51 - 6)/5
y = (45)/5
y = 9 -------------------------- equation 4.
7) The length of the rectangle will be 17 units and the breadth of the rectangle will
be 9 units.
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