Pair Of Linear Equations In Two Variables

Let the number of times Akhila played Hoopla be x and number of times she played Giant wheel be y.

Given, number of times she played Hoopla is half the number of rides she had on the Giant Wheel.

⇒ x = 2y

⇒ x – 2y = 0
For above equation, we have following table

Each ride costs Rs 3, and a game of Hoopla costs Rs 4 and she spent Rs 20 in the fair.

⇒ 4x + 3y = 20

Thus, algebraically it is represented by :

x – 2y = 0 and 4x + 3y = 0

Geometrically, it can be graphed as,

Let the present age of father be x and present age of daughter be y.

Aftab tells his daughter, ‘Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.”

∴ x – 7 = 7(y – 7)
⇒ x - 7 = 7y - 49
⇒ x – 7y + 42 = 0
⇒ x = 7y - 42
⇒ x = 7(y - 6) [1]
The coordinates satisfying above equation are

Also, x + 3 = 3(y + 3)
⇒ x + 3 = 3y + 9
⇒ x – 3y – 6 = 0
⇒ x = 3y + 6
⇒ x = 3(y + 2) [2]

Given equations are 3x + 4y – 12 = 0 and 6x + 8y – 48 = 0

Let us plot the given equations. For this we will need some points to plot for the equation.

3 x + 4 y - 12 = 0

4 y = 12 - 3 x

Now let us take random values of x and obtain the corresponding values of y by putting in the equation.

at x = 0,

4 y = 12 - 3 x 0
4 y = 12
y = 3

Now at y = 0,

3 x + 4 x 0 = 12
3 x = 12
x = 4



So we have two points for the equation 3 x + 4 y - 12 = 0 and those are (0, 3) and (4, 0)


6 x + 8 y - 48 = 0

Now let us find points for this equation

at x = 0,

6 x 0 + 8 y = 48
8 y = 48
y = 6

at y = 0,

6 x + 8 x 0 = 48
6 x = 48
x = 8

So we have two points for the equation 6 x + 8 y - 48 = 0 and those are (8, 0) and (0, 6)


3x + 4y – 12 = 0 passes through (4, 0) and (0, 3)

6x + 8y – 48 = 0 passes through (8, 0) and (0, 6)

Given, Gloria is walking along the path joining (-2, 3) and (2, -2), while Suresh is walking along the path joining (0, 5) and (4, 0).

Two lines, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

If , then the lines coincide

If , then the lines are parallel

If , then the lines intersect

(i)

Thus, .

The lines intersect.

(ii)

Thus,

The lines are coincident.

(iii)

Thus,

The lines are parallel.

Two lines, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

If , then the lines coincide

If , then the lines are parallel

If , then the lines intersect

Given the linear equation .

An intersecting line is x + 2y – 4 =0

A parallel line is 4x + 6y – 12 = 0

A coincident line is 4x + 6y – 16 = 0

Let the cost of 1 kg of apple be x and cost of 1 kg of grape be y.

Given, cost of 2 kg of apples and l kg of grapes on a day was found to be Rs 160.

∴ 2x + y = 160

Also, after a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300.

∴ 4x + 2y = 300
Now to present these equations graphically plot the points of respective lines,
For 2x + y = 160,
y = 160 - 2x
When x = 50, y = 60
When x=60, y = 40
Table is :

Plot the points A(50,60) and B(60,40)

For 4x + 2y = 300,

⇒ y = 150 - 2x
When x = 40, y = 70
When x = 60, y = 30
Table is :

Plot the points C(40,-30) and D(60,30)

Let the speed of car from A be ‘a’ and of car from B be ‘b’

Speed = distance/time

Relative speed of cars when moving in same direction = a + b

Relative speed of cars when moving in opposite direction = a – b

Given, if they travel in the same direction, they meet in 7 hours, but if they travel towards each other, they meet in one hour. A and B are 70 km a part on a highway

⇒ a + b = 70 ------ (1)

Also, a – b = 70/7 = 10 ------ (2)

Adding (1) and (2)

⇒ 2a = 80

⇒ a = 40 km/hr

Thus, b = 40 – 10 = 30 km/hr

Speed = distance/time

Let the speed of sailor in still water be ‘a’ and speed of the current be ‘b’.

The relative speed of sailor going upstream = a – b

The relative speed of sailor going downstream = a + b

Given, sailor goes 8 km downstream in 40 minutes and returns in 1 hour.

⇒ a + b = 12 ----- (1)

Also,

a – b = 8/1 = 8 ------ (2)

Adding (1) and (2).

⇒ 2a = 20

⇒ a = 10km/hr

Thus,

b = 12 – 10 = 2 km/hr

Given: The boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream.
To find: the speed of stream and that of the boat in still water.
Solution:
Speed = distance/time

Let the speed of boat be ‘a’ and speed of stream be ‘b’

Relative speed of boat going upstream = a – b

Relative speed of boat going downstream = a + b

Given, boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream
and we know,

------- (1)

Also, ------ (2)

⇒ a- b = 5 .... (4)
a + b = 11 .... (5)
Add eq 4 and 5 to get
a - b + a + b = 5 + 11
⇒ 2a = 16
⇒ a = 8
Substitute value of a in eq 4 we get,
8 - b = 5
⇒ b = 8 - 5
⇒ b = 3
Speed of boat = 8 km/hr and speed of stream = 3 km/hr

Speed = distance/time

Let the speed of boat be ‘a’ and speed of stream be ‘b’

Relative speed of boat going upstream = a – b

Relative speed of boat going downstream = a + b

Given, boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in hrs

------- (1)

Also, ------ (2)

Multiplying eq1 by 3 and eq2 by 4 and (1) – (2)

⇒ a - b = 6 ------- (3)

Substituting value of (a - b) into eq1

⇒ a + b = 14 ------ (4)

From (1) and (2), we get

a = 10 km/hr and b = 4 km/hr

Speed = distance/time

Let the speed of Ajeet be ‘a’ and speed of Amit be ‘b’.

Given, while covering distance of 30 km. Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit.

Also,

Adding the two equations

⇒ 30/a - 15/a = 3

⇒ a = 5 km/hr

Thus, 30/5 – 30/b = 2

⇒ b = 7.5 km/hr

Given: A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours.

To find: the distance covered by the man and his original rate of walking.

Solution:

Let the original speed, original time taken and distance be ‘a’, ‘t’ and ‘d’.

As we know,

distance = speed × time

⇒ d = at ------- (1)

Given, if he walks 1/2 km = 0.5 km an hour faster, he takes 1 hour less,

⇒ d = (a + 0.5)(t – 1)

⇒ d = at + 0.5t – a – 0.5

⇒ at = at + 0.5t - a - 0.5 [From 1]

⇒ 0.5t - a = 0.5

Multiply the above equation by 10 to get,

⇒ 5t - 10a = 5

⇒ 10a = 5t - 5 -------- (2)

Also if he walks 1 km an hour slower, her takes 3 more hours.,

⇒d = (a – 1)(t + 3)

⇒ d = at + 3a - t - 3

⇒ at = at + 3a - t - 3 [From 1]

⇒ 3a - t = 3

⇒ t = 3a - 3 ------- (3)

Put the value of t in eq (2)

⇒ 10a = 5(3a - 3) - 5

⇒ 10a = 15a - 15 - 5

⇒ 10a - 15a = - 15 - 5

⇒ - 5a = - 20

⇒ a = 4

Putting back in (3), we get

⇒ t = 3(4) - 3 = 9

Therefore,

d = at
= 4(9) = 36

Hence, speed of man = a = 4 km/hr

distance covered by man = d = 36 km

Let the speed of train be ‘a’ and speed of car be ‘b’.

Speed = distance/time

Given, Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours if he travels 160 km by train and the rest by car. He takes 12 minutes more if the travels 240 km by train and the rest by car.

------- (1)

Also,

⇒ 240/a + 520/b = 41/5 ----- (2)

Multiplying eq1 by 3 and eq2 by 2 and subtracting eq2 from eq1

⇒ b = 100 km/hr

Thus, 160/a + 6 = 8

⇒ a = 80 km/hr

Given: A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km by train and the rest by car, he takes half an hour longer.

To find: the speed of the train and that of the car.

Solution:


Let the speed of train be ‘a’ and speed of car be ‘b’.

Speed = distance/time

Given, man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes.

------- (1)


But, if he travels 200 km by train and the rest by car, he takes half an hour longer.

------- (2)

Multiplying eq1 by 2 and subtracting from eq2

⇒ a = 100 km/hr

Thus, 4 + 200/b = 6.5

200/b = 6.5-4

⇒ b = 80 km/hr

Given: places A and B are 80 km apart form each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes.
To find: the speeds of the cars.
Solution:
Let the speed of car from A be ‘a’ and of car from B be ‘b’.

