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

Last Updated : 03 Apr, 2024
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This exercise has been deleted as per new NCERT Syllabus

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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