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Learn Exercise 3.5 with Free Lessons & Tips

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

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

(iii) 3x – 5y = 20, 6x-10y=40

(iv) x – 3y – 7 = 0, 3x – 3y – 15 = 0

Solution 1:

$(i) Here, a_{1}=1, b_{_{1}}= -3, c_{1}=-3$

$a_{2}=3,b_{2}=-9,c_{2}=-2$

$\frac{a_{1}}{}a_{2}=\frac{1}{3}, \frac{b_{1}}{b_{2}}=\frac{-3}{-9}=\frac{1}{3}, \frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}$

$\frac{a_{1}}a_{2}{=\frac{b_{1}}b_{2}{}} eq \frac{c_{1}}{c_{2}}$

The equation has no solution.

$(ii)a_{1}=2, b_{1}=1, c_{1}=-5$

$a_{2}=3, b_{2}=2, c_{2}= -8$

$\frac{a_{1}}a_{2}{}=\frac{2}{3},\frac{b_{1}}{b_{2}}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{-5}{-8}$

$\frac{a_{1}}a_{2} eq \frac{b_{1}}{b_{2}}$

Hence a unique solution exists. By cross multiplication,

$(iii)$ $a_{1}=3, b_{1}=-5, c_{1}=-20$

$a_{2}=6, b_{2}=-10, c_{2}=-40$

$\frac{a_{1}}{a_{2}}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{1}{2}$

Thus, $\frac{a_{1}}a_{2}{=\frac{b_{1}}b_{2}{}}=\frac{c_{1}}{c_{2}}$

Thus, the given pair of equations has infinite solutions.

$(iv)a_{1}=1, b_{1}=-3, c_{1}=-7$

$a_{2}=3, b_{2}=-3,c_{2}=-15$

$\frac{a_{1}}a_{2}{=}\frac{1}{3},\frac{b_{1}}{b_{2}}=\frac{-3}{-3}=1, \frac{c_{1}}{c_{2}}=\frac{7}{15}$

$\frac{a_{1}}{a_{2}} eq \frac{b_{1}}{b_{2}}$

Thus, the given pair of equations has unique solution.

By cross-multiplication,

Comments

Tasneem

Question 2:

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 2:

Given equations have infinite number of solutions,

=> $\frac{a_{1}}a_{2}{=\frac{b_{1}}b_{2}{}}=\frac{c_{1}}{c_{2}}$

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

Taking the first 2 terms

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

Cross multiply we get,

2a +2b = 3a -3b

a-5b=0

a= 5b——(1)

Taking first and last term,

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

We get,

a- 9b +4=0——(2)

Substitute (1) in (2)

5b-9b+4=0

b= 1

Substitute b =1 in (1)

a=5

Thus the value of a and b are 5 and 1 respectively.

Comments

Bhavani Ramesh

Question 3:

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 3:

$Here, a_{1}=3, b_{1}=1, c_{1}=-1$

$a_{2}=2k-1, b_{2}=k-1, c_{2}=-(2k+1)$

$\frac{a_{1}}a_{2}{=\frac{3}{2k-1}}, \frac{b_{1}}{b_{2}}=\frac{1}{k-1}, \frac{c_{1}}{c_{2}}=\frac{-1}{-(2k+1)}$

$\frac{a_{1}}a_{2}{=\frac{b_{1}}b_{2}{}} eq \frac{c_{1}}{c_{2}}$

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

$k=2$

Comments

Tasneem

Question 4:

Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9, 3x + 2y = 4

Solution 4:

Substitution method:

Given 8x+5y=9—(1)

And 3x+2y=4 —(2)

Now rearrange (2) we get 2y=4-3x

=> $y=\frac{4-3x}{2}$

Substitute y =(4-3x)/2 in (1)

$\frac{8x+5(5-3x)}{2}=9$

Taking LCM we get $\frac{(16x+20-15x)}{2}=9$

=> x+20=18

=>x=18-20=-2

$\therefore x=-2$

Substitute x=-2 in y=(4-3x)/2

We get y=4+6/2=10/2=5 $\therefore y=5$

Cross multiplication method:

write the given equations as follows:

8x+5y-9=0

3x+2y-4=0

x/(-20+18)=y/(-28+32)=1/(16-15)

x/(-2)=y/5=1/1

Taking first and last term we get

x=-2 and y=5

Comments

Bhavani Ramesh

Question 5:

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.

Solution 5:

(i) Let the fixed charge of the food and the charge for food per day be x and y respectively.
According to the question,
x + 20y = 1000 ...(1)
x + 26y = 1180 ...(2)

Substituting this value of y in equation (1) from equation (2), we obtain
6y = 180
y = 30

Substituting this value of y in equation (1), we obtain:
x+ 20 x 30 = 1000
x = 1000 - 600
x = 400

Thus, the fixed charge of the food and the charge per day are Rs 400 and Rs 30 respectively.

(ii) Let the fraction be $\frac{x}{y}$ According to the question,

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

$3x-y=3.................(i)$

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

$4x-y=8.............(ii)$

Subtracting equation (1) from equation (2), we obtain:
x = 5
Putting the value of x in equation (1), we obtain:
15 - y = 3
y = 12

Hence the fraction is $\frac{5}{12}$

(iii) Let the number of right answers and wrong answers be x and y
respectively.
$3x-y=40............(i)$

$4x-2y=50$

$\Rightarrow 2x-y=25...............(ii)$

Subtracting equation (2) from equation (1), we obtain:
x = 15
Substituting the value of x in equation (2), we obtain:
30 - y = 25
y = 5

Thus, the number of right answers and the number of wrong answers is 15 and 5 respectively.
Therefore, the total number of questions is 20.

(iv) Let the speed of first car and second car be u km/h and v km/h respectively.

Speed of both cars while they are travelling in same direction = (u - v) km/h
Speed of both cars while they are travelling in opposite directions i.e., when they are travelling towards each other = (u + v) km/h

Distance travelled = Speed x Time

$5(u-v)=100$

$u-v=20...............(i)$

$1(u+v)=100$

$\Rightarrow u+v=100............(ii)$

adding the two equations, we get, $u=60$

Substituting the value of u in equation (2), we obtain:
v = 40
Hence, speed of the first car is 60 km/h and speed of the second car is 40 km/h.

(v) Let length and breadth of rectangle be x unit and y unit respectively.

Area = xy

$(x-5)(y+3)=xy-9$

$\Rightarrow 3x-5y-6=0..............(i)$

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

$\Rightarrow \Rightarrow 2x+3y-61=0.............(ii)$

By cross multiplication method,

Thus, the length and breadth of the rectangle are 17 units and 9 units respectively.

Comments

Tasneem

All Exercises - Chapter 3 - Pair of Linear Equations in Two Variable

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Exercise 3.5

Exercise 3.6

Exercise 3.7

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Learn Exercise 3.5 with Free Lessons & Tips

All Exercises - Chapter 3 - Pair of Linear Equations in Two Variable

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