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Class 10 NCERT Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.5

  • Last Updated : 12 Mar, 2021

Question 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0, 3x – 9y – 2 = 0

Solution:

Here, 

a1 = 1, b1 = -3, c1 = -3

a2 = 3, b2 = -9, c2 = -2

So,



\frac{a_1}{a_2} = \frac{1}{3}

\frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}

\frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}

As, \frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}

Hence, the given pairs of equations have no solution.

(ii) 2x + y = 5, 3x + 2y = 8

Solution:

Rearranging equations, we get

2x + y -5 = 0



3x + 2y -8 = 0

Here,

a1 = 2, b1 = 1, c1 = -5

a2 = 3, b2 = 2, c2 = -8

So,

\frac{a_1}{a_2} = \frac{2}{3}

\frac{b_1}{b_2} = \frac{1}{2}

As, \frac{a_1}{a_2} ≠ \frac{b_1}{b_2}

Hence, the given pairs of equations have unique solution.

For cross multiplication,



\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}

\frac{x}{(1)(-8)-(2)(-5)} = \frac{y}{(-5)(3)-(-8)(2)} = \frac{1}{(2)(2)-(3)(1)}

\frac{x}{(-8)+10} = \frac{y}{(-15)+16} = \frac{1}{4-3}

\frac{x}{2}  = y = 1

Hence, 

\frac{x}{2}  = 1

x = 2

and, y = 1

Hence, the required solution is x = 2 and y = 1.

(iii) 3x – 5y = 20, 6x – 10y = 40 

Solution:



Rearranging equations, we get

3x – 5y – 20 = 0

6x – 10y – 40 = 0

Here,

a1 = 3, b1 = -5, c1 = -20

a2 = 6, b2 = -10, c2 = -40

So,

\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}

\frac{b_1}{b_2} = \frac{-5}{-10} = \frac{1}{2}

\frac{c_1}{c_2} = \frac{-20}{-40} = \frac{1}{2}



As,

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

Hence, the given pairs of equations have infinitely many solutions.

(iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0

Solution:

Here,

a1 = 1, b1 = -3, c1 = -7

a2 = 3, b2 = -3, c2 = -15

So,

\frac{a_1}{a_2} = \frac{1}{3}

\frac{b_1}{b_2} = \frac{-3}{-3} = 1

As,

\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}

Hence, the given pairs of equations have unique solution.

For cross multiplication,

\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}

\frac{x}{(-3)(-15)-(-3)(-7)} = \frac{y}{(-7)(3)-(-15)(1)} = \frac{1}{(1)(-3)-(3)(-3)}

\frac{x}{45-21} = \frac{y}{(-21)+15} = \frac{1}{-3+9}

\frac{x}{24} = \frac{y}{-6} = \frac{1}{6}

Hence,

x = \frac{24}{6}



x = 4

and, y = \frac{-6}{6}

y = -1

Hence, the required solution is x = 4 and y = -1.

Question 2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

Solution:

Here,

a1 = 2, b1 = 3, c1 = -7

a2 = a-b, b2 = a+b, c2 = -(3a+b-2)

For having infinite number of solutions, it has to satisfy below conditions:

\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

\frac{2}{a-b} = \frac{3}{a+b} = \frac{7}{(3a+b-2)}

Now, on comparing

\frac{2}{a-b} = \frac{3}{a+b}

2(a+b) = 3(a-b)

2a+2b = 3a – 3b

a – 5b = 0 …………………(1)

And, now on comparing

\frac{3}{a+b} = \frac{7}{(3a+b-2)}

3(3a+b-2) = 7(a+b)

9a+3b-6 = 7a+7b



2a-4b-6=0 

Reducing form, we get

a-2b-3=0 …………………(2)

Now, new values for

a1 = 1, b1 = -5, c1 = 0

a2 = 1, b2 = -2, c2 = -3

Solving Eq(1) and Eq(2), by cross multiplication,

\mathbf{\frac{a}{b_1c_2-b_2c_1} = \frac{b}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}

\frac{a}{(-5)(-3)-(-2)(0)} = \frac{b}{(0)(1)-(-3)(1)} = \frac{1}{(1)(-2)-(1)(-5)}

\frac{a}{15-0} = \frac{b}{0+3} = \frac{1}{-2+5}

\frac{a}{15} = \frac{b}{3} = \frac{1}{3}

Hence,

a = \frac{15}{3}

a = 5

and, b = \frac{3}{3}

b = 1

Hence, For values a = 5 and b = 1 pair of linear equations have an infinite number of solutions

(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

Solution:

Here,

a1 = 3, b1 = 1, c1 = -1



a2 = (2k-1), b2 = k-1, c2 = -(2k+1)

For having no solution, it has to satisfy below conditions:

\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}

\frac{3}{2k-1} = \frac{1}{k-1} ≠ \frac{1}{2k+1}

Now, on comparing

\frac{3}{2k-1} = \frac{1}{k-1}

3(k-1) = 2k-1

3k-3 = 2k-1

k = 2

And, now on comparing

\frac{1}{k-1} ≠ \frac{1}{2k+1}

2k+1 ≠ k-1

k ≠ -2

Hence, for k = 2 and k ≠ -2 the pair of linear equations have no solution.

