7/4/2018 …...2x+6y−14=0 x+3y−7=0 hence this is the required relation between x and y. #465357...

7/4/2018 https://community.toppr.com/content/questions/print/?show_answer=1&show_topic=1&show_solution=1&page=1&qid=465389%2C+4653…

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#465350

Topic: Solution of Pair of Equations

Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution

The given points are collinear then the area of triangle formed by these points will be 0

∴1

2x1(y2 − y1) + x2(y3 − y1) + x3(y1 − y2) = 0

Here (x1, y1) = (x, y)

(x2, y2) = (1, 2)

(x3, y3) = (7, 0)

⟹1

2[x(2 − 0) + 1(0 − y) + 7(y − 2)] = 0

⟹1

2[2x − y + 7y − 14] = 0

⟹1

2[2x + 6y − 14] = 0

⟹ 2x + 6y − 14 = 0

⟹ x + 3y − 7 = 0

Hence this is the required relation between x and y.

#465357

Topic: Pair of Linear Equations

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." (Isnt this

interesting?) Represent this situation algebraically and graphically.

Solution

Consider Aftab's age as x and his daughter's age as y.

Then , seven years ago,

Aftab's age = x − 7

His daughter's age = y − 7

According to the question,

x − 7 = 7(y − 7)

x − 7 = 7y − 49

x − 7y = − 49 + 7

x − 7y = − 42 …(i)

After three years,

Aftab's age = x + 3

His daughter's age = y + 3


x + 3 = 3(y + 3)

x + 3 = 3y + 9

x − 3y = 9 − 3

x − 3y = 6 …(ii)

Representing equation (i) and (ii) geometrically, we plot these equations by finding points on the lines representing these two equations

x − 7y = − 42 ⟹ x = 7y − 42

x = 7y − 42 42 35 49 y 12 11 13

[ ]



x − 3y = 6 ⟹ x = 3y + 6

x = 3y + 6 42 36 48 y 12 10 14

From the graph we can see that two lines will intersect at a point.

x − 7y = − 42

x − 3y = 6

On subtracting the two equations, we get

4y = 48

or y = 12

Substituting value of y in (2),

x − 36 = 6

∴ x = 42

#465358


The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation

algebraically and geometrically.

Solution



Let the cost of a bat be Rs. x and cost of a ball be Rs. y.

Then algebraically representation of the question is -

3x + 6y = 3900 ⇒ x + 2y = 1300......(1)

x + 3y = 1300........(2)

To represent it geometrically, we plot the two equations

x + 3y = 1300 ⇒ y =1300 − x

3

x 100 1000 1300

y =1300 − x

3 400 100 0

x + 2y = 1300 ⇒ y =1300 − x

2

x 100 500 1300

y =1300 − x

2600 400 0

As we can see, the lines intersect at (1300, 0) which is the solution for this question.

#465361


The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the

situation algebraically and geometrically.

Solution



x 65 55 45 35

y 20 40 60 80

Let the cost of 1 kg of apples be x and that of 1 kg be y

So the algebraic representation can be as follows:

2x + y = 160

4x + 2y = 300 ⇒ 2x + y = 150

The situation can be represented graphically by plotting these two equations.

2x + y = 160 ⇒ y = 160 − 2x

x 70 60 50 40 y = 160 − 2x 20 40 60 80

2x + y = 150 ⇒ y = 150 − 2x

x 65 55 45 35 y = 150 − 2x 20 40 60 80

We can see that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

#465362


Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Solution



i) Consider number of boys = x

and number of girls = y.


x + y = 10

or, y = 10 − x ………(i)

y = x + 4 ………..(ii)

x 1 2 3 4

y 9 8 7 6

x 1 2 3 4 y 5 6 7 8

Graphical solution : Girls = 7, boys = 3

(ii) Consider price of a pencil is x

and price of a pen is y.


5x + 7y = 50.....(i)

7x + 5y = 46 .….(ii)

x 1 2 3 4 y 7.8 6.4 5 4.2

x 1 2 3 4

y 6.4 5.7 54.2

Graphical solution :

Price of a pencil is Rs 3

Price of a pen is Rs 5.

#465363

Topic: Consistency of Pair of Equations

On comparing the ratios a1

a2,

b1

b2, and

c1

c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x − 4y + 8 = 0 ; 7x + 6y − 9 = 0

(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

(iii) 6x − 3y + 10 = 0 ; 2x − y + 9 = 0

Solution



(i) a1

a2=

5

7 ,b1

b2=

−4

6 ,c1

c2=

8

−9

∵a1

a2≠

b1

b2, the lines intersect at a point.

(ii) a1

a2=

9

18 =1

2 ,b1

b2=

3

6 =1

2 ,c1

c2=

12

24 =1

2

∵a1

a2=

b1

b2=

c1

c2, the lines are coincident.

(iii) a1

a2=

6

2 =3

1 ,b1

b2=

−3

−1 =3

1 ,c1

c2=

10

9

∵a1

a2=

b1

b2≠

c1

c2, the lines are parallel.

#465364


On comparing the ratios a1

a2,

b1

b2 and

c1

c2, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5; 2x − 3y = 7

(ii) 2x − 3y = 8; 4x − 6y = 9

(iii) 3

2x +

5

3y = 7; 9x − 10y = 14

(iv) 5x − 3y = 11; − 10x + 6y = − 22

(v) 4

3x + 2y = 8; 2x + 3y = 12

Solution



(i) a1

a2=

3

2 ,b1

b2=

2

−3 ,c1

c2=

5

7

∵a1

a2≠

b1

b2≠

c1

c2,

the lines intersect and have an unique consistent solution.