Speed = distance/time

Relative speed of cars when moving in same direction = a + b

Relative speed of cars when moving in opposite direction = a – b

Given, places A and B are 80 km apart form each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8 hours and if they move in opposite directions, they meet in 1 hour and 20 minutes.


⇒ a – b = 80/8 = 10 ------ (1)

Adding (1) and (2)
⇒ a - b + a + b = 10 + 60

⇒ 2a = 70

⇒ a = 35 km/hr

Thus, b = 35 – 10 = 25 km/hr

Speed = distance/time

Let the speed of boat be ‘a’ and speed of stream be ‘b’

Relative speed of boat going upstream = a – b

Relative speed of boat going downstream = a + b

Given, boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time

------- (1)

Also, ------ (2)

Equating (1) and (2), we get

⇒ 8/(a + b) = 4/(a – b)

⇒ a = 3b

Substitute value of a in eq1

⇒ 12/2b + 40/4b = 8

⇒ 16/b = 8

⇒ b = 2 km/hr

Thus, a = 6 km/hr.

Let the speed of train be ‘a’ and speed of bus be ‘b’.

Speed = distance/time

Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer

------- (1)

Also,

------- (2)

Multiplying eq1 by 5 and eq2 by 3 and subtracting eq2 from eq1

⇒ b = 80 km/hr

Thus, 60/a + 3 = 4

⇒ a = 60 km/hr
Hence speed of train is 60 km/h and bus is 80 km/h.

Let the speed of her rowing in still water be ‘a’ and
speed of current be ‘b.’
We know,

Relative speed of boat going upstream = a – b

Relative speed of boat going downstream = a + b

Given, Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours

⇒ a + b = 20/2 = 10 ---- (1)
and a – b = 4/2 = 2 ------ (2)

Adding eq1 and eq2

⇒ 2a = 12

⇒ a = 6 km/hr

Also, b = a – 2 = 4 km/hr

Let the speed of A be ‘a’ and speed of B be ‘b’.

Given, A takes 3 hours more than B to walk a distance of 30 km. But, if A doubles his pace (speed) he is ahead of B by hours

-------- (1)

------ (2)

Adding (1) and (2)

⇒ 15/a = 4.5

⇒ a = 150/45 = 10/3 km/hr

Thus,

⇒ b = 5 km/hr

Let the speed of train be ‘a’ and speed of taxi be ‘b’.

Speed = distance/time

Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi he takes 6 minutes longer.

------- (1)

Also,

------- (2)

Multiplying eq1 by 6 and eq2 by 5 and subtracting eq2 from eq1

⇒ a = 100 km/hr

Thus, 3 + 200/b = 5.5

⇒ b = 80 m/hr

Let the speed of train be ‘s’, scheduled time be ‘t’ and distance be ‘d’.

Speed = distance/time

Given, covered a certain distance at a uniform speed. If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/hr, it would have taken 3 hours more than the scheduled time.

⇒ d = st ------- (1)

d = (s + 10)(t – 2) ------ (2)

d = (s – 10)(t + 3) ------ (3)

Solving the above three equation we get, d = 600km, s = 50 km/hr and t = 12 hours

Let the speed of car from A be ‘a’ and of car from B be ‘b’

Speed = distance/time

Relative speed of cars when moving in same direction = a + b

Relative speed of cars when moving in opposite direction = a – b

Given, 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

⇒ a – b = 100/5 = 20 (1)

Also, a + b = 100/1 = 100 (2)

Adding (1) and (2)

⇒ 2a = 120

⇒ a = 60 km/hr

Thus, b = 60 – 20 = 40 km/hr
Speed of two cars are 60 km/h and 40 km/h

Area of a rectangle = l × b

Given, if in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units.

⇒ (l + 2)(b – 2) = lb – 28

⇒ 2b – 2l = - 24 ------ (1)

Also, (l – 1)(b + 2) = lb + 33

⇒ 2l – b = 35 ------- (2)

Adding (1) and (2), we get

⇒ b = 11

Thus, 2l = 46, l = 23

Area of the rectangle = lb = 253 square units.

Area of a rectangle = l × b

Given, area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres.
If the length is decreased by 7 metres and breadth is increased by 4 metres, the area is decreased by 21 sq. metres.

⇒ (l + 7)(b – 3) = lb

lb - 3l + 7b - 21 = lb

⇒ 7b – 3l = 21 ------ (1)

Also, (l – 7)(b + 4) = lb - 21

⇒ 4l – 7b = 7 ------- (2)

Adding (1) and (2), we get

7b - 3l + 4l - 7b = 21 + 7

⇒ l = 28 m

Thus, 7b – 84 = 21

⇒ b = 15 m

Hence the length is 28 m and breadth is 15 m.

Area of a rectangle = l × b

Given, if the length is increased by 3 metres and breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. If length is reduced by 1 metre and breadth is increased by 4 metres, the area is increased by 89 sq. metres.

⇒ (l + 3)(b – 4) = lb – 67

⇒ 3b – 4l = -55 ------ (1)

Also, (l – 1)(b + 4) = lb + 89

⇒ 4l – b = 93 ------- (2)

Adding (1) and (2), we get

⇒2 b = 38

⇒ b = 19

Thus, 4l – 19 = 93

⇒ l = 28m

Given, incomes of X and Y are in the ratio of 8 : 7 and their expenditures are in the ratio 19 : 16.

Let the incomes of X and Y be 8a and 7a

Let the expenditures of X and Y be 19b and 16b

Each saves Rs. 1250

⇒ 8a – 19b = 1250 -------- (1) and 7a – 16b = 1250 ------- (2)

Equating both equation

⇒ 8a – 19b = 7a – 16b

⇒ a = 3b

Substituting value of a in eq1

⇒ 24b – 19b = 1250

⇒ b = 250

Thus, a = 750

X’s income = 8a = Rs. 6000

Y’s income = 7a = Rs. 5250

Let the amount of money which A has be ’a’ and which B has be ‘b’.

Given, if A gives Rs 30 to B, then B will have twice the money left with A.

⇒ b + 30 = 2(a – 30)

⇒ b + 30 = 2a – 60

⇒ b = 2a-60-30

⇒ b = 2a – 90 --------- (1)


But, if B gives Rs 10 to A, then A will have thrice as much as is left with B

a + 10 = 3(b - 10)

a+10=3b-30

a=3b-30-10

⇒ a = 3b - 40 --------- (2)

Substituting ‘a’ from eq2 in eq1

b = 2(3b-40) – 90

⇒ b = 6b - 80 – 90

⇒ b = Rs. 34

Thus, a = 102 – 40 = Rs. 62

Hence, A has Rs 62 and B has Rs 34.

Let the number of candidates in rooms A and B be ‘a’ and ‘b’ respectively.

Given, if 10 candidates are sent from A to B, the number of students in each room is same.

⇒ a – 10 = b + 10

⇒ a – b = 20 ----- (1)

Also, if 20 candidates are sent from B to A, the number of students in A is double the number of students in B.

⇒ a + 20 = 2(b – 20)

⇒ a – 2b = -60 ------- (2)

Subtracting eq2 from eq1

⇒ a – b – a + 2b = 20 + 60

⇒ b = 80

Thus, a = 80 + 20 = 100

Given:2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys.

To find: How long would it take one man and one boy to do it.

Solution:


Let the number of days in which 1 man and 1 boy can do the work be ‘a’ and ‘b’ respectively.

In 1 day,

In a days 1 man can work

⇒ In 1 day 1/a man can do work

⇒ 2 men can do work in 2/a days

Similarly

In b days 1 boy can work

⇒ In 1 day 1/b boy can do work

⇒ 7 boys can do work in 7/b days


Given, 2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys.

-------- (1)

-------- (2)

Multiplying eq1 by 2 and subtracting eq2 from eq1

⇒ 10/b = 1/6

⇒ b = 60 days

Thus, 4/a + 1/15 = 1/3

⇒ 4/a = 4/15

⇒ a = 15 days

Sum of angles of a triangle = 180°

Given,

.

⇒ x + 3x – 2 + y = 180

⇒ 4x + y = 182 ------ (1)

Also, .

