CBSE 10th Maths - Pair Of Linear Equations in Two Variables - Exercise 3.5

Solution:

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{1}{3}$ ,

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{-3}{-9}$ = $\dfrac{1}{3}$,

$\dfrac{c_{1}}{c_{2}}$= $\dfrac{-3}{-2}$ = $\dfrac{3}{2}$

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

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

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{2}{3}$ ,

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{1}{2}$ ,

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{-5}{-8}$

$\dfrac{a_{1}}{a_{2}}$ ≠ $\dfrac{b_{1}}{b_{2}}$

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

$\dfrac{x}{(b_{1}c_{2}-c_{1}b_{2})}$ = $\dfrac{y}{(c_{1}a_{2} – c_{2}a_{1})}$ = $\dfrac{1}{(a_{1}b_{2}-a_{2}b_{1})}$

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

$\dfrac{x}{2}$ = $\dfrac{y}{1}$ = 1

∴ x = 2 and y =1

(iii) Given, 3x – 5y = 20 and 6x – 10y = 40

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{-5}{-10}$ =$\dfrac{ 1}{2}$

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{20}{40}$ = $\dfrac{1}{2}$

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ = $\dfrac{c_{1}}{c_{2}}$

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{1}{3}$

$\dfrac{b_{1}}{b_{2}}$ =$\dfrac{ -3}{-3}$ = 1

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{-7}{-15}$

$\dfrac{a_{1}}{a_{2}}$ ≠$\dfrac{ b_{1}}{b_{2}}$

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

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

$\dfrac{x}{24}$ =$\dfrac{ y}{ -6}$ = $\dfrac{1}{6}$

$\dfrac{x}{24}$ = $\dfrac{1}{6}$ and$\dfrac{ y}{-6}$ = $\dfrac{1}{6}$

∴ x = 4 and y = 1.

Solution:

(i) 3y + 2x -7 =0

(a + b)y + (a-b)y – (3a + b -2) = 0

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{2}{(a-b)}$ ,

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{3}{(a+b)}$ ,

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{-7}{-(3a + b -2)}$

For infinitely many solutions,

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ =$\dfrac{ c_{1}}{c_{2}}$

Thus $\dfrac{2}{(a-b)}$ = $\dfrac{7}{(3a+b– 2)}$

6a + 2b – 4 = 7a – 7b

a – 9b = -4 ……………………………….(i)

$\dfrac{2}{(a-b)}$ = $\dfrac{3}{(a+b)}$

2a + 2b = 3a – 3b

a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4

b =1

Substituting this eq. in (ii), we get

a -5 x 1= 0

a = 5

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

(ii) 3x + y -1 = 0

(2k -1)x + (k-1)y – 2k -1 = 0

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{3}{(2k -1)}$ ,

$\dfrac{b_{1}}{b_{2}}$ = $\dfrac{1}{(k-1)}$,

$\dfrac{c_{1}}{c_{2}}$ = $\dfrac{-1}{(-2k -1)}$ = $\dfrac{1}{( 2k +1)}$

For no solutions

$\dfrac{a_{1}}{a_{2}}$ = $\dfrac{b_{1}}{b_{2}}$ ≠ $\dfrac{c_{1}}{c_{2}}$

$\dfrac{3}{(2k-1)}$ =$\dfrac{1}{(k -1)} $ ≠$\dfrac{ 1}{(2k +1)}$

$\dfrac{3}{(2k –1)}$ = $\dfrac{1}{(k -1)}$

3k -3 = 2k -1

k =2

Therefore, for k = 2 the given pair of linear equations will have no solution.

Solution:

8x + 5y = 9 …………………..(1)

3x + 2y = 4 ……………….….(2)

From equation (2) we get

x = $\dfrac{(4 – 2y )}{ 3}$ ……………………. (3)

Using this value in equation 1, we get

$\dfrac{8(4-2y)}{3}$ + 5y = 9

32 – 16y +15y = 27

-y = -5

y = 5 ……………………………….(4)

Using this value in equation (2), we get

3x + 10 = 4

x = -2

Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:

8x +5y – 9 = 0

3x + 2y – 4 = 0

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

$\dfrac{-x}{2}$ =$\dfrac{ y}{5}$ =$\dfrac{1}{1}$

∴ x = -2 and y =5.

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.

Let x be the fixed charge and y be the charge of food per day.

According to the question,

x + 20y = 1000……………….. (i)

x + 26y = 1180………………..(ii)

Subtracting (i) from (ii) we get

6y = 180

y = Rs.30

Using this value in equation (ii) we get

x = 1180 -26 x 30

x= Rs.400.

Therefore, fixed charges is Rs.400 and charge 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.

Let the fraction be x/y.

So, as per the question given,

$\dfrac{(x -1)}{y}$ = $\dfrac{1}{3}$ => 3x – y = 3…………………(1)

$\dfrac{x}{(y + 8)}$ = $\dfrac{1}{4}$ => 4x –y =8 ………………..(2)

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

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

(4×5)– y = 8

y= 12

Therefore, the fraction is $\dfrac{5}{12}$.

(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?

Let the number of right answers is x and number of wrong answers be y

According to the given question;

3x−y=40……..(1)

4x−2y=50

⇒2x−y=25…….(2)

Subtracting equation (2) from equation (1), we get;

x = 15 ….….(3)

Putting this in equation (2), we obtain;

30 – y = 25

Or y = 5

Therefore, 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?

Let x km/h be the speed of car from point A and y km/h be the speed of car from point B.

If the car travels in the same direction,

5x – 5y = 100

x – y = 20 …………………………………(i)

If the car travels in the opposite direction,

x + y = 100………………………………(ii)

Solving equation (i) and (ii), we get

x = 60 km/h………………………………………(iii)

Using this in equation (i), we get,

60 – y = 20

y = 40 km/h

Therefore, 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.

Let,

The length of rectangle = x unit

And breadth of the rectangle = y unit

Now, as per the question given,

(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,

$\dfrac{x}{(305 +18)}$ = $\dfrac{y}{(-12+183)}$ = $\dfrac{1}{(9+10)}$

$\dfrac{x}{323}$ =$\dfrac{ y}{171}$ = $\dfrac{1}{19}$

Therefore, x = 17 and y = 9.

Hence, the length of rectangle = 17 units

And breadth of the rectangle = 9 units