(ii) a1

a2=

2

4 =1

2 ,b1

b2=

−3

−6 =1

2 ,c1

c2=

8

9

∵a1

a2=

b1

b2≠

c1

c2,

the lines are parallel and have no solutions, i.e. the equations are an inconsistent pair

(iii) a1

a2=

3

2

9 =1

6 ,

b1

b2=

5

3

−10 = −1

6 ,

c1

c2=

7

14 =1

2

∵a1

a2≠

b1

b2≠

c1

c2

,

the lines intersect and have an unique consistent solution.

(iv) a1

a2=

5

−10 = −1

2 ,b1

b2=

−3

6 = −1

2 ,c1

c2=

11

−22 = −1

2

∵a1

a2=

b1

b2=

c1

c2,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

(v) a1

a2=

4

3

2 =2

3 ,b1

b2=

2

3 ,c1

c2=

8

12 =2

3

∵a1

a2=

b1

b2=

c1

c2,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

#465365


Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) x + y = 5, 2x + 2y = 10

(ii) x − y = 8, 3x − 3y = 16

(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0

(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0

Solution

(i) x + y = 5 ...(i)

2x + 2y = 10 ...(ii)

⇒ x + y = 5

⇒ y = 5 − x



x 0 3

y 5 2

Plot (0, 5) and (3, 5) on graph and join them to get equation x + y = 5.

2x + 2y = 10

⇒ 2y = (10 − 2x)

⇒ y =10 − 2x

2= 5 − x ...(iii)

x 5 2

y 0 3

So, the equation is consistent and has infinitely many solution

(ii) x − y = 8 ....(i)

3x − 3y = 16 ....(ii)

⇒ x − y = 8

⇒ − x + y = − 8

⇒ y = − 8 + x

⇒ y = x − 8

x 8 0

y 0 -8

3x − 3y = 16 ...(ii)

⇒ 3x = 16 + 3y

⇒ 3x − 16 = 3y

⇒ y =3x − 16

3⇒ y = x −

16

3

x 16

3 0

y 0 −16

3

Plotting both the equation in graph, we see that the lines are parallel , so inconsistent.

(iii) 2x + y − 6 = 0

4x − 2y − 4 = 0

2x + y = 6 ....(i)

4x − 2y = 4 ...(ii)

For equation (i), 2x + y = 6 ⇒ y = 6 − 2x

x 0 3

y 6 0

Plot point (0, 6) and (3, 0) on a graph and join then to get equation 3x + y = 6

For equation (ii), 4x − 2y = 4 ⇒4x − 4

2= y

x 1 0

y 0 −2

Plot point (1, 0) and (0, − 2) on a graph and join them to get equation 4x − 2y = 0

x = 2, y = 2 is the solution of the given pairs of equation . So. solution is consistent.

(iv) 2x − 2y = 2 ...(i)

4x − 4y = 5 ...(ii)

2x − 2y = 2 ⇒ 2x − 2 = 2y

y = x − 1



x 0 1

y −1 0

Plot point (0, 1) and (1, 0) and join to get the equation 2x − 2y = 2 on a graph 4x − 4y = 5 ⇒ 4x − 5 = 4y

⇒4x − 5

4= y

x 05

4

y −5

40

Plot point 0, −5

4 and 5

4, 0 and join them to get the equation 4x − 4y = 5 on a graph.

The two lines never intersect, so, the solution is inconsistent.

#465366


Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution

Consider the width of garden = x

Length of garden = y.


x + 4 = y ……..(i)

x + y = 36 ………(ii)

Putting the value of y from equation (i) in equation (ii),

x + x + 4 = 36

2x + 4 = 36

2x = 32

x = 16

y = x + 4 = 16 + 4 = 20

So, length = 20m and breadth = 16m

#465367


( ) ( )



Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines (ii) parallel lines (iii) coincident lines

Solution

a) Intersecting lines

Solution: For intersecting line, the linear equations should meet following condition:

a1

a2≠

b1

b2

For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can

be as follows:

4x + 9y − 8 = 0

(b) Parallel lines

Solution: For parallel lines, the linear equations should meet following condition:

a1

a2=

b1

b2≠

c1

c2

For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible

equation can be as follows:

4x + 6y– 24 = 0

(c) Coincident lines

Solution: For getting coincident lines, the equations should meet following condition;

a1

a2=

b1

b2=

c1

c2

For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:

4x + 6y– 16 = 0

#465369


Draw the graphs of the equations x − y + 1 = 0 and 3x + 2y − 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the

triangular region.

Solution



x − y + 1 = 0

x − y = − 1 ...(i)

3x + 2y − 12 = 0

3x + 2y = 12 ...(ii)

x − y = − 1

y = 1 + x

x 0 −1

y 1 0

Plot (0, 1) and (1, 2) on graph and join them to get eqn (i)

For 3x + 2y = 12

⇒ 2y = 12 − 3x

⇒ r =12 − 3x

2

x 0 4 y 6 0

Plot point (0, 6) and (4, 0) on graph and join them to get eqn (ii)

△ABC is the required triangle with coordinates of A as (2, 3), B( − 1, 0) and C(4, 0).

#465375


Solve the following pair of linear equations by the substitution method.