⇒ y – 3x + 2 = 9

⇒ y – 3x = 7 ------- (2)

(1) – (2)

⇒ 4x + y – y + 3x = 182 – 7

⇒ 7x = 175

⇒ x = 25°

Thus, y = 75 + 7 = 82°

∠A = 25°

∠B = 3X – 2 = 73°

∠C = 82°

Opposite angles of a cyclic quadrilateral are supplementary

∠A + ∠C = 180°

∠B + ∠D = 180°

Given, ,

2x + 4 + 2y + 10 = 180

⇒ x + y = 83 ------- (1)

y + 3 + 4x – 5 = 180

⇒ y + 4x = 182 -------- (2)

(1) – (2)

⇒ x – 4x = 83 – 182

⇒ x = 33

Thus, y = 50

∠A = 2x + 4 = 70°

∠B = y + 3 = 53°

∠C = 2y + 10 = 110°

∠D = 4x – 5 = 127°

Let the basic full fare be ‘a’, half far be ‘b’ and reservation charges be ‘r’.

Given, railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket.

a = 2b ----- (1)

Also, reserved first class ticket from Mumbai to Ahmedabad costs Rs 216 and one full and one half reserved first class tickets cost Rs 327.

⇒ a + r = 216 ------ (2)

Also, a + b + 2r = 327 ----- (3)

Substituting value of a in eq2 and eq3

⇒ 2b + r = 216 and 3b + 2r = 327

Solving the above equations, we get

r = Rs. 6 and b = Rs. 105

Thus a = Rs. 210

Sum of angles of a triangle = 180°

Given,

∴ x + 3x + y = 180

⇒ y + 4x = 180 ------- (1)

Also, 3y – 5x = 30 ------- (2)

Multiplying eq1 by 5 and eq2 by 4 and adding them

⇒ 5y + 12y = 900 + 120

⇒ y = 60

Thus, 60 + 4x = 180

⇒ x = 30

∠A = 30° , ∠B = 90° and ∠C = 60°

Triangle is right angled at B.

Let the fixed charge be ‘a’ and the charge per km travelled be ‘b’

Given, for a journey of 12 km, the charge paid is Rs 89 and for a journey of 20 km, the charge paid is Rs 145

⇒ a + 12b = 89 ----- (1)

and a + 20b = 145 ----- (2)

(1) – (2)

⇒ -8b = -56

⇒ b = 7

Thus, a + 84 = 89

⇒ a = 5

For a distance of 30km, amount paid = distance travelled x charge per kilometre + fixed charge
= 30 × 7 + 5 = Rs. 215

Let the fixed charges be ‘a’ and the cost of food per day be ‘b’

Given, when a student a takes food for 20 days, he has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges.

⇒ a + 20b = 1000

a + 26b = 1180

Subtracting one from another

⇒ 6b = 180

⇒ b = Rs. 30

Thus, a + 600 = 1000

⇒ a = Rs. 400

Given, half the perimeter of a garden, whose length is 4 more than its width is 36 m.

⇒ l = b + 4 and l + b = 36

Solving these two equations, we get

Length = 20 m, Width = 16 m

Given: The larger of two supplementary angles exceeds the smaller by 18 degrees.

To find: The measure of both angles.

solution:

Let the larger angle be ‘a’ and the smaller angle be ‘b’.

Given, larger of two supplementary angles exceeds the smaller by 18 degrees.

⇒ a = b + 18

⇒ a - b = 18 ...... (1)


The sum of supplementary angles is 180°.


⇒a + b = 180.....(2)

Adding the equations 1 and 2 we get,

⇒ 2a = 198

⇒ a = 99°

Put the value of b in eq. 1 to get,

99-b=18

⇒ -b=18-99

⇒ -b=-81

⇒ b=81

Thus, b = 81°

Hence the measure of smaller angle is 81° and larger angle is 99°.

Let the number of days in which 1 woman and 1 man can do the work be ‘a’ and ‘b’ days respectively.

In 1 day, 1 woman completes 1/a part while 1 man does 1/b part

Given, 2 Women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days.

------ (1)

-------(2)

Multiplying eq1 by 3 and eq2 by 2

⇒ 3/b = 1/12

⇒ b = 36 days

Thus, 3/a + 1/6 = 1/3

⇒ 3/a = 1/6

⇒ a = 18 days
Hence, A women can finish the embroidery in 18 days and A single man can finish the embroidery in 36 days

Let the stake money be ‘a’ and ‘b’ respectively.

Given, to one he said; if your bird wins, than you give me your stake-money, but if you do not win, I shall give you two third of that. Going to the other, he promised in the same way to give three fourths.

From both of them his gain would be only 12 gold coins

If the 1^st one wins

⇒ a - 3b/4 = 12 ----- (1)

If the 2^nd one wins

⇒ b – 2a/3 = 12 ----- (2)

Equating 1 and 2

⇒ 5a/3 = 7b/4

⇒ a = 21b/20

Thus, 21b/20 – 3b/4 = 12

⇒ 6b/20 = 12

⇒ b = 40

Thus, a = 21b/20 = Rs. 42

Given: Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all.

To find: how many notes of Rs 50 and Rs 100 she received.

Solution:

Let the number of Rs. 50 and Rs. 100 notes be ‘a’ and ‘b’.

Given, Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only.

⇒ 50a + 100b = 2000

⇒ a + 2b = 40 ------- (1)

⇒ a + b = 25 ----- (2)

(1) – (2)


a+2b-(a+b)=40-25

⇒ b = 15

Thus, a = 10

Therefore, Meena received 10 notes of denomination 50 and 15 notes of denomination 100.

Given: 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.
To find: How many questions were there in the test?
Solution:
Let the number of right answers be ‘a’ and number of wrong answers be ‘b’

Given, Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer.
Deducted marks are represented by "-" sign.

⇒ 3a – b = 40 ------ (1)
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

⇒ 4a – 2b = 50 ---- (2)

Multiplying eq1 by 2 and subtracting eq2 from it
⇒ 4a - 2b - 2(3a - b) = 50 - 2(40)
⇒ 4a - 2b - 6a +2b = 50 - 80
⇒ 4a - 6a = 50 - 80

⇒ 6a – 4a = 80 – 50

⇒ a = 15
Put the value of a in eq. 1 to get,
3(15) - b=40
⇒ 45 - b = 40
⇒ -b = 40 - 45
⇒ -b = -5
⇒ b = 5

Total number of questions in the test = right answers + wrong answers
= a + b
= 15 + 5
= 20

Let the number of students in a row be ‘a’ and number of rows be ‘b’.

Given, 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row there would be 2 rows more.

Number of students remain constant.

⇒ ab = (a + 3)(b – 1)
ab = (a – 3)(b + 2)

Solving the above equations we get , a = 9 and b = 4

Thus, number of students = ab = (9)(4) = 36

Given: One says, “give me hundred, friend! I shall then become twice as rich as you” The other replies, “If you give me ten, I shall be six times as rich as you.”

To find: the amount of their respective capital.

Solution:


Let the capitals be ‘a’ and ‘b’.

Given, one says, “give me hundred, friend! I shall then become twice as rich as you” .


Lets assume "b" gives hundred to "a".

According to given condition

a + 100 = 2(b – 100)

⇒ a + 100 = 2b - 200

⇒ a = 2b – 300


⇒ a - 2b = -300 ------ (1)


Now The other replies, “If you give me ten, I shall be six times as rich as you.”

So,


b + 10 = 6(a – 10)


⇒ b + 10 = 6a – 60


⇒ b = 6a – 60 - 10

⇒ b = 6a – 70


⇒ 6a - b = 70 ----- (2)

Multiplying eq1 by 6 and subtract from eq2

⇒ 6a - b - 6 ( a - 2b ) = 70 - 6 (- 300)

⇒ 6a - b - 6 a + 12b = 70 + 1800

⇒ 11b = 1870

⇒ b = 170

substitute the value of b in eq 1 to get,

a - 2(170) = -300

⇒ a - 340 = -300

⇒ a = -300 + 340

⇒ a = 40

The amount of their respective capital is Rs 40 and Rs 170.

In a cyclic quadrilateral sum of opposite angles is 180°.

Given, in a cyclic quadrilateral ABCD ,

∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°

⇒ 2x + 4 + 2y + 10 = 180 and y + 3 + 4x – 5 = 180

⇒ x + y = 83 --------- (1) and y + 4x = 182 ---------- (2)

Subtracting eq1 from eq2.

⇒ y + 4x – x – y = 182 – 83

⇒ 3x = 99

⇒ x = 33

Substituting in eq1.

