Pair Of Linear Equations In Two Variables

The given equations are -

6x - 3y + 10 = 0

Divide the equation by 3 -

…(i)

And 2x - y + 9 = 0 …(ii)

Let's find the x and y intercepts (point at which it touches x and y axis respectively) for both the lines -

For line = 0;

At x = 0, y = 10/3 (y - intercept),

So (0, 10/3) is a point in y axis.

At y = 0, x = - 5/3 (x - intercept)

And ( - 5/3, 0) is a point in x axis.

For line 2x - y + 9 = 0;

At x = 0 , y = 9 ( y - intercept ),

So (0,9) is a point in y axis.

At y = 0 , x = - 9/2 ( x - intercept )

And ( - 9/2,0) is a point in x axis.

Blue line represents

Red line represents 2x - y + 9 = 0

It is clear from the figure that the two lines are parallel.

(Also slope of the lines = = 2, which is the same)

Hence, the pair of equations represents two parallel lines.

Given, equations are x + 2y + 5 = 0 and - 3x - 6y + 1 = 0.

Comparing the equations with general equation of the form:

ax + by + c = 0;

a₁x + b₁y + c₁ = x + 2y + 5

a₂x + b₂y + c₂ = - 3x - 6y + 1

Here, a₁ = 1, b₁ = 2, c₁ = 5

And a₂ = - 3, b₂ = - 6, c₂ = 1

Taking the ratio of coefficients to compare -

a₁ /a₂ = - 1/3

b₁ /b₂ = - 1/3

c₁ /c₂ = 5/1;

here a₁/a₂ = b₁/b₂ c₁/c₂

This represents pair of parallel lines.

Hence, the pair of equations has no solution.

Alternative solution -

We know for a line y = mx +c;

m represents slope of line.

So, we can write this lines in that form first -

For line: x + 2y + 5 = 0

2y = - 5 - x

Hence slope of first line is - 1/2.

For line: - 3x - 6y + 1 = 0

6y = - 3x + 1

Hence slope of second line is - 1/2.

Slope for both the lines represents the lines are parallel and will never intersect and so will have no solution.

Condition for a consistent pair of linear equations -

Consistent equations means the equations must have solution so lines may either coincide(with infinitely many solutions) or with a unique solution {lines must not be parallel}.

1) Must not be parallel so,

a₁/a₂ b₁/b₂ [intersecting lines with unique solution]

2) Also the lines can coincide, so,

a₁/a₂ = b₁/b₂ = c₁/c₂ [coincident or dependent]

The equation y = 0 represents x axis and

Also y = - 7 is line parallel to x axis (no x - intercept).

The give pair of equations are y = 0 and y = - 7.

By graphically, both lines are parallel and thus will have no solution.

x = a and y = b represents a rectangle with lines x = a for a>0 and x = - a for a <0 ;

y = b for b>0 and y = - b for b<0;

As it is clear from the figure, the lines will intersect at x = a and y = b or (a,b) point.

Condition for coincident lines is -

a₁/a₂ = b₁/b₂ = c₁/c₂ …(i)

Given lines, 3x - y + 8 = 0

and 6x - ky + 16 = 0;

Comparing with ax + by + c = 0;

Here, a₁ = 3, b₁ = - 1, c₁ = 8; a₂ = 6, b₂ = - k, c₂ = 16;

and

From Eq. (i),

So, k = 2

Condition for parallel lines is

a₁/a₂ = b₁/b₂ c₁/c₂ …(i)

Given lines, 3x + 2ky - 2 = 0

and 2x + 5y - 1 = 0;

Comparing with ax + by + c = 0;

Here, a₁ = 3, b₁ = 2k, c₁ = - 2

and a₂ = 2, b₂ = 5, c₂ = - 1

From Eq. (i), 3/2 = 2k/5

k = 15/4

Condition for infinitely many solutions

a₁/a₂ = b₁/b₂ c₁/c₂ …(i)

The given lines are cx - y - 2 = 0 and 6x - 2y - 3 = 0;

Comparing with ax + by + c = 0;

Here, a₁ = c, b₁ = - 1, c₁ = - 2

and a₂ = 6, b₂ = - 2,c₂ = - 3

From Eq. (i),

Here, c/6 = -1/- 2

Also c/6 = -2/- 3

Solving, we get, c = 3 and c = 4.

Since, c has different values and so it’s not possible.

Hence, for no value of c the pair of equations will have infinitely many solutions.

Condition for dependent linear equations -

a₁ /a₂ = b₁/b₂ = c₁/c₂ …(i)

Given equation of line is, - 5x + 7y - 2 = 0;

Comparing with ax+ by +c = 0;

Here, a₁ = - 5, b₁ = 7, c₁ = - 2;

For second equation, let’s assume a₂x + b₂y + c₂ = 0;

From Eq. (i),

Where, k is any arbitrary constant.

Putting k = - 1/2 then

a₂ = 10, b₂ = - 14, c₂ = 4;

∴ The required equation of line becomes

a₂x + b₂y + c₂ = 0;

10x - 14y + 4 = 0;

If x = 2 and y = - 3 is a unique solution of any pair of equation, then these values must satisfy that pair of equations.

Putting the values in the equations for every option and checking it -

From option (b).