(i) x + y = 14; x − y = 4

(ii) s − t = 3;s

3+

t

2= 6

(iii) 3x − y = 3; 9x − 3y = 9

(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3

(v) √2x + √3y = 0; √3x − √8y = 0

(vi) 3x

2−

5y

3= − 2;

x

3+

y

2=

13

6

Solution

(i) x + y = 14 ⇒ y = 14 − x

Substituting this value in the second equation, we get

x − (14 − x) = 4

2x = 18 ⇒ x = 9

Substituting this value of x in the first equation, we get



9 + y = 14 ⇒ y = 5

(ii) s − t = 3 ⇒ s = t + 3

Substituting in 2nd equation

t + 3

3 +t

2 = 6

2t + 6 + 3t

6 = 6

5t + 6 = 36 ⇒ t = 6

Substituting value of t in 1st equation

s = t + 3 = 9

(iii) ∵a1

a2=

b1

b2=

c1

c2

Hence all the points lying on the line y = 3x − 3 like x = 2, y = 3; are a solution.

(iv) 0.2x + 0.3y = 1.3 ⇒ x =1.3 − 0.3y

0.2

Substituting this value in the second equation

0.4 ×1.3 − 0.3y

0.2+ 0.5y = 2.3

2.6 − 0.6y + 0.5y = 2.3

2.6 − 2.3 = 0.6y − 0.5y

0.1y = 0.3 ⇒ y = 3

Substituting this value of y,

x =1.3 − 0.3y

0.2=

1.3 − 0.3 × 3

0.2=

0.4

0.2

x = 2

(v) √2x + √3y = 0 ⇒ y = −√2

√3x

Substituting this in 2nd equation

√3x − √8 × −√2

√3x = 0

3x + 4x = 0 ⇒ x = 0

Substituting value of x,

y = −√2

√3× 0 = 0

(vi) 3x

2 −5y

3 = − 2

5y

3=

3x

2+ 2 ⇒ y =

9x + 12

10

Substituting this in 2nd equation

x

3+

9x + 12

20=

13

6

47x + 36

60=

13

6

47x + 36 = 130

47x = 94 ⇒ x = 2

Substituting this value of x in 1st equation

( )



y =9x + 12

10=

9 × 2 + 12

10= 3

#465376


Solve 2x + 3y = 11 and 2x − 4y = − 24 and hence find the value of m for which y = mx + 3.

Solution

2x + 3y = 11

2x = 11 − 3y

Substituting this in 2nd equation, we get

(11 − 3y) − 4y = − 24

11 − 7y = − 24

7y = 11 + 24

y =35

7= 5

Substituting value of y

2x = 11 − 3 × 5

x =−4

2= − 2

Now, to find the value of m, we substitute the values of x and y in the equation

5 = m × ( − 2) + 3

m =5 − 3

−2= − 1

#465379


Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ⊂ 3800. Later, she buys 3bats and 5 balls for ⊂ 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ⊂ 105 and for a journey of

15 km, the charge paid is ⊂ 155. What are the fixed charges and thecharge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 9

11, if 2 is added to both the numerator and the denominator.If, 3 is added to both the numerator and the denominator it becomes 5

6. Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacobs age was seven times that of his son. What are their present ages?

Solution

(i) Let the larger number be x and smaller number be y. Then,

x − y = 26 and x = 3y

Substituting the value of x from second equation in the first equation, we get;

x − y = 26

Or, 3y − y = 26

Or, 2y = 26

Or, y = 13

Substituting the value of y in second equation, we get;

x = 3y

Or, x = 3 × 13 = 39

Hence, x = 39 and y = 13

(ii) Let the larger angle be x and smaller angle be y. Then,

x − y = 18 ∘ and x + y = 180 ∘



Substituting the value of x from first equation in second equation, we get;

x + y = 180 ∘

Or, y + 18 ∘ + y = 180 ∘

Or, 2y = 180 ∘ − 18 ∘ = 162 ∘

Or, y = 81 ∘

Substituting the value of y in first equation, we get;

x = y + 18 ∘

Or, x = 81 ∘ + 18 ∘ = 99 ∘

Hence, x = 99 ∘and y = 81 ∘

(iii) Let the cost of one bat and one ball be Rs. x and Rs. y respectively. Then,

7x + 6y = 3800 .....(1)

and, 3x + 5y = 1750 ...(2)

From (2), 5y = 1750– 3x

y =1750 − 3x

5

Substituting y =1750 − 3x

5 in (1), we get

7x + 61750 − 3x

5 = 3800

⇒ 35x + 10500– 18x = 19000

⇒ 17x = 19000– 10500

⇒ 17x = 8500

⇒ x =8500

17 = 500

Putting x = 500 in (2), we get

3 × (500) + 5y = 1750

⇒ 5y = 1750– 1500

⇒ 5y = 250

⇒ y =250

5= 50

Hence, the cost of one bat is Rs. 500 and the cost of one ball is Rs 50.

(iv) Let the fixed charges of taxi be Rs. x per km and the running charges be Rs y km/hr.

According to the given condition, we have

x + 10y = 105 ...(1)

x + 15y = 155 ...(2)

From (1), x = 105– 10y

Substituting x = 105– 10y in (2), we get

105– 10y + 15y = 155

⇒ 105 + 5y = 155

⇒ 5y = 155– 105

⇒ 5y = 50

⇒ y = 10

Putting y = 10 in (1), we get

x + 10 × 10 = 105

⇒ x = 5

Total charges for travelling a distance of 25 km

= x + 25y = Rs(5 + 25 × 10)

= Rs. 255

Hence, the fixed charge is Rs 5, the charge per km is Rs. 10 and the total charge for travelling a distance of 25 km is Rs 255.