⇒ y = 50

∠A = 2 × 33 + 4 = 70°

∠B = 50 + 3 = 53°

∠C = 2 × 50 + 10 = 110°

∠D = 4 × 33 – 5 = 127°

Given:

Equation 1: kx – y = 2

Equation 2: 6x – 2y = 3

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have unique solution we must have

………(i)

According to the problem:

a₁ = k

a₂ = 6

b₁ = – 1

b₂ = – 2

c₁ = 2

c₂ = 3

Putting the above values in equation (i) we get

⇒ k ≠ ⇒ k ≠ 3

Thevalue of k for which the system of equations has unique solution isk≠3

Given:

Equation 1: x + y = 4

Equation 2: 2x + ky = 3

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have no solutions we must have

………(i)

According to the problem:

a₁ = 1

a₂ = 2

b₁ = 1

b₂ = k

c₁ = 4

c₂ = 3

Putting the above values in equation (i) we get:

⇒ k = 2

Also we find

Thevalue of k for which the system of equations has no solution is k = 2

Given:

Equation 1: 2x + 3y = 5

Equation 2: 4x + ky = 10

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 2

a₂ = 4

b₁ = 3

b₂ = k

c₁ = 5

c₂ = 10

Putting the above values in equation (i) and solving the extreme left and middle portion of the equality we get the value of k

⇒ 2k = 12 ⇒ k = 6

Also we find

Thevalue of k for which the system of equations has infinitely many solution is k = 6

Given:

Equation 1: 2x – y = 5

Equation 2: 6x + ky = 15

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 2

a₂ = 6

b₁ = – 1

b₂ = k

c₁ = 5

c₂ = 15

Putting the above values in equation (i) we get:

⇒ 2k = – 6 ⇒ k = – 3

Also we find

Thevalue of k for which the system of equations has infinitely many solution is k = – 3

Given:

Equation 1: x + 2y = 3

Equation 2: 5x + ky = – 7

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have no solutions we must have

………(i)

According to the problem:

a₁ = 1

a₂ = 5

b₁ = 2

b₂ = k

c₁ = 3

c₂ = – 7

Putting the above values in equation (i) and solving we get:

⇒ k = 10

Also we find

Thevalue of k for which the system of equations has no solution is k = 10

Given:

Equation 1: 3x – 2y = 0

Equation 2: kx + 5y = 0

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 3

a₂ = k

b₁ = – 2

b₂ = 5

c₁ = 0

c₂ = 0

In this problem since c₁ & c₂ = 0 so which is undefined.

So for this problemthe system of linear equations will have infinite solutions if

.......(ii)

Putting the above values in equation (ii) we get:

⇒ – 2k = 15

⇒ k =

Thevalue of k for which the system of equations has infinitely many solution isk =

Given:

Equation 1: 3x + 5y = 0

Equation 2: kx + 10y = 0

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

According to the problem:

a₁ = 3

a₂ = k

b₁ = 5

b₂ = 10

c₁ = 0

c₂ = 0

The equation will have a non zero solution only when it will satisfy a non trivial solution i.e. the equations should satisfy with values other than x = 0 & y = 0 .For the given system of equations the equation will have a non zero solution if

.......(i)

Putting the above values in equation (i) we get:

⇒ 5k = 30 ⇒ k = 6

Thevalue of k for which the system of equations has non zero solution isk = 6

Given:

Equation 1: x + ky = 0

Equation 2: 2x – y = 0

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

According to the problem:

a₁ = 1

a₂ = 2

b₁ = k

b₂ = – 1

c₁ = 0

c₂ = 0

So for this problemthe system of linear equations will have unique solution if

.......(i)

Putting the above values in equation (i) we get:

⇒ k≠

Thevalue of k for which the system of equations has unique solution isk≠

Given:

Equation 1: 2x + 3y = 7

Equation 2: 2ax + (a + b)y = 28

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 2

a₂ = 2a

b₁ = 3

b₂ = (a + b)

c₁ = 7

c₂ = 28

Putting the above values in equation (i) we get:

…(ii)

To obtain the value of a & b we need to solve the above equality. First we solve the extreme left and extreme right of the equality to obtain the value of a.

⇒⇒ 2a*7 = 2*28 ⇒ 14a = 56 ⇒ a = 4

After obtaining the value of a we again solve the extreme left and middle portion of the equality (ii)

⇒ 2*(4 + b) = 3*2*4 ⇒ b + 4 = 12 ⇒ b = 8

Thevalue of a & b for which the system of equations has infinitely many solution is a = 4 & b = 8

Given:

Equation 1: 2x + 3y = 7

Equation 2: (a + b)x + (2a –b)y = 21

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 2

a₂ = (a + b)

b₁ = 3

b₂ = (2a – b)

c₁ = 7

c₂ = 21

Putting the above values in equation (i) we get:

On cross multiplication and solving the above equalities we get two sets of linear equation with the variables a & b.

⇒ 7(a + b) = 15*2 ⇒ 7a + 7b = 42 .......(ii)

⇒ 7(2a – b) = 15*3 ⇒ 14a – 7b = 63 ………(iii)

Equation (ii) and (iii) are two linear equations with a and b as variables. To solve this two set of linear equations we use the elimination technique.

In the elimination technique one variable is eliminated by equating it’s coefficient with the other equation. From equation (ii) and (iii) we first eliminate the variable b and find the value of a. Since the coefficient of b are equal but of opposite signs so we add equations (ii) and (iii) On adding we get

(14 + 7)a = 42 + 63 ⇒ 21a = 105 ⇒ a = 5

Putting the value of a in equation (ii) we get

7*5 + 7b = 42 ⇒ 7b = 42 – 35 ⇒ b = 1

Thevalue of a & b for which the system of equations has infinitely many solutions is a = 5 & b = 1.

Given:

Equation 1: 10x + 5y = (k – 5) Equation 2: 20x + 10y = k

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 10

a₂ = 20

b₁ = 5

b₂ = 10

c₁ = k – 5

c₂ = k

Putting the above values in equation (i) we get:

…(ii)

On solving the equality (ii) we get

5k = 10( k – 5 ) ⇒ 5k = 10k – 50 ⇒ 5k = 50 ⇒ k = 10

Thevalue of k for which the system of equations has infinitely many solution is k = 10

Given:

Equation 1: 3x + y = 1

Equation 2: (2k – 1)x + (k – 1)y = (2k + 1)

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have no solutions we must have

………(i)

According to the problem:

a₁ = 3

a₂ = (2k – 1)

b₁ = 1

b₂ = (k – 1)

c₁ = 1

c₂ = (2k + 1)

Putting the above values in equation (i) and solving we get:

⇒ 3 (k – 1) = 2k – 1 ⇒ 3k – 3 = 2k – 1 ⇒ k = 3 – 1 ⇒ k = 2

Therefore = ⇒ =

Putting the value of k we calculate

After comparing the ratio we find

So the given system of equations are inconsistent.

Thevalue of k for which the system of equations is inconsistent is k = 2

Given:

Equation 1: x + 2y = 8 Equation 2: 2x + 4y = 16

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

The system of linear equations needs to be analyzed by checking the nature of ratios of each coefficients in the above two equations.

According to the problem:

a₁ = 1

a₂ = 2

b₁ = 2

b₂ = 4

c₁ = 8

c₂ = 16

Comparing the ratios of the coefficients we see:

⇒ …(i)

⇒

⇒ …(ii)

⇒

⇒ …(iii)

On seeing equation (i), (ii) and (iii) we find

Conclusion: The system of linear equations have infinite number of solution.

The given system of linear equations will have infinite number of solutions for all values of x and y

Given:

Equation 1: ax + by = c

Equation 2: lx + my = n

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

According to the problem:

a₁ = a

a₂ = l

b₁ = b

b₂ = m

c₁ = c

c₂ = n

According to the question the condition given is

am _≠ bl …(i)

To develop a relationship between the coefficients we divide both sides of the equation by l*m

After dividing we get

Since a₁ = a ,a₂ = l ,b₁ = b ,b₂ = m

So

We know from our properties of linear equations that if the ratio of the coefficients of x and y are not equal then their exists a unique solution.

The given system of equation has a unique solution for all values of x and y

Given:

Equation 1: x + 3y = 4 Equation 2: 2x + 6y = 7

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

According to the problem:

a₁ = 1

a₂ = 2

b₁ = 3

b₂ = 6

c₁ = 4

c₂ = 7

Comparing the ratios of the coefficients we see:

⇒ …(i)

⇒ ⇒ …(ii)

⇒ …(iii)

On seeing equation (i), (ii) and (iii) we find

Conclusion: The system of linear equations has no solutions.

The given system of linear equations will have no solution for all values of x and y

Given:

Equation 1: 2x + 3y = 7

Equation 2: 2ax + (a + b)y = 28

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 2

a₂ = 2a

b₁ = 3

b₂ = (a + b)

c₁ = 7

c₂ = 28

Putting the above values in equation (i) and solving the extreme left and extreme right portion of the equality we get the value of a

⇒ 14a = 56 ⇒ a = 4

We now put the value of a and solve for b

⇒ ⇒ a + b = 12⇒ b = 8

So b = 2a

Thecorrect relationship between a & b for which the system of equations has infinitely many solution is b = 2a

Given:

Equation 1: x + 2y = 5

Equation 2: 3x + ky = – 15

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have no solutions we must have

………(i)

According to the problem:

a₁ = 1

a₂ = 3

b₁ = 2

b₂ = k

c₁ = 5

c₂ = – 15

Putting the above values in equation (i) and solving we get:

⇒ k = 6

Also we find

Thevalue of k for which the system of equations has no solution is k = 6

Given:

Equation 1: 2x – 3y = 7

Equation 2:(a + b)x + (a + b – 3)y = 4a + b

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

When two sets of linear equations which are coincident then they will have infinite number of solutions since both the equations represent the same line .So we have to use the conditions for the infinitely many number of solution.

