- Sudha went to market with her friends. They wanted to eat `gol - gappa' as well…
- Romila went to a stationary shop and purchased 2 pencils and 3 erasers for Rs.…
- Present age of father is 30 years more than twice that of his son. After 10…
- The path of a wheel of train A is given by the equation x + 2y — 4 = 0 and the…
- The path of highway number 1 and 2 are given by the equations x — y = 1 and 2x +…
- Person A walks along the path joining points (0, 3) and (1, 3) and person B…
- Examine which of the following pair of values of x and y is a solution of…
- Examine which of the following points lie on the graph of the linear equation 5x…
- 3x + y = 2 6x + 2y = 1 Solve graphically the following system of linear…
- 2x — 3y + 13 = 0 3x — 2y + 12 = 0 Solve graphically the following system of…
- 3x + 2y = 14 x — 4y = — 14 Solve graphically the following system of linear…
- 2x — 3y = 1 3x — 4y = 1 Solve graphically the following system of linear…
- 2x — y = 9 5x + 2y = 27 Solve graphically the following system of linear…
- 3y = 5 — x 2x = y + 3 Solve graphically the following system of linear…
- 3x — 5y = —1 2x — y = —3 Solve graphically the following system of linear…
- 2x — 6y + 10 = 0 3x — 9y + 15 = 0 Solve graphically the following system of…
- 3x + y — 11 = 0 x — y — 1 = 0 Solve graphically the following system of linear…
- 3x — 5y = 19, 3y — 7x + 1 = 0 Does the point (4, 9) lie on any of the lines?…
- Solve the following system of linear equations graphically: 2x — 3y = 1, 3x —…
- 2x + 3y = 7, (a + b) x + (2a - b) y = 3(a + b + 1) Find the values of a and b for which…
- x — 2y = - 3 2x + y = 4 Solve the following system of equations graphically.…
- 2x + 3y = 8 x — 2y = — 3 Solve the following system of equations graphically.…
- x + 2y = 5 2x — 3y = — 4 Solve the following system of equations graphically.…
- x — y + 1 = 0 4x + 3y = 24 Solve the following system of equations…
- x + 2y = 1 x— 2y = 7 Solve the following system of equations graphically. Also…
- x + 2y = 1 x — 2y = —7 Solve the following system of equations graphically.…
- 2x — y = 4 3y — x = 3 Solve the following system of equations graphically.…
- 2x + 3y — 12 = 0 2x — y — 4 = 0 Solve the following system of equations…
- 2x — y — 5 = 0 x — y — 3 = 0 Solve the following system of equations…
- 2x — y — 4 = 0 x + y + 1 = 0 Solve the following system of equations…
- 3x + y — 5 = 0 2x — y — 5 = 0 Solve the following system of equations…
- 3x + 2y — 4 = 0 2x — 3y — 7 = 0 Shade the region bounded by the lines and the…
- 3x — 2y - 1 = 0 2x — 3y + 6 = 0 Shade the region bounded by the lines and the…
- 2x + y = 6 2x — y = 0 Solve the following pair of linear equations…
- 2x + 3y = —5 3x — 2y = 12 Solve the following pair of linear equations…
- 4x — 3y + 4 = 0 4x + 3y — 20 = 0 Solve the following pair of linear equations…
- 2x + y = 6 2x — y + 2 = 0 Solve the following pair of linear equations…
- x — y = 1 2x + y = 8 Solve the following pair of linear equations graphically…
- 3x + y — 11 = 0 x — y — 1 = 0 Solve the following pair of linear equations…
- x + 2y — 7 = 0 2x — y — 4 = 0 Solve the following pair of linear equations…
- 4x — y = 4 3x + 2y = 14 Solve the following system of linear equations…
- x — y = 1 2x + y = 8 Solve the following system of linear equations…
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- cx + 3y - (c - 3) = 0, 12x + cy - c = 0 For what value of c, the following system of…
- 2x + 3y = 2, (c + 2) x + (2c + 1) y = 2 (c - 1) For what value of c, the following system…
- x + (c + 1) y = 5, (c + 1) x + 9y = 8c - 1 For what value of c, the following system of…
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- Solve the following system of equations graphically. Also determine the…
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###### Exercise 3.1

Question 1.Sudha went to market with her friends. They wanted to eat `gol - gappa' as well as `dahi - bhalla'. The number of plates of gol - gappa taken by them is half that of dahi - bhalla. The cost of one plate of gol - gappa was Rs. 10 and cost of one plate of dahi - bhalla was Rs. 5. She spent Rs. 60. Represent the situation algebraically and graphically.