(v) Let the fraction be x

y

( )



Given, x + 2

y + 2=

9

11

Or, 11x + 22 = 9y + 18 ⇒ 9y − 11x = 22 − 18

Or, 9y − 11x = 4 .....(1)

Also, x + 3

y + 3=

5

6

Or, 6x + 18 = 5y + 15

Or, 5y − 6x = 3 .....(2)

Or, y =6x + 3

5 .....(3)

Substituting this in (1), we get

9 ×6x + 3

5− 11x = 4

54x + 27 − 55x

5= 4

54x + 27 − 55x = 20

x = 7

Substituting value of x in (3),

y =6 × 7 + 3

5= 9

Hence, the fraction is 7

9

(vi) Let the present age of Jacob be x years and the present age of his son be y years.

Five years hence, Jacob's age = (x + 5) years

Son's age = (y + 5) years

Five years ago, Jacob's age = (y– 5) years

Son's age = (y– 5) years

As per question, we get

(x + 5) = 3(y + 5)

⇒ x + 5 = 3y + 15

⇒ x– 3y = 10

⇒ x– 3y = 15– 5

and, (x– 5) = 7(y– 5)

⇒ x– 5 = 7y– 35

⇒ x– 7y = – 30

⇒ x– 7y = – 35 + 5

From (1), x = 3y + 10

Substituting x = 3y +10 in (2), we get

3y + 10– 7y = – 30

⇒ – 4y = – 30– 10

⇒ – 4y = – 40

⇒ y = 10

Puttingy = 10 in (1),we get

x– 3 × 10 = 10

⇒ x = 10 + 30 = 40

Hence, present age of Jacob is 40 years and that his son is 10 years.



#465380


Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x + y = 5 and 2x − 3y = 4

(ii) 3x + 4y = 10 and 2x − 2y = 2

(iii) 3x − 5y − 4 = 0 and 9x = 2y + 7

(iv) x

2+

2y

3= − 1 and x −

y

3= 3

Solution

(i) x + y = 5 ...(i)

2x − 3y = 4 ...(ii) (elimination method)

Multiply eqn (i) by (ii)

2x + 2y = 10

2x − 3y = 4

− + −

____________

5y = 6

⇒ y =6

5

Substitute y =6

5in eqn (i)

x + y = 5

x +6

5= 5

⇒ x =5 × 5

5−

6

5

=19

5

Substitution method

xy = 5 ....(i)

2x − 3y = 4 ....(ii)

y = 5 − x ...(iii)

Putting eqn. (3) in (2)

2x − 3(5 − x) = 4

2x − 15 + 3x = 4

x =19

5

y =5 × 5

5−

19

5

y =6

5

(ii) 3x + 4y = 10 ...(i)

2x − 2y = 2 ...(ii)

Multiply eqn (2) by 2

4x − 4y = 4

3x + 4y = 10

____________

7x = 14

x = 2

2 × 2 − 2y = 2

⇒ 4 − 2 = 2y



y = 1

Substitution method

3x + 4y = 10 ...(i)

2x − 2y = 2 ⇒ x − y = 1

x − 1 = y ...(ii)

Putting eqn (ii) in (i)

3x + 4(x − 1) = 10

3x + 4x − 4 = 10

_________________

7x = 14

x = 2

y = x − 1

y = 2 − 1

⇒ y = 1

(iii) 3x − 5y = 4 ...(i)

9x − 2y = 7 ...(ii) (elimination method)

Multiply eqn (i) by (ii)

9x − 2y = 7

3x − 5y = 4

_____________

− 13y = 5

⇒ y =−5

13

3x − 5 ×−5

134= 4

⇒ 3x = 4 −25

13

3x =52 − 25

13

x =27

13 × 3

⇒ x =9

13

Substitution method

3x − 5y = 4 ...(i)

9x − 2y = 7

⇒ 9x − 7 = 2y

9x − 7

2= y

3x − 5y = 4

⇒ 3x − 5x9x − 7

2 = 4

⇒ 3x −45

2x +

35

2= 4

⇒6x

2−

45

2x = 4 −

35

2

( )



⇒−39x

2=

8 − 35

2

x =−27

39

⇒ x =9

13

Putting x =9

13 in 3x − 5y = 4

3 × 9

13− 5y = 4

27

13− 5y = 4

⇒27

13− 4 = 5y

⇒27 − 52

13= 5y

−25

13= 5y

⇒ y =−5

13

(iv) x

2+

2y

3= − 1

3x + 4y

6= − 1

⇒ 3x + 4y = − 6 ...(i)

x −y

3= 3

⇒ 3x − y = 9 ...(ii)

3x + 4y = − 6

3x − y = − 9

− + −

______________

5y = − 15

⇒ y = − 3

Putting y=-3 in eqn (i)

3x − 12 = − 6

⇒ 3x = − 6 + 12

⇒ 3x = 6

⇒ x = 2

Substitution method

3x + 4y = − 6 ...(i)

3x − y = 9 ...(ii)

3x − 9 = y ....(iii)

Putting enq (iii) in (i)

3x + 4(3x − 9) = − 6

⇒ 3x + 12x − 36 = − 6

⇒ 15x = − 6 + 36



⇒ 15x = 30

⇒ x = 2

Putting x=2 in enq (iii)

6 − 9 = y

⇒ y = − 3

#465385


Form the pair of linear equations in the following problems, and find their solutions(if they exist) by the elimination method :

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1

2 if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) 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. Find how many notes of Rs. 50 and

Rs. 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while

Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution

(i) Let the fraction be x

y

Given,

x + 1

y − 1= 1

x + 1 = y − 1

x − y = − 2 .....(1)

Also,

x

y + 1=

1

2

2x = y + 1

2x − y = 1 .....(2)

Substracting (1) from (2)

x = 3


6 − y = 1

y = 5

Hence, the fraction is 3

5

(ii) Let the age of Nuri be x and that of Sonu be y.