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 2

a₂ = a + b

b₁ = – 3

b₂ = a + b – 3

c₁ = 7

c₂ = 4a + b

Putting the above values in equation (i) and solving the extreme left and middle portion of the equality

⇒ – 3(a + b) = 2(a + b – 3) ⇒ 5a + 5b = 6 …(ii)

Again We Solve for the extreme left and right side of the equality

⇒ 8a + 2b = 7a + 7b ⇒ a = 5b (iii)

We solve for a & b from Equation (ii) & (iii).We substitute the value of a from equation (iii) in equation (ii)

After substituting we get

5*5b + 5b = 6 ⇒ 30b = 6⇒ b =

Putting the value of b in equation (iii) we get a = 1

So 5b = a

The relationship between a and b for which the two equations represent coincident line is a = 5b or a – 5b = 0

A pair of linear equations is called inconsistent when the lines doesn’t have any solution. It means both the lines are parallel to each other.

A pair of linear equations is called consistent when they have infinite number of solutions or they have a unique solution.

An intersecting line will always have a unique solution.

A coincident line will have infinite number of solutions.

So the line represented by a pair of linear equations in two variables is always intersecting or coincident if the system of equation is consistent.

Let a₁x + b₁y = c₁ & a₂x + b₂y = c₂ be two lines where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

The system of linear equations will have unique solution if

.......(ii)

For the system of linear equations to be consistent either condition (i) or (ii) must be satisfied.

If the equations are consistent then they are either intersecting or coincident

Given:

The given linear equation is in the slope intercept form. Intercept means the distance at which the given equation cuts or meets the coordinate axis. In this problem a & b are the intercepts on the x and y axis respectively.

The triangle formed by a straight line with the coordinate axis is a right angled triangle where the angle subtended at origin is 90⁰. So the length of the x intercepts becomes the perpendicular and y intercept becomes the base of the triangle.

We know Area Of a triangle = * (perpendicular length) * (base length)

So Area of the triangle becomes ab

The Area of the triangle isab

Given:

Equation 1: y = x Equation 2: x = 6 Equation 3: y = 0

According to the given question Equation 1 , 2 & 3 cuts each other at three different points creating a triangle and we need to calculate the area of this triangle formed by these lines.

Now Equation 2 is a line parallel to Y axis at a distance of 6 units. Equation 3 is the equation of the x axis.

So we can say that equation 2 & 3 are mutually perpendicular to each other and the triangle formed by these 3 equations is a right angled triangle .

We solve equation 1 & 2 by substitution method

After solving we get x = 6 & y = 6

So the perpendicular height of the triangle turns out to be 6 units.

Since Equation 2 is at a distance of 6 units from the origin so the length of the base turns out to be 6 units.

Perpendicular height = 6 units

Base Length = 6 units.

Area of the Triangle = * (perpendicular length) * (base length)

Area of the triangle becomes = * 6 *6 = 18 sq.units

The Area of the triangle is 18 sq. units

Given:

Equation 1: 2x + 3y = 5

Equation 2: 4x + ky = 10

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have infinitely many solutions we must have

………(i)

According to the problem:

a₁ = 2

a₂ = 4

b₁ = 3

b₂ = k

c₁ = 5

c₂ = 10

Putting the above values in equation (i) and solving the extreme left and extreme right portion of the equality we get the value of a

⇒ 2k = 12 ⇒ k = 6

Thevalue of k for which the system of equations has infinitely many solution is k = 6

Given:

Equation 1: kx – 5y = 2

Equation 2: 6x + 2y = 7

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have no solutions we must have

………(i)

According to the problem:

a₁ = k

a₂ = 6

b₁ = – 5

b₂ = 2

c₁ = 2

c₂ = 7

Putting the above values in equation (i) and solving we get:

⇒ k ⇒ k = – 15

Also we find

Thevalue of k for which the system of equations has no solution is k = – 15

Given :

Equation 1: x = 3

Equation 2: y = 4

Equation 3: x = y

Equation 1 is a line parallel to y axis

Equation 2 is a line parallel to x axis

So Equation 1 & 2 are mutually perpendicular to each other.

Hence the triangle formed is a right angled triangle.

First we solve the three lines simultaneously by method of substitution and get the three points of intersection or three coordinates of the triangle.

Solving Equation 1 & 2 we get the coordinate ( 3, 4 ). Let this Coordinate name be P₁

Solving Equation 2 & 3 we get the coordinate ( 4 ,4 ). Let this Coordinate name be P₂

Solving Equation 3 & 1 we get the coordinate ( 3 ,3 ). Let this Coordinate name be P₃

We now use the formula for Area of a triangle through 3 given points

Area = *| x₁ * (y₂ – y₃) + x₂ * (y₃ – y₁) + x₃ * (y₁ – y₂) |

Where x_{1 ,}y₁ are the coordinates of P₁

x_2, y₂ are the coordinates of P₂

x_{3 ,}y₃ are the coordinates of P₃

Area of the Given Triangle = *| 3* (4– 3) + 4 * (3 – 4) + 3* (4– 4) |

Area = *| 3* (1) + 4 * ( – 1) + 3* (0) |

Area = *| 3– 4 |⇒ Area = sq. units

The Area of the triangle issq. units

Given

Equation 1: 2x + 3y = 12 Equation 2: x – y = 1 Equation 3: x = 0

To calculate the Area at first we solve the Equation 1 & 2 Simultaneously by method of substitution.

We substitute the value of x from Equation 2 in Equation 1 to get the value of y

Equation 2: x – y = 1 ⇒ x = y + 1

Equation 1: 2x + 3y = 12

Substituting the value from equation 2 we get

2(y + 1) + 3y = 12 ⇒ 2y + 2 + 3y = 12 ⇒ 5y = 10 ⇒ y = 2

Putting the value in Equation 1 we get

x = 2 + 1 ⇒ x = 3

So both this lines passes through ( 3 , 2) Let this Coordinate name be P₁

Equation 3 is the equation for y axis

Equation 1 meets Y axis at (0 ,4) which is calculated by substituting x = 0 in Equation 1. Let this Coordinate name be P₂

Equation 2 meets Y axis at (0 , – 1) which is calculated by substituting x = 0 in Equation 2. Let this Coordinate name be P₃

So Area of the triangle = *| x₁ * (y₂ – y₃) + x₂ * (y₃ – y₁) + x₃ * (y₁ – y₂) |

Where x_{1 ,}y₁ are the coordinates of P₁

x_2, y₂ are the coordinates of P₂

x_{3 ,}y₃ are the coordinates of P₃

⇒ Area of the Given Triangle = *| 3* (4+ 1) + 0 * ( – 1 – 2) + 0* (2– 4) |

⇒ Area of the Given Triangle = *| 3* 5 |

⇒ Area = ⇒ Area = 7.5 Sq. Units

Area of the triangle is 7.5 sq. Units

For the solutions first we make graph for the equations.

For x+y=3,

x + y = 3 passes through (3,0) and (0,3)
For 2x+5y=12,

2x + 5y = 12 passes through (6, 0) and (0, 2.4)
Now we plot the points and join them to make graph.
Wherever, they intersect that is common solution to both equations.

They meet at point x = 1 and y = 2.
So (1,2) is solution.

Given: The system of equations:

To find: The solution of above system.
Solution:

For x - 2y = 5,
Put x = 5, we get y = 0
Put x = 1, we get y = -2
Table for x - 2y = 5

x - 2y = 5, passes through (0,5) and (1,-2)

For 2x + 3y =10
Put x = -1 , we get y = 4
Put x = 2, we get y = 2
Table for 2x + 3y =10

2x + 3y =10 passes through (0,5) and (2,2)
Plot the points in the graph,

The lines meet at the point (5,0)
So x=5 and y=0.