LHS = 2x + 5y = 2(2+(- 3) = 4 + (- 15) = - 11 = RHS

and

LHS = 4x + 10y = 4(2) +10(- 3) = 8 + (- 30) = - 22 = RHS

It satisfies the pair of linear equation and hence is the unique solution for the equation.

Solving the equations -

x - y = 2 and

x + y = 4;

adding them,

2x = 6

x = 3;

subtracting them,

2y = 2

y = 1;

so, a = 3 and b = 1 is the solution of the equations.

Let number of Re 1 coins = x

and number of Rs 2 coins = y

Now, by given conditions:

Total number of coins = x + y = 50 …(i)

Also, Amount of money with her = (Number of Re 1 1) + (Number of Rs2 coin 2)

= x(1)+ y(2) = 75

= x+2y = 75 …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

(x + 2y) - (x + y) = (75 - 50)

So, y = 25

Putting y = 25 we get x = 25.

Hence he has 25 Re 1 coins and 25 Rs 2 coins.

Let ‘x’ year be the present age of father and ‘y’ year be the present age of son.

Four years hence, it has relation by given condition,

(x+4) = 4(y+4)

x+4 = 4y+16

x- 4y-12 = 0 …(i)

and initially, x = 6y …(ii)

On putting the value of from Eq. (ii) in Eq. (i), we get

6y - 4y - 12 = 0

2y = 12

Hence, y = 6

Putting y = 6, we get x = 36.

Hence, present age of father is 36 years and age of son is 6 years.

The Condition for no solution is : (parallel lines)

(i) Yes.

Given pair of equations are,

2x+4y - 3 = 0 and 6x + 12y - 6 = 0

Comparing with ax+ by +c = 0;

Here, a₁ = 2, b₁ = 4, c₁ = - 3;

And a₂ = 6, b₂ = 12, c₂ = - 6;

a₁ /a₂ = 2/6 = 1/3

b₁ /b₂ = 4/12 = 1/3

c₁ /c₂ = - 3/ - 6 = 1/2

Here, a₁/a₂ = b₁/b₂ c₁/c₂, i.e parallel lines

Hence, the given pair of linear equations has no solution.

(ii) No.

Given pair of equations,

x = 2y and y = 2x

or x - 2y = 0 and 2x - y = 0;

Comparing with ax + by + c = 0;

Here, a₁ = 1, b₁ = - 2, c₁ = 0;

And a₂ = 2, b₂ = - 1, c₂ = 0;

a₁ /a₂ = 1/2

b₁ /b₂ = -2/-1 = 2

Here, a₁/a₂ b₁/b_2.

Hence, the given pair of linear equations has unique solution.

(iii) No.

Given pair of equations,

3x + y - 3 = 0

and

Comparing with ax + by + c = 0;

Here, a₁ = 3, b₁ = 1, c₁ = - 3;

And a₂ = 2, b₂ = 2/3, c₂ = - 2;

a₁ /a₂ = 2/6 = 3/2

b₁ /b₂ = 4/12 = 3/2

c₁ /c₂ = - 3/-2 = 3/2

Here, a₁/a₂ = b₁/b₂ = c₁/c₂, i.e coincident lines

Hence, the given pair of linear equations is coincident and having infinitely many solutions.

Condition for coincident lines,

a₁/a₂ = b₁/b₂ = c₁/c₂;

(i) No.

Given pair of linear equations

and 7x + 3y = 7

Comparing with ax + by + c = 0;

Here, a₁ = 3, b₁ = 1/7, c₁ = - 3;

And a₂ = 7, b₂ = 3, c₂ = - 7;

a₁ /a₂ = 3/7

b₁ /b₂ = 1/21

c₁ /c₂ = - 3/ - 7 = 3/7

Here, a₁/a₂ b₁/b_2.

Hence, the given pair of linear equations has unique solution.

(ii) Yes, given pair of linear equations.

- 2x - 3y - 1 = 0 and 4x + 6y + 2 = 0;

Comparing with ax + by + c = 0;

Here, a₁ = - 2, b₁ = - 3, c₁ = - 1;

And a₂ = 4, b₂ = 6, c₂ = 2;

a₁ /a₂ = - 2/4 = - 1/2

b₁ /b₂ = - 3/6 = - 1/2

c₁ /c₂ = - 1/2

Here, a₁/a₂ = b₁/b₂ = c₁/c₂, i.e. coincident lines

Hence, the given pair of linear equations is coincident.

(ii) No, given pair of linear equations are

and = 0

Comparing with ax + by + c = 0;

Here, a₁ = 1/2, b₁ = 1, c₁ = 2/5;

And a₂ = 4, b₂ = 8, c₂ = 5/16;

a₁ /a₂ = 1/8

b₁ /b₂ = 1/8

c₁ /c₂ = 32/25

Here, a₁/a₂ = b₁/b₂ c₁/c₂, i.e. parallel lines

Hence, the given pair of linear equations has no solution.

Conditions for pair of linear equations are consistent

a₁/a₂ b₁/b_2. [unique solution]

and a₁/a₂ = b₁/b₂ = c₁/c₂ [coincident or infinitely many solutions]

(i) No.