Given,

(x − 5) = 3(y − 5)

3y − x = 10 ..... (1)

And,



(x + 10) = 2(y + 10)

2y − x = − 10 ....(2)

Subtracting (2) from (1), we get

y = 20

Substituting value of y in (1)

60 − x = 10

x = 50

Hence, the ages of Nuri and Sonu are 50 and 20 respectively.

(iii) Let the two digit number be 10x + y

Given, x + y = 9 .....(1)

And, 9(10x + y) = 2(10y + x)

90x + 9y = 20y + 2x

88x − 11y = 0 .....(2)

Multiplying (1) by 11, we get

11x + 11y = 99 ....(3)

Adding (1) and (3)

99x = 99

x = 1

Substituting this value of x in (1),

1 + y = 9

y = 8

Hence, the number is 18.

(iv) Let the number of Rs 100 notes be x and the number of Rs 50 notes be y.

Given,

x + y = 25 ...(1)

And,

100x + 50y = 2000 ...(2)


50x + 50y = 1250 ....(3)

Subtracting (3) from (2),

50x = 750

x = 15


15 + y = 25

y = 10

(v) Let the fixed charge be Rs x and the charge for each day be Rs y.

Given,

x + 7y = 27 ....(1)

And,

x + 5y = 21 ...(2)


2y = 6

y = 3

Substituting value of y in (1)



x + 7 × 3 = 27

x + 21 = 27

x = 6

Hence, the fixed charge is Rs 6 and the per day charge is Rs 3.

#465386


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

(i) x– 3y– 3 = 0

3x– 9y– 2 = 0

Comparing these with pair of equations of the form a1x + b1y + c1 = 0

and a2x + b2y + c2 = 0,

a1

a2=

1

3 ,b1

b2=

−3

−9 =1

3 ,

c1

c2=

3

2

∵a1

a2=

b1

b2≠

c1

c2, the lines have no solution.

(ii) 2x + y = 5

3x + 2y = 8

a1

a2=

2

3 ,b1

b2=

1

2 ,c1

c2=

5

8

∵a1

a2≠

b1

b2, the lines have unique solution.

(iii) 3x − 5y = 20

6x − 10y = 40

a1

a2=

3

6 =1

2 ,

b1

b2=

−5

−10 =1

2 ,

c1

c2=

20

40 =1

2

∵a1

a2=

b1

b2=

c1

c2

, the lines have infinitely many solution.

(iv) x − 3y − 7 = 0; 3x − 3y − 15 = 0

a1

a2=

1

3 ,b1

b2=

−3

−3 = 1,c1

c2=

7

15

∵a1

a2≠

b1

b2, the lines have unique solution.

#465387




(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

(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

i) For the pair of linear equations to have infinite number of solutions, a1

a2=

b1

b2=

c1

c2

Given equation is 2x + 3y = 7

Thus a1 = 2, b1 = 3, c1 = − 7

We know a2 = a − b, b2 = a + b, c2 = − 3a − b + 2

Therefore, 2

a − b=

3

a + b=

7

(3a + b − 2)

⇒ − 6a − 2b + 4 = − 7a + 7b

⇒ a − 9b = − 4 ...(1)

and −9a + 3b + 6 = − 7a − 7b

⇒ − 2a + 4b = − 6 ...(2)

Solving (1) and (2), we get

a = 5, b = 1

ii) For the pair of linear equations to have infinite number of solutions, a1

a2=

b1

b2=

c1

c2

Given equation is 3x + y = 1

Thus a1 = 3, b1 = 1, c1 = 1

We know a2 = 2k − 1, b2 = k − 1, c2 = 2k + 1

Therefore, 3

2k − 1=

1

(k − 1)=

1

(2k + 1)

⇒ − 3k − 3 = 2k − 1 ⇒ 5k = − 2 ⇒ k = −2

5 .

#465388


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

8x + 5y = 9; 3x + 2y = 4

Solution



Substitution method:

8x + 5y = 9

∴ y =9 − 8x

5 ....(1)

3x + 2y = 4 ...(2)

Substitute eq(1) in eq (2), we get

3x + 29 − 8x

5 = 4

Multiply by 5 on both sides, we get

15x + 18 − 16x = 20

⇒ − x = 2

⇒ x = − 2

Substitute x = − 2 in eq (1), we get

y =9 − 8( − 2)

5

⇒ y =25

5= 5

So, (x, y) = ( − 2, 5)

Cross-multiplication method:

8x + 5y = 9 ...(1)

So, a1 = 8, b1 = 5, c1 = − 9

3x + 2y = 4 ...(2)

So, a2 = 3, b2 = 2, c2 = − 4

∴ x =b1c2 − b2c1

a1b2 − a2b1

⇒ x =5( − 4) − 2( − 9)

8(2) − 3(5)=

−20 + 18

16 − 15= − 2

y =c1a2 − c2a1

a1b2 − a2b1

⇒ y =−9(3) − ( − 4)8

8(2) − 3(5) =

−27 + 32

16 − 15= 5

So, (x, y) = ( − 2, 5)

#465389


( )



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 14 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

i) Let the fixed monthly hostel charges be Rs. x and the cost of food per day taken from the mess be y.