3x + y + 1= 0 passes through (-1/3, 0) and (0, -1)

2x – 3y + 8 = 0 passes through (-4, 0) and (0, 8/3)

They meet at point x = -1 and y = 2

2x + y – 3 = 0 passes through (3/2, 0) and (0, 3)

2x – 3y – 7 = 0 passes through (7/2, 0) and (0, - 7/3)

They meet at point x = 2, y = -1

Given:
To show: The given pair of equations graphically.
Solution:
For x + y = 6,
Put y = 0, we get x = 6
Put x = 0, we get y = 6
Table for x + y = 6

x + y = 6 passes through (6, 0) and (0, 6)

For x – y = 2
Put y = 0, we get x = 2
Put x = 0, we get y = -2
Table for x – y = 2

x – y = 2 passes through (2, 0) and (0, - 2)
Plot the points in the graph,

They meet at point x = 4, y = 2

X – 2y = 6 passes through (6, 0) and (0, -3)

3x – 6y = 0 passes through origin

Both lines are parallel, thus it has no solution.

For x + y = 4,
Put y = 0, we get x = 4
Put x = 0, we get y = 4
Table for x + y = 4

x + y = 4 passes through (4, 0) and (0, 4)

For 2x – 3y = 3

Put y = 1, we get x = 3
Put x = 0, we get y = -1
Table for x – y = 2

2x – 3y = 3 passes through (3, 1) and (0, - )
Plot the points in the graph,

They meet at point (3,1)
Hence x = 3, y = 1.

2x + 3y = 4 passes through (2, 0) and (0, 4/3)

X – y + 3 = 0 passes through (-3, 0) and (0, 3)

They meet at point x = -1, y = 2

On solving graphically,

2x – 3y + 13 = 0 passes through (-6.5, 0) and (0, 13/3)

3x – 2y + 12 = 0 passes through (-4, 0) and (0, 6)

Thus the meeting point of the straight lines is, (-2, - 3)

For the equation 2x + 3y + 5 = 0,

When x = -1 , y =- 1
When x = 5 , y =- 5
Table is :

Plot the points A(-1,-1) B(-5,5) on the graph

For the equation 3x – 2y – 12 = 0 ,

When x = 0, y = -6
When x = 2 , y = -3
Table is :

Plot the points C(0,-6) D(2,-3) on the graph

Since the two lines meet at the point (2,-3)
Therefore, x = 2 and y = -3

Given: the system of equations:

To show:systems of equations has infinitely many solutions.
Solution:
consider the equation 2x + 3y = 6
To plot its graph,
we have

Putting x = 0 we get y = 2
putting y = 0 we get x = 3
The table for points of 2x + 3y = 6 is:

Plot A(0,2) and B(3,0) in the graph
Consider the equation 4x + 6y = 12
To plot its graph,
we have

Putting x = 6 we get y = -2
Putting y = 4 we get x = -3
The table for the points 4x + 6y = 12 is:

Plot C(6,-2) and D(-3, 4) in the graph

Now plot the graphs for these equations as:

As Both the lines pass through the same points and are coinciding.
There can be infinite points lying on both the lines.
Hence,systems of equations has infinitely many solutions.

Both pass through the same points and are coinciding.

Given: the system of equations:

To show:systems of equations has infinitely many solutions.
Solution:
consider the equation 3x+y=8
To plot its graph,
we have y=8-3x
Putting x=0 we get y=8
putting y=2 we get x=2
The table for points of 3x+y=8 is:

Plot A(0,8) and B(2,2) in the graph
Consider the equation 6x+2y=16
To plot its graph,
we have
Putting x=2 we get y=2
Putting y=0 we get x=8
The table for the points 6x+2y=16 is:

Now plot the graphs for these equations as:

As Both the lines pass through the same points and are coinciding.
There can be infinite points lying on both the lines.
Hence,systems of equations has infinitely many solutions.

Both pass through the same points and are coinciding.

3x – 5y = 20 passes through (20/3, 0) and (0, -4)

6x – 10y = - 40 passes through (-20/3, 0) and (0, 4)

Both are parallel lines, thus system of equations is in-consistent.

X – 2y = 6 passes through (6, 0) and (0, -3)

3x – 6y = 0 passes through origin

Both are parallel lines, thus system of equations is in-consistent.

2y – x = 9 passes through (-9, 0) and (0, 9/2)

6y – 3x = 21 passes through (-7, 0) and (0, 7/2)

Both are parallel lines, thus system of equations is in-consistent.

3x – 4y – 1 = 0 passes through (1/3, 0) and (0, -1/4)

2x – 8y/3 + 5 = 0 passes through (-5/2, 0) and (0, 15/8)

Both are parallel lines, thus system of equations is in-consistent.

(i)

2y – x = 8 passes through (-8, 0) and (0, 4)

5y – x = 14 passes through (-14, 0) and (0, 14/5)

Y – 2x = 1 passes through (0, 1) and (-1/2, 0)

Vertices of the triangle are (-4, 2), (1, 3) and (2, 5)

(ii)

3x + 3y = 10 passes through (10/3, 0) and (0, 10/3)

Vertices of the triangle are (0, 0), (10/3, 0) and (5/3, 5/3)

X – 2y = 2 passes through (2, 0) and (0, -1)

4x – 2y = 5 passes through (5/4, 0) and (0, - 5/2)

Thus the system of equations is consistent and it has a unique solution.

(i) 2x – 3y = 6 passes through (3, 0) and (0, -2)

X + y = 1 passes through (0, 1) and (1, 0)

Thus it has unique solution

(ii) 2y = 4x -6 passes through (3/2, 0) and (0, -3)

2x = y + 3 passes through (3/2, 0) and (0, -3)

The lines are coincident, it has infinite solutions

(i)

It has a unique solution at x = 3 and y = 2

Coordinates where lines meet y axis is (0, 4/5) and (0, 8)

(ii)

Unique solution at x = 2 and y = 3.

Lines meet the y axis at (0, 6) and (0, - 2)

(iii)

Unique solution at x = 4 and y = 3

Lines meet the y axis at (0, 11) and (0, -1)

(iv)

Unique solution at x = 3 and y = 2

Lines meet y axis at (0, 7/2) and (0,- 4)

(v)

Unique solution at x = 2 and y = -1

Lines meet the y-axis at (0, 5) and (0, - 5)

(vi)

Unique solution at x = 2 and y = -1

Lines meet y-axis at (0, -5) and (0, -3)

(i)

The triangle formed has vertices

(0, 0), (2, 4) and (3,3)

(ii)

The triangle formed has vertices

(0, 0), (4, 4) and (6, 2)

i) 2x + 3y = 12 passes through (6, 0) and (0, 4)

x – y = 1 passes through (1, 0) and (0, -1)

They meet at x = 3, y = 2

ii) 3x + 2y – 4 = 0 passes through, (4/3, 0) and (0, 2)

2x – 2y – 7 = 0 passes through (7/2, 0) and (0, -7/2)

They meet at x = 2, y = -1

iii) 3x + 2y – 11 = 0 passes through (11/3, 0) and (0, 11/2)

2x – 3y + 10 = 0 passes through (-5, 0) and (0, 10/3)

They meet at x = 1, y = 4

Lines meet at x = 3, y = 2

Coordinates of the vertices of the triangle formed with y –axis

(0, - 1), (0, 4) and (3, 2)

and

Coordinates of the vertices of triangle formed with x-axis

(2, 3), (-1, 0) and (4, 0)

Height of the triangle = 3 units

Base of the triangle = 5 units

Area = 0.5 × base × height = 7.5 sq units

They meet at x = 2, y = 4

Height of the triangle formed = 4 units

Base of the triangle formed = 5 + 1 = 6

Area of the region bounded by these lines and x-axis

= 0.5 × 4 × 6

= 12 sq. units

They meet at x = 3, y = 2

Height of the triangle formed with y-axis = 3 units

Base of the triangle formed = 11 + 1 = 12 units

Area of the bounded region = 0.5 × 3 × 36 = 18 sq. units

i)

Unique solution at x = 2, y = 2

Lines meet x-axis at (3, 0) and (-2, 0)

ii)

Unique solution at x = 3, y = 3

Lines meet x-axis at (1, 0) and (2, 0)

iii)

Unique solution at x =1, y = 2

Lines meet x-axis at (5, 0) and (-2, 0)

iv)

Unique solution at x =1, y = 2

Lines meet x-axis at (4, 0) and (-3, 0)

For equation, 2x - 3y + 6 = 0
Coordinates satisfying graph are,

and
y - 2 = 0
⇒ y = 2, which is a straight line parallel to x-axis intersecting the y-axis at 2
On plotting all the three line on graphs, we get the graph as follows

Vertices of the triangle obtained are

(0, 2), (3, 4) and (6, 2)

Height of triangle = 4 – 2 = 2

Base of the triangle = 6

Area of the triangle

= 6 square units

Unique solution at x = 3 and y = 4

Vertices of triangle formed with y-axis are

(0, 2), (3, 4) and (0, 6)

Height of the triangle obtained = 3 units

Base of the triangle obtained = 6 – 2 = 4 units

Area of the triangle = 0.5 × 3 × 4 = 6 sq. units

Unique solution at x = 5 and y = 0

Vertices of the triangle with y axis are

(0, 3), (0, - 4) and (5, 0)

Coordinates of the vertices of the triangle formed by these lines and y-axis

(1, 0), (0, -3), (0, -5)

Height of triangle = 1 unit

Base of triangle = 5 – 3 = 2 unit

Area of triangle = 0.5 × 1 × 2 = 1 sq. units

i) Let the number of boys and girls be ‘a’ and ‘b’.