The given pair of linear equations -

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

Comparing with ax + by + c = 0;

Here, a₁ = - 3, b₁ = - 4, c₁ = - 12;

And a₂ = 3, b₂ = 4, c₂ = - 12;

a₁ /a₂ = - 3/3 = - 1

b₁ /b₂ = - 4/4 = - 1

c₁ /c₂ = - 12/ - 12 = 1

Here, a₁/a₂ = b₁/b₂ c₁/c₂

Hence, the pair of linear equations has no solution, i.e., inconsistent.

(ii) Yes.

The given pair of linear equations

and

Comparing with ax + by + c = 0;

Here, a₁ = 3/5, b₁ = - 1, c₁ = - 1/2;

And a₂ = 1/5, b₂ = 3, c₂ = - 1/6;

a₁ /a₂ = 3

b₁ /b₂ = - 1/ - 3 = 1/3

c₁ /c₂ = 3

Here, a₁/a₂ b₁/b_2.

Hence, the given pair of linear equations has unique solution, i.e., consistent.

(iii) Yes.

The given pair of linear equations -

2ax + by –a = 0 and 4ax + 2by - 2a = 0

Comparing with ax + by + c = 0;

Here, a₁ = 2a, b₁ = b, c₁ = - a;

And a₂ = 4a, b₂ = 2b, c₂ = - 2a;

a₁ /a₂ = 1/2

b₁ /b₂ = 1/2

c₁ /c₂ = 1/2

Here, a₁/a₂ = b₁/b₂ = c₁/c₂

Hence, the given pair of linear equations has infinitely many solutions, i.e., consistent or dependent.

(iv) No.

The given pair of linear equations

x + 3y = 11 and 2x + 6y = 11

Comparing with ax + by + c = 0;

Here, a₁ = 1, b₁ = 3, c₁ = 11

And a₂ = 2, b₂ = 6, c₂ = 11

a₁ /a₂ = 1/2

b₁ /b₂ = 1/2

c₁ /c₂ = 1

Here, a₁/a₂ = b₁/b₂ c₁/c₂.

Hence, the given pair of linear equations has no solution.

No.

The given pair of linear equations

and 2x + 6y - 14 = 0.

Here, Comparing with ax + by + c = 0;

Here, a₁ = , b₁ = 3, c₁ = 7;

And a₂ = 2, b₂ = 6, c₂ = - 14;

a₁ /a₂ = /2

b₁ /b₂ = 1/2

c₁ /c₂ = - 1/2

If a₁/a₂ = b₁/b₂ = c₁/c₂, then system has infinitely many solutions.

So /2 = 1/2

= 1

Also /2 = - 1/2

Since, does not have a unique value.

So, for no value of, the given pair of linear equations has infinitely many solutions.

The answer is False.

The given pair of linear equations

x - 2y - 8 = 0

and 5x - 10y - c = 0

On Comparing with ax + by + c = 0;

Here,

Here, a₁ = 1, b₁ = - 2, c₁ = - 8;

And a₂ = 5, b₂ = - 10, c₂ = - c;

a₁ /a₂ = 1/5

b₁ /b₂ = 1/5

c₁ /c₂ = 8/c

But if c = 40 (real value), then the ratio c₁/c₂ becomes 1/5 and then the system of linear equations has an infinitely many solutions.

Hence, at c= 40, the system of linear equations does not have a unique solution.

From the figure, the line x = 7 is parallel to the y-axis and not the x-axis.

The given pair of linear equations -

x + y - ² = 0and x + y - 1 = 0

Comparing with ax + by + c = 0;

Here, a₁ = , b₁ = 1, c₁ = - ²;

And a₂ = 1, b₂ = , c₂ = - 1;

a₁ /a₂ = λ/1

b₁ /b₂ = 1/λ

c₁ /c₂ = λ²

(i) For no solution,

a₁/a₂ = b₁/b₂ c₁/c₂

i.e. = 1/²

so, ² = 1;

and ²

Here, we take only = - 1 because at = 1, the system of linear equations has infinitely many solutions.

(ii) For infinitely many solutions,

a₁/a₂ = b₁/b₂ = c₁/c₂

i.e. = 1/ = ²

so = 1/ gives = 1;

= ² gives = 1,0;

Hence satisfying both the equations

= 1 is the answer.

(iii) For a unique solution,

a₁/a₂ b₁/b₂

so 1/

hence, ² 1;

1;

So, all real values of except

The given pair of linear equations is

kx + 3y = k - 3 …(i)

and 12x + ky = k …(ii)

On comparing with ax + by = c = 0, we get

a₁ = , b₁ = 3, c₁ = -(k - 3)

And a₂ = 12, b₂ = , c₂ = - k

a₁ /a₂ = /12

b₁ /b₂ = 3/

c₁ /c₂ =

For no solution of the pair of linear equations,

a₁/a₂ = b₁/b₂ c₁/c₂

So =

Taking first two parts, we get

/12 = 3/

k² = 36

k = 6

Taking last two parts, we get

3k k(k - 3)

k² - 6k 0

so, k 0,6

Hence, required value of k for which the given pair of linear equations has no solution is k = - 6.