As per the given condition,

x + 20y = 1000 ...(1)

x + 26y = 1180 ...(2)

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

⇒ 6y = 180

⇒ y = 30

Substitute y = 30 in eq (1), we get

⇒ x + 20(30) = 1000

⇒ x + 600 = 1000

⇒ x = 400

So, the fixed monthly hostel charge is Rs. 600 and the cost of food per day is Rs. 30.

ii) Let the numerator be x and the denominator be y, so the fraction will be x

y.

x − 1

y=

1

3

⇒ 3x − y = 3 ...(1)

x

y + 8= 14

⇒ 14y + 112 = x ...(2)

Substituting for x in (1) we get,

3(14y + 112) − y = 3 42y + 336 − y = 3 41y = − 333 y = −333

41

Substituting the value of y in (1), we get x as −70

41.

x=−70

41 , y=−

333

41

iii) Let x be the number of correct answers and y be the number of wrong answers.

So, 3x − y = 40 ...(1)



4x − 2y = 50 ...(2)


6x − 2y = 80 ...(3)

4x − 2y = 50 ...(4)

Subtracting eq(4) from eq(3), we get

⇒ 2x = 30

⇒ x = 15

Putting x = 15 in eq (1), we get

⇒ 3(15) − y = 40

⇒ y = 5

So, Correct answers = 15 and wrong answers = 5.

Total number of questions in the test = x + y = 20.

iv) Let the speeds of the first and second cars be x km/hr and y km/hr

respectively.

Distance between places A and B = 100km.

Moving in the same direction, relative speed will be = x − y

Moving in opposite directions, relative speed will be = x + y

5(x − y) = 100

⇒ x − y = 20 ...(1)

⇒ x + y = 100 ...(2)

Adding (1) and (2), we get

⇒ 2x = 120

⇒ x = 60

Substitute in eq (1), we get

⇒ y = 40

The speeds of the first and second cars are 60 km/hr and 40 km/hr respectively.

v) Let the length of rectangle be l and the breadth be b.

So, the area will be A = l × b.

Case 1:

l ′ = l − 5

b ′ = b + 3

A ′ = A − 9 = lb − 9

⇒ (l − 5)(b + 3) = lb − 9 ⇒ lb + 3l − 5b − 15 = lb − 9 ⇒ 3l − 5b = 6 . . . . (1)

Case 2:

l ″ = l + 3

b ″ = b + 2

A ″ = A + 67 = lb + 67

(l + 3)(b + 2) = lb + 67 ⇒ lb + 2l + 3b + 6 = lb + 67 ⇒ 2l + 3b = 61 . . . (2)

Multiplying (1) by 3 and (2) by 5 and adding both equations, we get

9l − 15b = 18 ...(3)

10l + 15b = 305 ...(4)

Adding eq (3) and eq (4), we get



19l = 323 ⇒ l = 17

Putting l = 17 in eq(1), we get

⇒ 3(17) − 5b = 6

⇒ 45 = 5b

⇒ b = 9

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

#465404


Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1

2x+

1

3y= 2;

1

3x+

1

2y=

13

6

(ii) 2

√x+

3

√y= 2;

4

√x−

9

√y= − 1

(iii) 4

x+ 3y = 14;

3

x− 4y = 23

(iv) 5

x − 1+

1

y − 2= 2;

6

x − 1−

3

y − 2= 1

(v) 7x − 2y

xy= 5;

8x + 7y

xy= 15

(vi) 6x + 3y = 6xy; 2x + 4y = 5xy

(vii) 10

x + y+

2

x − y= 4;

15

x + y−

5

x − y= − 2

(viii) 1

3x + y+

1

3x − y=

3

4;

1

2(3x + y)−

1

2(3x − y)=

−1

8

Solution

(i)

1

2x+

1

3y= 2

1

3x+

1

2y=

13

6

Lets assume 1

x= p and

1

y= q, both equations will become

p

2+

q

3= 2

⇒ 3p + 2q = 12 ... (A)

p

3+

q

2=

13

6

⇒ 2p + 3q = 13 ...(B)

Multiply (A) by 3 and (B) by 2, we get

9p + 6q = 36

4p + 6q = 26

Subtracting these both, we get

5p = 10

⇒ p = 2

⇒ x =1

p=

1

2



∴ q =26 − 8

6= 3

⇒ y =1

q=

1

3

(ii)

2

√x+

3

√y= 2

4

√x−

9

√y= − 1

Lets assume 1

√x= p and

1

√y= q, both equations will become

2p + 3q = 2 ...(A)

4p − 9q = − 1 ...(B)

Multiply (A) by 3, we get

6p + 9q = 6

4p − 9q = − 1

Adding these both, we get

10p = 5

⇒ p =1

2

⇒ x =1

p2 = 4

∴ q =1

3

⇒ y =1

q2 = 9

(iii)

4

x+ 3y = 14

3

x− 4y = 23

Lets assume 1

x= p, both equations become

4p + 3y = 14 ...(A)

3p − 4y = 23 ...(B)

Multiply (A) by 4 and (B) by 3, we get

16p + 12y = 56

9p − 12y = 69

Adding these both, we get

25p = 125

⇒ p = 5

⇒ x =1

p=

1

5

⇒ y = − 2

(iv)

5

x − 1+

1

y − 2= 2



6

x − 1−

3

y − 2= 1

Lets assume 1

x − 1= p and

1

y − 2= q, both equations become

5p + q = 2

6p − 3q = 1

Multiply (A) by 3, we get

15p + 3q = 6

6p − 3q = 1

Adding both of these, we get

21p = 7

⇒ p =1

3

⇒ x = 4

∴ q =1

3

⇒ y = 5

(v)