10 students of class X took part in Mathematics quiz. If the number of girls is 4 more than the number of boys

⇒ a + b = 10 and b = a + 4

Thus, a = 3 and b = 7

ii ) Let the cost of 1 pen be a and 1 pencil be b.

5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46

⇒ 7a + 5b = 50 and 5a + 7b = 46

Solving by multiplying 1^st by 7 and 2^nd by 5 and subtracting

⇒ 24a = 120

⇒ a = Rs. 5

Thus, 35 + 5b = 50

⇒ b = Rs. 3

iii) Let the number of pants and skirts bought be ‘a’ and ‘b’

Champa went to a ‘sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased.”

⇒ b = 2a – 2 and b = 4a – 4

⇒ 2a – 2 = 4a – 4

⇒ a = 1

Thus, b = 2 – 2 = 0

i)

Unique solution at x = 1, y = -1

ii)

Unique solution at x = 2, y = 4

Lines meet y-axis at (0, 2) and (0, -4)

Two lines, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

If , then the lines coincide

If , then the lines are parallel

If , then the lines intersect

Given line

i) intersecting line : 3x + 2y – 6 = 0

ii) parallel line : 4x + 6y = 15

iii) coincident line : 4x + 6y – 16 = 0

[1]

[2]

Multiplying the 1^st equation by 2 and 2^nd equation by 15, we get
22x + 30y + 46 = 0 [3]
105x - 30y - 300 = 0 [4]

Adding [3] and [4]

⇒ 22x + 30y + 46 + 105x – 30y – 300 = 0

⇒ 127x = 254

⇒ x = 2

Substituting value of x in eq1.

⇒ 11(2) + 15y + 23 = 0

⇒ 15y = -45

⇒ y = -3

_{Multiplying eq1 by 2 and eq2 by 3 and adding.}

_{⇒6x – 14y + 20 + 3y – 6x – 9 = 0}

_{⇒-11y = -11}

_{⇒y = 1}

_{Substituting value of y in eq1, we get x = -1}

Multiplying eq1 by 0.2 and eq2 by 0.3 and adding

⇒ 0.08x + 0.06y + 0.21x – 0.06y = 0.34 + 0.24

⇒ 0.29x = 0.58

⇒ x = 2

Substituting value of x in eq1, 0.8x + 0.3y = 1.7

⇒ y = 3

= 10

⇒ x + y/2 = 0.7

Multiplying eq1 by 2 and subtracting eq2 from it.

⇒ x + 2y – x – y/2 = 1.6 – 0.7

⇒ 3y/2 = 0.9

⇒ y = 0.6

Thus, x/2 + 0.6 = 0.8

⇒ x = 0.4

7y + 21 - 2x - 4 = 14

7y - 2x = 14 - 21 + 4

⇒ 7y – 2x = -3 ---- (1)

4y - 8 + 3x - 9 = 2

4y + 3x = 2 + 8 + 9

⇒ 4y + 3x = 19 ------ (2)

Multiplying eq1 by 3 and eq2 by 2 and adding them

⇒ 21y – 6x + 8y + 6x = -9 + 38

⇒ 29 y = 29

⇒ y = 1

7 – 2x = -3

- 2x = -3 - 7

-2x = -10

⇒ x = 5

⇒ 3x + 7y = 105

⇒ 9x – 2y = 108

Multiplying eq1 by 3 and subtracting eq2 from it

⇒ 9x + 21y – 9x + 2y = 315 – 108

⇒ 23y = 207

⇒ y = 9

Thus, 9x – 18 = 108

⇒ x = 14

⇒ 4x + 3y = 132

⇒ 5x – 2y = -42

Multiplying eq1 by 2 and eq2 by 3 and adding them

⇒ 8x + 6y + 15x – 6y = 264 – 126

⇒ 23x = 138

⇒ x = 6

Thus, 24 + 3y = 132

⇒ y = 36

Multiplying eq1 by 4 and eq2 by 3 and adding

⇒ 16/x + 12y + 18/x – 12y = 32 – 15
⇒ 34/x = 17

⇒ x = 2

Thus, 2 + 3y = 8
⇒ 3y = 6

⇒ y = 2

⇒ 2x + y = 8 …(i)

⇒ x + 6y = 15 …(ii)

Multiplying eq2 by 2 and subtracting from eq1

⇒ 2x + y - 2x -12y = 8 – 30

⇒ -11y = -22

⇒ y = 2

Thus, 2x + 2 = 8

⇒ x = 3

⇒ 2x + 4y = 3 …(i)

⇒ 4x + 2y = 3 …(ii)

Multiplying eq2 by 2 and subtracting from eq1

2x + 4y – 8x – 4y = 3 – 6

⇒ -6x = - 3

⇒ x = 1/2

Thus, 2 × 1/2 + 4y = 3 ⇒ y = 1/2

Given: The following systems of equations:

To find : The values of x and y.
Solution:

2x = 3y

⇒
Squaring both sides to get,
3x = 8y ..... (1)
Substitute the value of x in (1),

Substitute the value of y in 2x = 3y
⇒2x=3(0)
⇒2x=0
⇒x=0

Solution is x=0,y=0.

Consider ,

⇒ 14y + x + 11 = 70

⇒ 14y + x = 59 .... (2)

Put the value of y from (1) in (2) to get,

14(33x-95) + x = 59

462x - 1330 + x = 59

463x = 1389

x = 3

Put the value of x in (1) to get,

33(3) - 95 = y

99 - 95 = y

4 = y

⇒ x = 3 and y = 4

Multiplying eq1 by 3 and eq2 by 2 and subtracting eq1 from eq2

⇒ 6x + 21/y – 6x + 6/y = 4 – 27

⇒ 23/y = - 23

⇒ y = -1

Thus, 2x + 3 = 9

⇒ x = 3

Multiplying eq1 by 0.3 and eq2 by 0.5 and subtracting eq1 from eq2

⇒ 0.15x + 0.25y – 0.15x – 0.21y = 0.25 – 0.222

⇒ 0.04y = 0.028

⇒ y = 0.7

Thus, x = 0.5

systems of equations:

To find: The values of x and y.
Solution:
Let's take,

The equations become:

6u + 7v = 126 ...... (1)

3u - 2v = 30 ...... (2)
Now multiply eq. (2) with 2 and subtract it from eq. (1)
⇒ 6u + 7v - 2(3u - 2v) = 126 - 2(30)
⇒ 6u + 7v - 6u +4v = 126 - 60
⇒ 11v = 66
⇒ v = 6
Now put the value of v in eq. (2) to get,
⇒ 3u-2(6)=30
⇒ 3u-12=30
⇒ 3u=42
⇒ u=14
Now

And

Multiplying eq 1 by 1/2 and eq2 by 1/3 and subtracting

⇒ 1/4x – 1/9x = 1 – 13/18

⇒ 5/36x = 5/18

⇒ x = 1/2

Thus, 1 + 1/3y = 2

⇒ y = 1/3

Adding both equation

⇒ 2x/xy = 8

⇒ y = 1/4

Thus,

⇒ x + 1/4 = x/2

⇒ x = -1/2

Multiplying eq2 by 2 and subtracting

13/u = 17 – 72/5

⇒ u = 5

Thus, 3 + 2/v = 17

⇒ v = 1/7

Multiplying eq1 by 3 and adding to eq1

⇒ 11/x = -27 +5

⇒ x = -1/2

Thus, -4 + 3/y = 5

⇒ y = 1/3

Multiplying eq1 by 8 and subtracting from eq2

⇒ 44/x = 19 – 8

⇒ x = 4

Thus, 2/4 + 5/y = 1

⇒ 5/y = 1/2

⇒ y = 10

Multiplying eq1 by 1/3 and eq2 by 1/5 and subtracting

⇒ 1/18y + 3/35y = 4 - 8/5

⇒ 89/630y = 12/5

⇒ y = 89/1512

Thus, 1/5x + 1512/534 = 12

⇒ 1/5x = 816/89

⇒ x = 89/4080

⇒ 2y + 3x = 9 ----- (1)

⇒ 4y + 9x = 21 ---- (2)