The given pair of linear equations are:

x + 2y = 1 …(i)

and (a-b)x + (a + b)y = a + b - 2 …(ii)

On comparing withax + by = c = 0 we get

a₁ = , b₁ = 2, c₁ = - 1

And a₂ = (a - b), b₂ = (a + b), c₂ = - (a + b - 2)

a₁ /a₂ =

b₁ /b₂ =

c₁ /c₂ =

For infinitely many solutions of the, pair of linear equations,

a₁/a₂ = b₁/b₂c₁/c₂(coincident lines)

so, = =

Taking first two parts,

=

a + b = 2(a - b)

a = 3b …(iii)

Taking last two parts,

=

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

a + b = 4 …(iv)

Now, put the value of a from Eq. (iii) in Eq. (iv), we get

3b + b = 4

4b = 4

b = 1

Put the value of b in Eq. (iii), we get

a = 3

So, the values (a,b) = (3,1) satisfies all the parts. Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.

(i) Given pair of linear equations is

3x - y - 5 = 0 …(i)

and 6x - 2y - p = 0 …(ii)

On comparing with ax + by + c = 0 we get

Here, a₁ = , b₁ = - 1, c₁ = - 5;

And a₂ = 6, b₂ = - 2, c₂ = - p;

a₁ /a₂ = /6 = 1/2

b₁ /b₂ = 1/2

c₁ /c₂ = 5/p

Since, the lines represented by these equations are parallel, then

a₁/a₂ = b₁/b₂ c₁/c₂

Taking last two parts, we get

So, p

Hence, the given pair of linear equations are parallel for all real values of p except 10 i.e.,

P

(ii) Given pair of linear equations is

- x + py = 1 …(i)

and px - y - 1 = 0 …(ii)

On comparing with ax + by + c = 0, we get

Here, a₁ = , b₁ = p, c₁ =- 1;

And a₂ = p, b₂ = - 1, c₂ =- 1;

a₁ /a₂ =

b₁ /b₂ = - p

c₁ /c₂ = 1

Since, the lines equations has no solution i.e., both lines are parallel to each other.

a₁/a₂ = b₁/b₂ c₁/c₂

= - p 1

Taking last two parts, we get

p

Taking first two parts, we get

p² = 1

p = 1

Hence, the given pair of linear equations has no solution for p = 1.

(iii) Given, pair of linear equations is

- 3x + 5y = 7

and 2px - 3y = 1

On comparing with ax + by + c = 0, we get

Here, a₁ = , b₁ = 5, c₁ = - 7;

And a₂ = 2p, b₂ = - 3, c₂ = - 1;

a₁ /a₂ =

b₁ /b₂ = - 5/3

c₁ /c₂ = 7

Since, the lines are intersecting at a unique point i.e., it has a unique solution

a₁/a₂ b₁/b₂

so,

p 9/10

Hence, the lines represented by these equations are intersecting at a unique point for all real values of p except

(iv) Given, pair of linear equations is

2x + 3y - 5 = 0

and px - 6y - 8 = 0

On comparing with ax + by + c = 0 we get

Here, a₁ = , b₁ = 3, c₁ = - 5;

And a₂ = p, b₂ = - 6, c₂ = - 8;

a₁ /a₂ =

b₁ /b₂ = - 3/6 = - 1/2

c₁ /c₂ = 5/8

Since, the pair of linear equations has a unique solution.

a₁/a₂ b₁/b₂

so - 1/2

p - 4

Hence, the pair of linear equations has a unique solution for all values of p except - 4

i.e., p

(v) Given pair of linear equations is

2x + 3y = 7

and 2px + py = 28 - qy

or 2px + (p + q)y - 28 = 0

On comparing with ax + by + c = 0 we get

Here, a₁ = , b₁ = 3, c₁ = - 7;

And a₂ = 2p, b₂ = (p + q), c₂ = - 28;

Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.

a₁/a₂ = b₁/b₂ = c₁/c₂

= =

Taking first and third parts, we get

p = 4

Again, taking last two parts, we get

=

p + q = 12

Since p = 4

So, q = 8

Here, we see that the values of p = 4 and q = 8 satisfies all three parts.

Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.

Given linear equations are

x - 3y - 2 = 0 …(i)

and -2x + 6y - 5 = 0 …(ii)

On comparing with ax + by c=0, we get

Here, a₁ =1, b₁ =-3, c₁ =- 2;

And a₂ = -2, b₂ =6, c₂ =- 5;

i.e., a₁/a₂ = b₁/b₂ c₁/c₂ [parallel lines]

Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other.

Condition for the pair of system to have unique solution

a₁/a₂ b₁/b₂

Let the equations are,

a₁x + b₁y + c₁ = 0

and a₂x + b₂y + c₂ = 0

Since, x = - 1 and y = 3 is the unique solution of these two equations, then

It must satisfy the equations -

a₁(-1) + b₁(3) + c₁ = 0

- a₁ + 3b₁ + c₁ = 0 …(i)

and a₂(- 1) + b₂(3) + c₂ = 0

- a₂ + 3b₂ + c₂ = 0 …(ii)

Since for the different values of a₁, b₁, c₁ and a₂, b₂, c₂ satisfy the Eqs. (i) and (ii).

Hence, infinitely many pairs of linear equations are possible.

Given equations are

2x + y = 23 …(i)

and 4x - y = 19 …(ii)

On adding both equations, we get

6x = 42

So, x = 7

Put the value of x in Eq. (i), we get

2(7) + y = 23

y = 23 - 14

so, y = 9

Hence 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31

and

Hence, the values of (5y - 2x) and are 31 and respectively.