Answer:

Let no. of plates of gol - gappa = x

and no. of plates of dhai - bhalla = y

Cost of 1 plate gol - gappa = Rs. 10

Cost of 1 plate dhai - bhalla = Rs. 5

Total money spent = Rs. 20

According to the question,

…(1)

10x + 5y = 60 …(2)

From eqn (1), we get

2x – y = 0 …(3)

Now, table for 2x – y = 0

Now, table for 10x + 5y = 60

On plotting points on a graph paper and join them to get a straight line representing .

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing 10x + 5y = 60.

Here, the lines representing Eq. (1) and Eq. (2) intersecting at point A i.e. (3,6).

Question 2.

Romila went to a stationary shop and purchased 2 pencils and 3 erasers for Rs. 9. Her friend Sonali saw the new variety of pencils and erasers with Romila and she also bought 4 pencils and 6 erasers of the same kind for Rs. 18. Represent this situation algebraically and graphically.

Answer:

Let the cost of one pencil = Rs x

and cost of one eraser = Rs y

Romila spent = Rs. 9

Sonali spent = Rs. 18

According to the question

2x + 3y = 9 …(1)

4x + 6y = 18 …(2)

Now, table for 2x + 3y = 9

Now, table for 4x + 6y = 18

On plotting points on a graph paper and join them to get a straight line representing 2x + 3y = 9.

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing 4x + 6y = 18.

Here, we can see that both the lines coincide. This is so, because, both the equations are equivalent, i.e.2(2x + 3y) = 2×9 equation (2) is derived from the other.

Question 3.

Present age of father is 30 years more than twice that of his son. After 10 years, the age of father will be thrice the age of his son. Represent this situation algebraically and geometrically.

Answer:

Let the present age of son = x year

and the age of his father = y year

According to the question

y = 2x + 30

or, 2x – y = – 30 …(1)

After 10 years,

Age of son = (x + 10)year

Age of father = (y + 10)year

So, According to the question

y + 10 = 3(x + 10)

y + 10 = 3x + 30

y = 3x + 20

or, 3x – y = – 20 …(2)

Now, table for y = 2x + 30

Now, table for y = 3x + 20

On plotting points on a graph paper and join them to get a straight line representing y = 2x + 30.

Similarly, on plotting the points on the same graph paper and join them to get a straight line representing y = 3x + 20.

Here, the lines representing Eq. (1) and Eq. (2) intersecting at point A i.e. (10,50).

So, the age of son is 10years and age of his father is 50years.

Question 4.

The path of a wheel of train A is given by the equation x + 2y — 4 = 0 and the path of a wheel of another train B is given by the equation 2x + 4y — 12 = O. Represent this situation geometrically.

Answer:

The given equation is

x + 2y – 4 = 0

and

2x + 4y – 12 = 0

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for x + 2y – 4 = 0

Now, table for 2x + 4y – 12 = 0

From the graph, it is clear that lines represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0 are parallel.

Question 5.

The path of highway number 1 and 2 are given by the equations x — y = 1 and 2x + 3y = 12 respectively. Represent these equations geometrically.

Answer:

The given equation is

x – y = 1

and

2x + 3y = 12

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for x – y = 1 or y = x – 1

Now, table for 2x + 3y = 12

From the graph, it is clear that lines represented by the equations x – y = 1 and 2x + 3y – 12 = 0 are intersecting at a point A i.e. (3,2)

Question 6.

Person A walks along the path joining points (0, 3) and (1, 3) and person B walks along the path joining points (0, 4) and (1, 5). Represent this situation geometrically.

Answer:

The given points are at which Person A walks (0,3) and (1,3)

and the points at which person B walks (0,4) and (1,5)

Now, we plot these points on a same graph as shown in the following figure.

Question 7.

Examine which of the following pair of values of x and y is a solution of equation 4x — 3y + 24 = 0.