Multiplying eq1 by 2 and subtracting it from eq 2 we get,

⇒ x = 1

Thus, 2y + 3(1) = 9

2y + 3 = 9

2y = 9 - 3

2y = 6

y = 3

hence x = 1 and y = 3

Given : the following systems of equations:

...... (1)

Dividing the two equations

⇒ (y –x )/(x + y) = 1/5

⇒ 5y – 5x = x + y

⇒ y = 3x/2

Thus,

⇒ 3x/5 = 6/5

⇒ x = 2

Thus, y = 3

Multiplying eq1 by 3 and subtracting from eq2

⇒ -11/(x + y) = -1

⇒ x + y = 11 ---------- (3)

Multiplying eq1 by 5 and eq2 by 2 and subtracting

⇒ 15/(x –y) = 3

⇒ x – y = 5 ------ (4)

(3) + (4)

⇒ 2x = 16

⇒ x = 8

Thus, y = 3

Multiplying eq1 by 3 and subtracting from eq2

⇒ 13/(x – y) = 13

⇒ x – y = 1 ------ (3)

Multiplying eq1 by 7 and eq2 by 2 and adding

⇒ 65/(x + y) = 13

⇒ x + y = 5 ------- (4)

Thus, 2x = 6

⇒ x = 3

Y = x – 1

⇒ y = 2

Multiplying eq1 by 2 and adding to eq2

⇒ 15/(x + y) = 5

⇒ x + y = 3 ------- (3)

Multiplying eq1 by 3 and subtracting eq2 from it

⇒ 10/(x – y) = 5

⇒ x – y = 2 ---- (4)

Adding the two equation

⇒ 2x = 5

⇒ x = 5/2

Thus, y = 1/2

Multiplying eq1 by 5/2 and subtracting from eq2

⇒ 3x – 2y = - 1 ------- (3)

Multiplying eq1 by 12/25 and adding to eq2

⇒ x + 2y = 3 ------- (4)

Solving (3) and (4)

We get, x = 1/2 and y = 5/4

Adding eq1 and eq2

⇒ 15/(x + 1) = 3

⇒ x + 1 = 5

⇒ x = 4

Thus, 5/5 – 2/(y – 1) = 12

⇒ y – 1 = 4

⇒ y = 5

Subtracting the two eq.

⇒ -y = -8xy

⇒ x = 1/8

Thus, 1/8 + y = 5y/8

⇒ y = 1/3

x + y = 2xy

x – y = 6xy

Adding the two equation

⇒ 2x = 8xy

⇒ y = 1/4

Thus, x + 1/4 = 2 × x × 1/4

⇒ x/2 = -1/4

⇒ x = -1/2

Equating both equations

⇒ 6u – 2v = 2u + 6v

⇒ u = 2v

Substituting value of u

⇒ 2(6v – v) = 5 × 2v × v

⇒ v = 1

Thus, 2(3u – 1) = 5u

⇒ 6u – 2 = 5u

⇒ u = 2

Multiplying eq1 by 5 and eq2 by 2 and subtracting

⇒ 13/(3x – 2y) = 13

⇒ 3x – 2y = 1------- (3)

Multiplying eq2 by 3 and subtracting eq1 from it

⇒ 13/(3x +2y) = 13/5

⇒ 3x + 2y = 5 ------ (4)

(3) + (4)

⇒ 6x = 6

⇒ x = 1

Thus, 3 + 2 = 5

⇒ y = 1

Multiply eq1 by 4 and eq2 by 3 and adding

⇒ 25/x = 125

⇒ x = 1/5

Thus, 20 + 3y = 14

⇒ y =- 2

Adding both equations

⇒ 200x + 200y = 1000

⇒ x + y = 5 ------ (1)

Subtracting both equation

⇒ 2y – 2x = - 2

⇒ x – y = 1 ----- (2)

Adding 1 and 2

⇒ 2x = 6

⇒ x = 3

Thus, 3 + y =5

⇒ y = 2

Adding both equation

⇒ 52x – 52y = 208

⇒ x – y = 4 ----- (1)

Subtracting both equation

⇒ -6x - 6y = -12

⇒ x + y = 12 ----- (2)

(1) + (2)

⇒ 2x = 16

⇒ x = 8

Thus, y = 4

Subtracting eq2 and eq3

⇒ x + 3y = 0

⇒ x = -3y

Substituting in eq1

⇒ -3y – y + z = 4

⇒ z = 4 + 4y

Substituting in eq2

⇒ -3y – 2y + 12 + 12y = 1

⇒ 7y = -11

⇒ y = -11/7

Thus, x = 33/7

And z = 4 – 44/7 = - 16/7

Adding eq 1 and eq2

⇒ 2(x + z) = 6

⇒ x = 3 – z

(3) – (2) – (1)

⇒ y – 5z = - 4

⇒ y = -4 + 5z

Substituting in eq2

⇒ 3 – z - 4 + 5z + z = 2

⇒ 5z = 3

⇒ z = 3/5

Thus, x = 3 – z

⇒ x = 12/5

Y = 5z – 4

⇒ y = -1

Multiplying eq1 by 5 and eq2 by 4 and subtracting

⇒ -10/(x – y) = -2

⇒ x – y = 5

Multiply eq1 by 4 and eq2 by 3 and subtracting

⇒ 11/(x + y) = 1

⇒ (x + y) = 11

Thus, 2x = 16

⇒ x = 8

∴ y = 8 – 5 = 3

Multiply eq1 by 4 and eq2 by 5 and subtracting

⇒ 1/x = 3

⇒ x = 1/3

Thus, 12 + 5y = 7

⇒ y = 1

Multiplying eq1 by 4 and eq2 by 3 and adding

⇒ 23/x = 46

⇒ x = 1/2

Thus, 4 + 3/y = 13

⇒ y = 1/3

Given : The system of equations:

To find: The values of x and y
Solution:
Let
Now system of equations become:
5u + v = 2 ....... (1)
6u - 3v = 1 ....... (2)
Now multiply equation (1) by 3 and add to equation (2),
3(5u + v)+ 6u - 3v = 3(2) +1
15u +3v + 6u -3v = 6+1
21u = 7

so,

⇒x - 1 =3
⇒x = 3 + 1
⇒x = 4
Now put in the equation (1) to get v,



So,
⇒y - 2 = 3
⇒y = 3 + 2
⇒y=5
Hence the values are x= 4 and y=5.

Multiplying eq1 by 9 and eq2 by 2 and adding

⇒ 120/(x + y) = 32

⇒ x + y = 15/4 ---- (1)

Multiplying eq1 by 3 and eq2 by 2 and subtracting

⇒ 24/(x – y) = 16

⇒ x – y = 3/2 ---- (2)

Adding (1) and (2)

⇒ 2x = 21/4

⇒ x = 21/8

Thus, 21/8 – y = 3/2

⇒ y = 9/8

Multiplying eq2 by 2 and adding to eq1

⇒ 2/(3x + y) = 3/4 - 2/8

⇒ 3x + y = 4 ---- (3)

Multiplying eq2 by 2 and subtracting from eq1

⇒ 2/(3x – y ) = 1

⇒ 3x – y = 2 ----- (4)

Adding (3) and (4)

⇒ 6x = 6

⇒ x = 1

Thus, y = 3 – 2 = 1

Multiplying eq1 by 3 and adding to eq2

⇒ 10/√x = 5

⇒ √x = 2

⇒ x = 4

Thus,

⇒ 2/√4 + 3/√y = 2

⇒ y = 9

⇒ 7/y – 2/x = 5 ---- (1)

⇒ 8/y + 7/x = 15 ---- (2)

Multiplying eq1 by 7 and eq2 by 2 and adding

⇒ 65/y = 65

⇒ y = 1

Thus, 7 – 2/x = 5

⇒ x = 1

Adding (1) and (2)

⇒ -226x – 226y = -678

⇒ x + y = 3----- (3)

(1) – (2)

⇒ 530x – 530y = 530

⇒ x – y = 1 ------ (4)

Adding (3) and (4)

⇒ 2x = 4

⇒ x = 2

Thus, y = 1

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

⇒ x = -21/-7 = 3

⇒ y = 14/-7 = -2

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

⇒ x= 5 and y = -20

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

⇒ x = 75/5 = 15

⇒ y = 25/5 = 5

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

⇒ x = -4/-1 = 4

⇒ y = -2/-1 = 2

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

⇒ 1/x = -2

⇒ x = -1/2

Also, 1/y = 4

⇒ y = 1/4

Method of cross multiplication

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c_{2 =} 0

Given,

ax + by = a – b

⇒ x = 1 and y = -1

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

⇒ x = a and y = b

Method of cross multiplication

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Given,

⇒ x = a², y = b²