7x − 2y

xy= 5

⇒7

y−

2

x= 5

8x + 7y

xy= 15

⇒ 8

y+

7

x= 15

Let t =1

x and r =

1

y equation will be

7r − 2t = 5 ; 8r + 7t = 15

On solving we get

t = 1 r = 1

∴ x = 1 and y = 1

(vi)

Dividing both side by xy then we have

6

y+

3

x= 6 ;

2

y+

4

x= 5

Let t =1

xand r =

1

y equation will be

6r + 3t = 6 ; 2r + 4t = 5 on solving we get

t = 1 r =1

2

∴ x = 1 and y = 2

(vii)

Let t =1

x + yand r =

1

x − y then equation will be

10t + 2r = 4 ; 15t − 5r = − 2 on solving we have

t =1

5and r = 1

∴ x + y = 5 and x − y = 1

Hence, x = 4, y = 1



(viii)

Let t =1

3x + yand r =

1

3x − y then equation will be

t + r =3

4;

t

2−

r

2=

−1

8

On solving we have

t =1

4and r =

1

2

∴ 3x + y = 4 and 3x − y = 2

Hence, x = 1, y = 1

#465405


Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the

work, and also that taken by 1 man alone.

(iii) 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. Find the speed of the train and the bus separately.

Solution



(i) Let her speed in still water be x and speed of stream be y

Now, According to question and using time = dist

speed

20

x + y= 2 and

4

x − y= 2

x + y = 10; x − y = 2

On solving we get x = 6 and y = 4

speed of ritu in still water is 6km/hr and that of stream is 4km/hr.

(ii)

Let the one woman do x unit work in one day and man do y unit

Now according to question,

2 × x × 4 + 5 × y × 4 = 1 (i)

3 × x × 3 + 6 × y × 3 = 1 (ii)

8x + 20y = 1; 9x + 18y = 1

Multiply equation (i) by 9 and (ii) by 8 then we have

72x + 180y = 1 ; 72x + 144y = 1

On solving we have

y =1

36 that is one man can finish that work in 36 days.

x =1

18 that is one woman can finish that work in 18 days.

(iii)

Let the speed of bus be x and that of train be y

Now, according to question and using time = dist

speed

60

y+

240

x= 4 ⇒

15

y+

60

x= 1 (i)

100

y+

200

x=

25

6⇒

24

y+

48

x= 1 (ii)

From (i)

1

x=

1

601 −

15

y putting this in eq (ii)

24

y+

48

601 −

15

y = 124

y+

4

5−

12

y= 1 ∴ y = 60km/hr

then x = 80km/hr

#465406


The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam

differ by 30 years. Find the ages of Ani and Biju.

Solution

( )( )



Let the age of Ani and Biju be x and y respectively.

Then according to the question-

x − y = 3..........(i)

Dharam is twice the age of Ani

Hence, age of Dharam = 2x

Age of Cathy is half the age of Biju

Then age of Cathy =y

2

Then according to the question

2x −y

2= 30

⇒ 4x − y = 60......(ii)

Subtracting (ii) from (i)

⇒ x − y − 4x + y = 3 − 60

⇒ − 3x = − 57

⇒ x =−57

−3= 19

Substitute the vale of x in (i)

⇒ 19 − y = 3

⇒ − y = − 16

⇒ y = 16

∴ Age of Ani = 19 years and age of Biju = 16 years.

#465407


One says, "Give me a 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." Tell me what is the

amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

[Hint : x + 100 = 2(y − 100), y + 10 = 6(x − 10)].

Solution



Let the initial amount with them be Rs x and Rs y respectively.

Then according to the question-

⇒ x + 100 = 2(y − 100)

⇒ x + 100 = 2y − 200

⇒ x − 2y = − 300........ (i)

And,

⇒ 6(x − 10) = (y + 10)

⇒ 6x − 60 = y + 10

⇒ 6x − y = 70...... (ii)

Multiplying equation (ii) by 2, we obtain

12x − 2y = 140....... (iii)

Subtracting equation (i) from equation (iii), we gets

⇒ 11x = 140 + 300

⇒ 11x = 440

⇒ x = 40

Substitute the vale of x in equation (i), we gets

⇒ 40 − 2y = − 300

⇒ 40 + 300 = 2y

⇒ 2y = 340

y =340

2= 170

∴ The friends had Rs 40 and Rs 170 with them respectively.

#465408


A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were

slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Solution



Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km. We know that,

Speed =Distance

Time

x =d

t

∴ d = xt.......(i)

Case 1

(x + 10) × (t − 2) = d

xt + 10t − 2x − 20 = d

d + 10t − 2x − 20 = d

−2x + 10t = 20...... (ii)

Case 2

(x − 10) × (t + 3) = d

xt − 10t + 3x − 30 = d

d − 10t + 3x − 30 = d

3x − 10t = 30......... (iii)

Adding equations (ii) and (iii), we gets

⇒ x = 50

Substitute the value of x in (ii) we gets

⇒ ( − 2) × (50) + 10t = 20

⇒ − 100 + 10t = 20

⇒ 10t = 120

⇒ t = 12 hours

Substitue the value of t and x in equation (i), we gets

Distance to travel = d = xt

⇒ d = 12 × 50 = 600Km

Hence, the distance covered by the train is 600 km.

#465409


The students of a class are made to stand in rows. If 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. Find the

number of students in the class.

Solution



Let the number of rows be x and number of students in a row be y.