(i) x = 0, y = 8 (ii) x = — 6, y = 0

(iii) x = 1, y = — 2 (iv) x = – 3,y = 4

(v) x = 1, y = — 2 (vi) x = — 4, y = 2

Answer:

Given equation is

i) Justification

On substituting x = 0, y = 8 in LHS of given equation, we get

LHS = 4(0) – 3(8) + 24 = 0 – 24 + 24 = 0 = RHS

Hence, x = 0, y = 8 is a solution of the equation 4x – 3y + 24 = 0

ii) Justification

On substituting x = – 6, y = 0 in LHS of given equation, we get

LHS = 4( – 6) – 3(0) + 24 = – 24 + 24 = 0 = RHS

Hence, x = – 6, y = 0 is a solution of the equation 4x – 3y + 24 = 0

iii) Justification

On substituting x = 1, y = – 2 in LHS of given equation, we get

LHS = 4(1) – 3( – 2) + 24 = 4 + 6 + 24 = 34 ≠ RHS

Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0

iv) Justification

On substituting x = – 3, y = 4 in LHS of given equation, we get

LHS = 4( – 3) – 3(4) + 24 = – 12 – 12 + 24 = 0 = RHS

Hence, x = – 3, y = 4 is a solution of the equation 4x – 3y + 24 = 0

v) Justification

On substituting x = 1, y = – 2 in LHS of given equation, we get

LHS = 4(1) – 3( – 2) + 24 = 4 + 6 + 24 = 34 ≠ RHS

Hence, x = 1, y = – 2 is not a solution of the equation 4x – 3y + 24 = 0

vi) Justification

On substituting x = – 4, y = 2 in LHS of given equation, we get

LHS = 4( – 4) – 3(2) + 24 = – 16 – 6 + 24 = – 22 + 24 = 2 ≠ RHS

Hence, x = – 4, y = 2 is not a solution of the equation 4x – 3y + 24 = 0

Question 8.

Examine which of the following points lie on the graph of the linear equation 5x — 3y + 30 = 0.

(i) A (— 6, 0) (ii) B (0, 10)

(iii) C (3, — 5) (iv) D (4, 2)

(v) E (— 9, 5) (vi) F (— 3, 5)

(vii) G (— 9, — 5)

Answer:

The given equation is 5x – 3y + 30 = 0

(i) Given A ( – 6,0). Here x = – 6 and y = 0

On substituting x = – 6, y = 0 in LHS of given equation, we get

LHS = 5( – 6) – 3(0) + 30 = – 30 + 30 = 0 = RHS

So, x = – 6, y = 0 is a solution of the equation 5x – 3y + 30 = 0.

Hence, point A lies on the graph of the linear equation 5x – 3y + 30 = 0.

(ii) Given B (0,10). Here x = 0 and y = 10

On substituting x = 0, y = 10 in LHS of given equation, we get

LHS = 5(0) – 3(10) + 30 = – 30 + 30 = 0 = RHS

So, x = 0, y = 10 is a solution of the equation 5x – 3y + 30 = 0

Hence, point B lies on the graph of the linear equation 5x – 3y + 30 = 0.

(iii) Given C (3, – 5). Here x = 3 and y = – 5

On substituting x = 3, y = – 5 in LHS of given equation, we get

LHS = 5(3) – 3( – 5) + 30 = 15 + 15 + 30 = 60 ≠ RHS

So, x = 3, y = – 5 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point C does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(iv) Given D (4,2). Here x = 4 and y = 2

On substituting x = 4, y = 2 in LHS of given equation, we get

LHS = 5(4) – 3(2) + 30 = 20 – 6 + 30 = 44 ≠ RHS

So, x = 4, y = 2 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point D does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(v) Given E ( – 9,5). Here x = – 9 and y = 5

On substituting x = – 9, y = 5 in LHS of given equation, we get

LHS = 5( – 9) – 3(5) + 30 = – 45 – 15 + 30 = – 30 ≠ RHS

So, x = – 9, y = 5 is not a solution of the equation 5x – 3y + 30 = 0

Hence, point E does not lie on the graph of the linear equation 5x – 3y + 30 = 0.

(vi) Given F ( – 3,5). Here x = – 3 and y = 5

On substituting x = – 3, y = 5 in LHS of given equation, we get

LHS = 5( – 3) – 3(5) + 30 = – 15 + 15 + 30 = 0 = RHS

So, x = – 3, y = 5 is a solution of the equation 5x – 3y + 30 = 0

Hence, point F lies on the graph of the linear equation 5x – 3y + 30 = 0.

(vii) Given G ( – 9, – 5). Here x = 3 and y = – 5

On substituting x = – 9, y = – 5 in LHS of given equation, we get

LHS = 5( – 9) – 3( – 5) + 30 = – 45 + 15 + 30 = 0 = RHS

So, x = – 9, y = – 5 is a solution of the equation 5x – 3y + 30 = 0

Hence, point G lies on the graph of the linear equation 5x – 3y + 30 = 0.

Or __Graphically__

Here, we can see through the graph also that Point A, B, F and G lie on the graph of the linear equation 5x – 3y + 30 = 0

Question 9.