By property of rectangle,

Lengths are equal, i.e., CD = AB

So, x + 3y = 13 …(i)

Breadth are equal, i.e., AD = BC

So, 3x + y = 7 …(ii)

On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get

3(3x + y) = 3(7)

- x + 3y = 13

8x = 8

So, x = 1

On putting x = 1 in Eq. (i), we get

y = 4

Hence, the required values of x and y are 1 and 4, respectively.

Given pair of linear equations are is

and

Now, multiplying Eq. (i) by 2 and then adding with Eq. (ii), we get

5x = 6

Now, put the value of x in Eq. (i), we get

y = 2.1

Hence, the required values of x and y are 1.2 and 2.1, respectively.

Given, pair of linear equations is

On multiplying both sides by LCM (3, 4) = 12, we get

…(i)

and

On multiplying both sides by LCM
(6, 8) = 24, we get

…(ii)

Now, adding Eqs. (i) and (ii), we get

24x = 144

x = 6

Now, put the value of x in Eq. (i), we get

Hence, the required values of x and y are 6 and 8, respectively.

Given pair of linear equations are

…(i)

and …(ii)

Let then above equation becomes

…(iii)

and …(iv)

On multiplying Eq. (iii) by 8 and Eq. (iv) by 6 and then adding both of them, we get

68x = 204

x = 3

Now, put the value of x in Eq. (iii), we get

Hence, the required values of x and y are 3 and 2, respectively.

Given pair of linear equations is

…(i)

and …(ii)

Let and then the above equations become:

…(iii)

and

…(iv)

On, multiplying Eq. (iv) by 2 and then adding with Eq. (iii), we get

( + (u - 2v) = (-2 + 32)

5u = 30

Now, put the value of u in Eq. (iv), we get

Hence, the required values of x and y are and respectively.

Given pair of linear equations is

…(i)

and …(ii)

On multiplying Eq. (i) by 43 and Eq. (ii) by 67 and then subtracting both of them, we get

Now, put the value of x in Eq. (i), we get

Hence, the required values of x and y are 1 and - 1, respectively.

Given pair of linear equations is

…(i)

and …(ii)

On multiplying Eq. (i) by and then subtracting from Eq. (ii), we get

-

Now, put the value of y in Eq. (ii), we get

Hence, the required values of x and y are and respectively.

Given pair of equations is

And
=

Now, put and then the pair of equations becomes

…(iii)

and …(iv)

On adding both equations, we get

Now, put the value of v in Eq. (iii), we get

and

Hence, the required values of x and y are and respectively.

Given pair of equations is

…(i)

and …(ii)

Now, multiplying both sides of Eq. (i) by LCM (10, 5) = 10, we get

…(iii)

Again, multiplying both sides of Eq. (iv) by LCM (8, 6) = 24, we get

…(iv)

On, multiplying Eq. (iii) by 2 and then subtracting from Eq., (iv), we get

Put the value of x in Eq. (iii), we get

Given that the linear relation between x, y and is

Now, put the values of x and y in above relation, we get

Hence, the solution of the pair of equations is x = 340, y = - 165 and the required value of is

Given pair of equations is

3x + y + 4 = 0 …(i)

and 6x -2y + 4 = 0…(ii)

comparing with ax + by + c = 0

Here, a₁ = , b₁ = 1, c₁ = 4;

And a₂ = 6, b₂ = - 2, c₂ = 4;

a₁ /a₂ = 1/2

b₁ /b₂ = -3/6 = -1/2

c₁ /c₂ = 1

since a₁/a₂ b₁/b₂

so system of equations is consistent with a unique solution.

We have,

When x = 0, then y = - 4

When x = - 1, then y = - 1

When x = - 2, then y = 2

and

When x = 0, then y = 2

When x = - 1,then y = - 1

When x = 1,then y = 5

Plotting the points B(0, - 4) and A( - 2,2),we get the straight tine AB. Plotting the points Q(0,2) and P(1,5) we get the straight line PQ. The lines AB and PQ intersect at C (-1, -1).

Given pair of equations is

x -2y = 6 …(i)

and 3x-6y = 0 …(ii)

On comparing with ax + by + c = 0

we get

Here, a₁ = , b₁ = - 2, c₁ = - 6;

And a₂ = 3, b₂ = - 6, c₂ = 0;

a₁ /a₂ = 1/3

b₁ /b₂ = - 2/ - 6 = 1/3

here a₁/a₂ = b₁/b₂ c₁/c₂ (parallel lines).

Hence, the lines represented by the given equations are parallel. Therefore, it has no solution. So, the given pair of lines is inconsistent.

Given pair of equations is

x + y = 3 …(i)

and 3x + 3y = 9 …(ii)

On comparing with ax + by + c = 0

Here, a₁ = 1, b₁ = 1, c₁ = - 3;

And a₂ = 3, b₂ = 3, c₂ = - 9;

a₁ /a₂ = 1/3

b₁ /b₂ = 1/3

c₁ /c₂ = 1/3

Here, a₁/a₂ = b₁/b₂ = c₁/c₂, i.e. coincident lines

Hence, the given pair of linear equations is coincident and having infinitely many solutions.

The given pair of linear equations is consistent.

Now,

If x = 0 then y = 3, If x = 3, then y = 0.

and

If x = 0 then y = 3, if x = 1, then y = 2, and if x = 3, then y = 0.