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9

3x + 2y = 4

Solution:

8x + 5y – 9 = 0  …………………(1)

3x + 2y – 4 = 0  …………………(2)

Substitution Method

From Eq(2), we get

x = 4-2y/3

Now, substituting it in Eq(1), we get

8(4-2y/3) + 5y – 9 = 0

32-16y/3 + 5y – 9 = 0

32 – 16y + 15y – 27 = 0

y = 5

Now, substituting y = 5 in Eq(2), we get

3x + 2(5) – 4 = 0

3x = -6

x = -2

Cross Multiplication Method

Here,

a1 = 8, b1 = 5, c1 = -9

a2 = 3, b2 = 2, c2 = -4

So,

\frac{a_1}{a_2} = \frac{8}{3}

\frac{b_1}{b_2} = \frac{5}{2}

As,

\frac{a_1}{a_2} ≠ \frac{b_1}{b_2}

Hence, the given pairs of equations have unique solution.

For cross multiplication,



\mathbf{\frac{x}{b_1c_2-b_2c_1} = \frac{y}{c_1a_2-c_2a_1} = \frac{1}{a_1b_2-a_2b_1}}

\frac{x}{(5)(-4)-(2)(-9)} = \frac{y}{(-9)(3)-(-4)(8)} = \frac{1}{(8)(2)-(3)(5)}

\frac{x}{(-20)+18} = \frac{y}{(-27)+32} = \frac{1}{16-15}

\frac{x}{-2} = \frac{y}{5}  = 1

Hence,

\frac{x}{-2}  = 1

x = -2

and, \frac{y}{-5}  = 1

y = 5

Hence, the required solution is x = -2 and y = 5.

Question 4. 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 ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution:

Let’s take,

Fixed charge = x

Charge of food per day = y

According to the given question,

x + 20y = 1000 ……………….. (1)

x + 26y = 1180 ………………..(2)

Subtracting Eq(1) from Eq(2) we get

6y = 180

y = 30



Now, substituting y = 30 in Eq(2), we get

x + 20(30) = 1000

x = 1000 – 600

x= 400.

Hence, fixed charges is ₹ 400 and charge per day is ₹ 30.

(ii) A fraction become \frac{1}{3}  when 1 is subtracted from the numerator, and it becomes \frac{1}{4}  when 8 is added to its denominator. Find the fraction.

Solution:

Let the fraction be \frac{x}{y} .

So, as per the question given,

\frac{x-1}{y} = \frac{1}{3}

3x – y = 3  …………………(1)

\frac{x}{y+8} = \frac{1}{4}

4x –y =8  ………………..(2)

Subtracting Eq(1) from Eq(2) , we get

x = 5

Now, substituting x = 5 in Eq(2), we get

4(5)– y = 8

y = 12

Hence, the fraction is \mathbf{\frac{5}{12}} .

(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?

Solution:

Let’s take 

Number of right answers = x 

Number of wrong answers = y

According to the given question;

3x−y=40 ……….……..(1)

4x−2y=50

2x−y=25 ……………….(2)

Subtracting Eq(2) from Eq(1), we get

x = 15

Now, substituting x = 15 in Eq(2), we get

2(15) – y = 25

y = 30-25

y = 5

Hence, number of right answers = 15 and number of wrong answers = 5

Hence, total number of questions = 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?

Solution:

Let’s take

Speed of car from point A = x km/he 

Speed of car from point B = y km/h

When car travels in the same direction,

5x – 5y = 100



x – y = 20 ………………(1)

When car travels in the opposite direction,

x + y = 100  ………………..(2)

Subtracting Eq(1) from Eq(2), we get

2y = 80

y = 40

Now, substituting y = 40 in Eq(1), we get

x – 40 = 20

x = 60

Hence, the speed of car from point A = 60 km/h

Speed of car from point B = 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.

Solution:

Let’s take

Length of rectangle = x unit

Breadth of the rectangle = y unit

Area of rectangle will be = xy sq. units

According to the given conditions,

(x – 5) (y + 3) = xy -9

3x – 5y – 6 = 0 ……………………(1)

(x + 3) (y + 2) = xy + 67

2x + 3y – 61 = 0 ……………………..(2)

Using cross multiplication method, we get,

\frac{x}{(-5)(-61)-(3)(-6)} = \frac{y}{(-6)(2)-(-61)(3)} = \frac{1}{(3)(3)-(2)(-5)}

\frac{x}{305 +18} = \frac{y}{-12+183} = \frac{1}{9+10}

\frac{x}{323} = \frac{y}{171} = \frac{1}{19}

Hence, 

\frac{x}{323} = \frac{1}{19}

x = 17

and, \frac{y}{171} = \frac{1}{19}

y = 9

Hence, the required solution is x = 17 and y = 9.

Length of rectangle = 17 units

Breadth of the rectangle = 9 units

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