Total students of the class= Number of rows × Number of students in a row

= x × y = xy

Case 1

Total number of students = (x − 1)(y + 3)

⇒ xy = (x − 1)(y + 3) = xy − y + 3x − 3

⇒ 3x − y − 3 = 0

⇒ 3x − y = 3..... (i)

Case 2

Total number of students = (x + 2)(y − 3)

⇒ xy = xy + 2y − 3x − 6

⇒ 3x − 2y = − 6..... (ii)

Subtracting equation (ii) from (i),

⇒ (3x − y) − (3x − 2y) = 3 − ( − 6)

⇒ − y + 2y = 3 + 6

⇒ y = 9

By substituting value of y in (i), we get

⇒ 3x − 9 = 3

⇒ 3x = 9 + 3 = 12

⇒ x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Number of total students in a class = x × y = 4 × 9 = 36

#465410


In a △ABC; ∠C = 3∠B = 2(∠A + ∠B). Find the three angles.

Solution



As given that,

∠C = 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B)

3∠B = 2∠A + 2∠B

∠B = 2∠A

2∠A − ∠B = 0 … (i)

As the sum of the measures of all angles of a triangle is 180°. Therefore,

∠A + ∠B + ∠C = 180 ∘

∠A + ∠B + 3∠B = 180 ∘

∠A + 4∠B = 180 ∘ ... (ii)

Multiplying equation (i) by 4, we gets

⇒ 8∠A − 4∠B = 0 … (iii)

Adding equations (ii) and (iii), we obtain

⇒ 9∠A = 180 ∘

⇒ ∠A =180 ∘

9 = 20 ∘

From equation (ii), we obtain

⇒ 20 ∘ + 4∠B = 180 ∘

⇒ 4∠B = 160 ∘

⇒ ∠B = 40 ∘

⇒ ∠C = 3∠B

⇒ 3 × 40 ∘ = 120 ∘

∴ ∠A, ∠B, ∠C are 20 ∘ , 40 ∘ , 120 ∘

#465411


Draw the graphs of the equations 5x − y = 5 and 3x − y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Solution



Plotting 3x − y = 3

Data points are:

x 0 1 2

y −3 0 3

Plotting 5x − y = 5

Data points are:

x 0 1 2

y −5 0 5

Plotting both on the graph, we get the following figure.

ΔACE is formed.

with A(1, 0), C(0, − 3), and E(0, − 5)

#465412


Solve the following pair of linear equations:

(i) px + qy = p − q; qx − py = p + q

(ii) ax + by = c; bx + ay = 1 + c

(iii) x

a−

y

b= 0; ax + by = a2 + b2

(iv) (a − b)x + (a + b)y = a2 − 2ab − b2; (a + b)(x + y) = a2 + b2

(v) 152x − 378y = − 74; − 378x + 152y = − 604

Solution

(i) px + qy = p − q ....(1)

qx − py = p + q ....(2)

Multiplying (1) by p and (2) by q we get (3) and (4) respectively as

p2x + pqy = p2 − pq ....(3)

q2x − pqy = pq + q2 ...(4)

Adding (3) and (4) we get

(p2 + q2)x = (p2 + q2)

⇒ x = 1

Substituting value of x in (1), we get

p + qy = p − q

qy = − q



y = − 1

Hence, x = 1, y = − 1

(ii) ax + by = c ...(1)

bx + ay = 1 + c ...(2)

Multiplying (1) by b and (2) by a we get (3) and (4) respectively as

abx + b2y = bc ...(3)

abx + a2y = a + ac ...(4)


(b2 − a2)y = (b − a)c − a

y =(b − a)c − a

b2 − a2

Substituting value of y in (i), we get

ax + b ×(b − a)c − a

b2 − a2 = c

ax = c −(b − a)bc − ab

b2 − a2

ax =

c b2 − a2 − b2c + abc + ab

b2 − a2

x =cb2 − ca2 − b2c + abc + ab

a b2 − a2

x =ab + abc − ca2

a b2 − a2

x =b − bc − ca

b2 − a2 =b + c(b − a)

b2 − a2

(iii) x

a−

y

b= 0 ...(1)

ax + by = a2 + b2 ....(2)

Multiplying (1) by b2

b2x

a− by = 0 ....(3)


b2x

a+ ax = a2 + b2

b2x + a2x

a= a2 + b2

x a2 + b2 = a a2 + b2

⇒ x = a

Substituting value of x in (1)

a

a−

y

b= 0

y

b= 1

( )

( )

( )( )

( )

( )

( )

( ) ( )

( ) ( )



⇒ y = b

(iv) (a − b)x + (a + b)y = a2 − 2ab − b2 ....(1)

(a + b)(x + y) = a2 + b2

⇒ (a + b)x + (a + b)y = a2 + b2 ...(2)


(a − b − (a + b))x = − 2ab − 2b2

−2bx = − 2b(a + b)

⇒ x = (a + b)

Substituting value of x in (1), we get

(a − b) × (a + b) + (a + b)y = a2 − 2ab − b2

a2 − b2 + (a + b)y = a2 − 2ab − b2

(a + b)y = − 2ab

⇒ y = −2ab

a + b

(v) 152x − 378y = − 74 ...(1)

−378x + 152y = − 604 ...(2)

Multiplying (1) by 378 and (2) by 152, we get

57456x − 142884y = − 27972 ...(3)

−57456x + 23104y = − 91808 ....(4)


−119780y = − 119780

⇒ y = 1

Putting this value of y in (1)

152x − 378 = − 74

152x = 378 − 74

x =304

152= 2

7/4/2018 …...2x+6y−14=0 x+3y−7=0 hence this is the required relation between x and y. #465357...

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