Plotting the points A(0, 3) and B(3, 0), we get the line AB. Again, plotting the points C(0, 3) and D(1, 2) and E(3, 0), we get the line CDE.

The given pair of linear equations

2x + y = 4

and 2x-y = 4

Table for line 2x + y = 4

and table for line

Graphical representation of both lines.

Here, both lines and Y - axis from a .

Hence, the vertices of a ABC are A (0,4), B(2,0) and C(0,- 4).

∴ Required area of ABC Area of

ABC = sq. units.

Hence, the required area of the triangle is 8 sq units.

Given pair of linear equations is

x + y-2 = 0 …(i)

and 2x-y-1 = 0 …(ii)

On comparing with ax + by + c = 0

Here, a₁ = 1, b₁ = 1, c₁ = - 2;

And a₂ = 2, b₂ = - 1, c₂ = - 1;

a₁ /a₂ = 1/2

b₁ /b₂ = - 1

c₁ /c₂ = 2

since a₁/a₂ b₁/b₂

So, both lines intersect at a point. Therefore, the pair of equations has a unique solution. Hence, these equations are consistent.

Now, x + y = 2 or y = 2-x

If x = 0 then y = 2 and if x = 2 then y = 0

and

If x = 0 then y = - 1 if x = then y = 0 and if x = 1 then y = 1

Plotting the points A (2,0) and B(0,2), we get the straight line AB. Plotting the points C(0,- 1) and D(1/2, 0), we get the straight line CD.

The lines AB and CD intersect at E (1, 1).

Hence, infinite lines can pass through the intersection point of linear equations and i.e. E (1, 1) like as y = x, x + 2y = 3, x + y = 2 and so on.

Given that (x + 1) is a factor of f(x)= (2x³ + ax² + 2bx + 1), then f(-1) = 0

[if is a factor of f(x)= ax² + bx + c then f( - α) = 0]

-2 + a-2b + 1 = 0
a - 2b - 1 = 0 …(i)

Also, 2a - 3b = 4

3b = 2a - 4

Now, put the value of b in Eq. (i), we get

3a - 2(2a - 4) - 3 = 0
3a - 4a + 5 = 0
a = 5

Now, put the value of a in Eq. (i), we get

5 - 2b - 1 = 0
2b = 4
b = 2

Hence, the required values of a and b are 5 and 2, respectively.

Given that, x, y and are the angles of a triangle.

So, x + y + 40 = 180

[since, the sum of all the angles of a triangle is 180^o]

And hence x + y = 140 …(i)

Also, difference of angles = x – y = 30 …(ii)

On adding Eqs. (i) and (ii), we get

2x = 170

So, x = 85

On putting in Eq. (i), we get

y = 55

Hence, the required values of x and y are 85 and 55, respectively.

Let Salim and his daughter’s age be x and y years respectively.

Now, by first condition

Two years ago, Salim was thrice as old as his daughter.

i.e., x - = 3(y - 2)

so, x - 3y = - 4 …(i)

and by second condition, six years later. Salim will be four years older than twice her age.

(x + 6) = 2(y + 6) + 4

x + 6 = 2y + 16

x – 2y = 10 …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

x – 2y = 10

- (x–3y =- 4)

We get, y = 14

Put the value of y in Eq. (ii), we get

x - 2(14) = 10

so, x = 38

Hence, Salim and his daughter’s age are 38 year and 14 year, respectively.

Let the present age (in year) of father and his two children be x, y and z year, respectively.

Now by given condition, x = 2(y + z) …(i)

and after 20 year, (x + 20) = (y + 20) + (z + 20)

so, x = y + z + 20 or y + z = x - 20 …(ii)

On putting the value of (y + z) in Eq, (i) and get the present age of father

So, x = 2(x -20)

x = 40

Hence, the father’s age is 40 yr.

Let the two numbers be x and y.

Then, by first condition, ratio of these two numbers = 5 :6

and by second condition, then, 8 is subtracted from each of the numbers, then ratio becomes 4:5

=

5x – 4y = 8

Now, put the value of y in Eq. (ii), we get

5x – 4 ( ) = 8

So

And x = 40

Put the value of x in Eq. (i), we get

So, y = = 48

Hence, the required numbers are 40 and 48.

Let the number of students in halls A and B are x and y, respectively.

Now, by given condition, x – 10 = y + 10

x - y = 20 …(i)

and (x + 20) = 2(y - 20)

x–2y = -60 …(ii)

On subtracting Eq. (ii) from Eq. (i), we get

(x - y) - (x -2y) = 80

y = 80

On putting y = 80 in Eq. (i), we get

x – 80 = 20

so, x = 100

Hence, 100 students are in hall A and 80 students are in hall B.

Let Latika takes a fixed charge for the first two day is Rs x and additional charge for each day thereafter is Rs y.

Now by first condition.

Latika paid Rs 22 for a book kept for six days i.e., for first two days as Rs x and remaining Rs 4y.

x + 4y = 22 …(i)

and by second condition,

Anand paid Rs 16 for a book kept for four days i.e., for first two days as x’ and remaining Rs 2y.

x + 2y = 16 …(ii)

Now, subtracting Eq. (ii) from Eq. (i), we get

2y = 6

y = 3

On putting the value of y in Eq. (ii), we get

x + 2(3) = 16

x = 10

Hence, the fixed charge = Rs 10

and the charge for each extra day Rs 3.

Let x be the number of correct answers of the questions in a competitive examination, then (120 - x) be the number of wrong answers of the questions.

Hence, Jayanti answered correctly 100 questions.

We know that, by property of cyclic quadrilateral,

Sum of opposite angles = 180^o

Since
and

So, 7x + y = 170 …(i)

and

Since
and

So, 5x + 3y = 190 …(ii)

On multiplying Eq. (i) by 3 and then subtracting, we get

3(7x + y) – (5x + 3y) = 3(170) – 190

16x = 320

x = 20

On putting x = 20 in Eq. (i), we get

7(20) + y = 170

So, y = 30

And hence

= 20 + 30 = 50

Hence, the required values of x and y are 20 and 30 respectively and the values of the four angles i.e., and, are 130, 100, 50, and 80, respectively.

Given equations are 2x + y = 6 and 2x – y + 2 = 0

Table for equation 2x + y - 6 = 0, for x = 0, y = 6, for y = 0, x = 3.

Table for equation 2x – y + 2 = 0, for x = 0, y = 2, for y = 0,x = - 1

Let A₁ and A₂ represent the areas of and & respectively where E is the intersection of lines.

Now, Area of triangle formed with x -axis =

=

And Area of triangle formed with y - axis =

Hence, the pair of equations intersect graphically at point E(1,4)

i.e., x = 1 and y = 4.

Given linear equations are

y = x …(i)

3y = x …(ii)

and x + y = 8 …(iii)

For equation y = x,

If x = 1, then y = 1

If x = 0, then y = 0

If x = 2, then y = 2

Table for line y = x,

For equation x = 3y,

If x = 0, then y = 0. if x = 3, then y = 1, and if x = 6, then y = 2

Table for line x = 3y,

For equation x + y = 8 or x = 8 - y

If x = 0, then y = 8, if x = 8, then y = 0 and if x = 4, then y = 4.

Table for line x + y = 8

Plotting the points A (1, 1) and B(2,2) we get the straight line AB.

Plotting the points C (3, 1) and D(6, 2), we get the straight line CD.

Plotting the point P (0, 8), Q(4, 4) and R(8, 0), we get the straight line PQR.

We see that lines AB and CD intersecting the line PR on Q and D, respectively.

So, is formed by these lines. Hence, the vertices of the formed by the given lines are O(0, 0), Q(4, 4) and D(6,2).

Given equation of lines x = 3, x = 5 and 2x-y-4 = 0.

Now, the table for line 2x - y - 4 = 0

for x = 0, y = - 4 and for y = 0, x = 2

Draw the points P(0, - 4) and Q(2, 0) and join these points and form a line PQ also draw the lines x = 3 and x = 5.

Area of Quadrileral ABCD = 1/2×(distance between parallel lines)

[since, quadrilateral ABCD is a trapezium]

{since AB = OB-OA = 5-3 = 2, AD = 2 and BC = 6}
= 8 sq units

Let the cost of a pen be Re x and the cost of a pencil box be Re y.

4x + 4y = 100

or x + y = 25 …(i)

and 3x = y + 15

3x-y = 15 …(ii)

On adding Eqs. (i) and (ii), we get

4x = 40

So, x = 10

By substituting x = 10, in Eq. (i) we get

Y = 25-10 = 15

Hence, the cost of a pen and a pencil box are Rs 10 and Rs15, respectively.

Given equation of lines are

3x - y = 2 …(i)

2x -3y = 2 …(ii)

and x + 2y = 8 …(iii)

Let lines (i), (ii) and (iii) represent the side of a ∆ABC i.e., AB, BC and CA respectively.

On solving lines (i) and (ii), we will get the intersecting point B.

On multiplying Eq. (i) by 3 in Eq. (i) and then subtracting, we get

(9x-3y)-(2x-3y) = 9-2

7x = 7

x = 1

On putting the value of x in Eq. (i), we get

3×1-y = 3
y = 0

So, the coordinate of point or vertex B is (1, 0)

On solving lines (ii) and (iii), we will get the intersecting point C.

On multiplying Eq. (iii) by 2 and then subtracting, we get

(2x + 4y)-(2x-3y) = 16-2
7y = 14
y = 2

On putting the value of y in Eq. (iii), we get

Hence, the coordinate of point or vertex C is (4, 2).

On solving lines (iii) and (i), we will get the intersecting point A.

On multiplying in Eq. (i) by 2 and then adding Eq. (iii), we get

(6x-2y) + (x + 2y) = 6 + 8
7x = 14
x = 2

On putting the value of x in Eq. (i), we get

3×2 - y = 3
y = 3

So, the coordinate of point or vertex A is (2, 3).

Hence, the vertices of the ∆ABC formed by the given lines are A (2, 3), B(1, 0) and C (4, 2).

Let the speed of the rickshaw and the bus are x and y km/h, respectively.

Now, she has taken time to travel 2 km by rickshaw,

she has taken time to travel remaining distance i.e., (14 - 2) = 12km

By bus

By first condition,

…(i)

Now, she has taken time to travel 4 km by rickshaw,

and she has taken time to travel remaining distance i.e., (14 - 4) = 10km, by bus =

By second condition,

…(ii)

Let and then Eqs. (i) and (ii) becomes

…(iii)

and …(iv)

On multiplying in Eq. (iii) by 2 and then subtracting, we get

Now, put the value of v in Eq. (iii), we get

Hence, the speed of rickshaw and the bus are 10 km/h and 40 km/h, respectively.

Let the speed of the stream be v km/h.

Given that, a person rowing in still water = 5 km/h

The speed of a person rowing in downstream =

and the speed of a person has rowing in upstream =

Now, the person taken time to cover 40 km downstream,

and the person has taken time to cover 40 km upstream,

By condition,

Hence, the speed of the stream is 2.5 km/h.

Let the speed of the motorboat in still water and the speed of the stream are u km/h and v km/h, respectively.

Then, a motorboat speed in downstream

and a motorboat speed in upstream

Motorboat has taken time to travel 30 km upstream,

and motorboat has taken time to travel 28 km downstream,

By first condition, a motorboat can travel 30 km upstream and 38 km downstream in 7 h i.e.,

…(i)

Now, motorboat has taken time to travel 21 km upstream and return i.e., [for upstream]

and [for downstream]

By second condition,

…(ii)

Let

Eqs. (i) and (ii) becomes …(iii)

and

…(iv)

Now, multiplying in Eq. (iv) by 28 and then subtracting from Eq. (iii), we get

On putting the value of x in Eq. (iv), we get

…(v)

and
…(vi)

Now, adding Eqs. (v) and (vi), we get

2u = 20

So u = 10

On putting the value of in Eq. (v), we get

10 + v = 6

So v = - 4

Hence, the speed of the motorboat in still water is 10 km/h and the speed of the stream 4 km/h.

Let the digits be x and y, then
two - digit number = 10x + y

Case I Multiplying the sum of the digits by 8 and then subtracting 5 = two digit number

Case II Multiplying the difference of the digits by 16 and then adding 3 = two - digit number

Now, multiplying in Eq. (i) by 3 and then subtracting from Eq. (ii), we get

Now, put the value of y in Eq. (i), we get

Hence, the required two - digit number

Let the cost of full and half first class fare be ₹ x and ₹ x/2 respectively and reservation charges be ₹ y per ticket.

Case I The cost of one reserved first class ticket from the stations A to B

= ₹ 2530

…(i)

Case II The cost of one reserved first class ticket and one reserved first class half ticket from stations A to B = ₹ 3810

3x + 4y = 7620 ..(ii)

Now, multiplying Eq. (i) by 4 and then subtracting from Eq. (ii), we get

On putting the value of x in Eq. (i), we get

Hence, full first class fare from stations A to B is ₹ 2500 and the reservation for a ticket is ₹ 30.

Let the cost price of the saree and the list price of the sweater be ₹ x and ₹ y, respectively.

Case I Sells a saree at 8% profit + Sells a sweater at 10% discount = ₹ 1008

…(i)

Case II Sold the saree at 10% profit + Sold the sweater at 8% discount = ₹ 1028

(100 + 10)% of x + (100-8%) of y = 1028
110% of x + 92% of y = 1028
1.10x + 0.92 y = 1028 …(ii)

On putting the value of y from Eq. (i) into Eq. (ii), we get

1.1 × 0.9x + 0.92 × (1008-1.08x) = 1028×0.9
0.99x-0.9936x = 9252-927.36
-0.0036x = -2.16

On putting the value of x in Eq. (i), we get

Hence, the cost price of the saree and the list price (price before discount) of the sweater are ₹ 600 and ₹ 400, respectively.

Let the amount of investments in schemes A and B ₹ x and ₹ y, respectively.

Case I Interest at the rate of 8% per annum on scheme A + Interest at the rate of 9% per annum on scheme B = Total amount received

₹

Since,

so, 8x + 9y = 186000 …(i)

Case II Interest at the rate of 9% per annum on scheme A + Interest at the rate of 8% per annum on scheme B = ₹ 20 more as annual Interest

So, 9x + 8y = 188000 …(ii)

On multiplying Eq. (i) by 9 and Eq. (ii) by 8 and then subtracting them, we get

(72x + 81y) – (72x + 64y) = 186000 – 188000

17y = 170000

y = 10000

On putting the value of y in Eq. (i), we get

8x + 9(10000) = 186000

So x = 12000

Hence, she invested ₹ 12000 and ₹ 10000 in two schemes A and B, respectively.

Let the number of bananas in lot A and B be x and y, respectively

Case I :

Cost of the first lot at the rate of ₹ 2 for 3 bananas + Cost of the second lot at the rate of ₹ 1 per banana = Amount received

…(i)

Case II :

Cost of the first lot at the rate of ₹ 1 per banana + Cost of the second lot at the rate of ₹ 4 for 5 bananas = Amount received

…(ii)

On multiplying in Eq. (i) by 4 and Eq. (ii) by 3 and then subtracting them, we get

(8x + 12y) – (15x + 12y) = 4800 – 6900

- 7x = - 2100

x = 300

Now, put the value of in Eq. (i), we get

2(300) + 3y = 1200

3y = 600

y = 200

∴ Total number of bananas = Number of bananas in lot Number of bananas in lot B = x + y

x + y = 300 + 200 = 500

Hence, he had 500 bananas.