Solve graphically the following system of linear equations if it has unique solution:

3x + y = 2

6x + 2y = 1

Answer:

The given pair of linear equations is

3x + y = 2 or 3x + y – 2 = 0

and 6x + 2y = 1 or 6x + 2y – 1 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get

a1 = 3, b1 = 1 and c1 = – 2

and a2 = 6, b2 = 2 and c2 = – 1

The lines representing the given pair of linear equations are parallel.

Question 10.

Solve graphically the following system of linear equations if it has unique solution:

2x — 3y + 13 = 0

3x — 2y + 12 = 0

Answer:

The given pair of linear equations is

2x – 3y + 13 = 0

and 3x – 2y + 12 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get

a1 = 2, b1 = – 3 and c1 = 13

and a2 = 3, b2 = – 2 and c2 = 12

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point B i.e. ( – 2,3)

Hence, the unique solution is x = – 2 and y = 3.

Question 11.

Solve graphically the following system of linear equations if it has unique solution:

3x + 2y = 14

x — 4y = — 14

Answer:

The given pair of linear equations is

3x + 2y = 14 or 3x + 2y – 14 = 0

and x – 4y = – 14 or x – 4y + 14 = 0

On comparing the given equations with standard form of pair of linear equations i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get

a1 = 3, b1 = 2 and c1 = – 14

and a2 = 1, b2 = – 4 and c2 = 14

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point B i.e. (2,4)

Hence, the unique solution is x = 2 and y = 4.

Question 12.

Solve graphically the following system of linear equations if it has unique solution:

2x — 3y = 1

3x — 4y = 1

Answer:

The given pair of linear equations is

2x – 3y = 1 or 2x – 3y – 1 = 0

and 3x – 4y = 1 or 3x – 4y – 1 = 0

a1 = 2, b1 = – 3 and c1 = – 1

and a2 = 3, b2 = – 4 and c2 = – 1

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point C i.e. ( – 1, – 1)

Hence, the unique solution is x = – 1 and y = – 1.

Question 13.

Solve graphically the following system of linear equations if it has unique solution:

2x — y = 9

5x + 2y = 27

Answer:

The given pair of linear equations is

2x – y = 9 or 2x – y – 9 = 0

and 5x + 2y = 27 or 5x + 2y – 27 = 0

a1 = 2, b1 = – 1 and c1 = – 9

and a2 = 5, b2 = 2 and c2 = – 27

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point C i.e. (5,1)

Hence, the unique solution is x = 5 and y = 1.

Question 14.

Solve graphically the following system of linear equations if it has unique solution:

3y = 5 — x

2x = y + 3

Answer:

The given pair of linear equations is

x + 3y = 5 or x + 3y – 5 = 0

and 2x – y = 3 or 2x – y – 3 = 0

a1 = 1, b1 = 3 and c1 = – 5

and a2 = 2, b2 = – 1 and c2 = – 3

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point C i.e. (2,1)

Hence, the unique solution is x = 2 and y = 1.

Question 15.

Solve graphically the following system of linear equations if it has unique solution:

3x — 5y = —1

2x — y = —3

Answer:

The given pair of linear equations is

3x – 5y = – 1 or 3x – 5y + 1 = 0

and 2x – y = – 3 or 2x – y + 3 = 0

a1 = 3, b1 = – 5 and c1 = 1

and a2 = 2, b2 = – 1 and c2 = 3

The lines representing the given pair of linear equations will intersect at a point.

Now, table for or

Now, table for or

Here, the lines intersecting at point B i.e. ( – 2, – 1)

Hence, the unique solution is x = – 2 and y = – 1.

Question 16.

Solve graphically the following system of linear equations if it has unique solution:

2x — 6y + 10 = 0

3x — 9y + 15 = 0

Answer:

The given pair of linear equations is

2x – 6y + 10 = 0

and 3x – 9y + 15 = 0

a1 = 2, b1 = – 6 and c1 = 10

and a2 = 3, b2 = – 9 and c2 = 15

The lines representing the given pair of linear equations will coincide.

Question 17.

Solve graphically the following system of linear equations if it has unique solution:

3x + y — 11 = 0

x — y — 1 = 0

Answer:

The given pair of linear equations is

3x + y – 11 = 0

and x – y – 1 = 0

a1 = 3, b1 = 1 and c1 = – 11

and a2 = 1, b2 = – 1 and c2 = – 1

The lines representing the given pair of linear equations will intersect at a point.

Now, table for 3x + y – 11 = 0 or y = 11 – 3x

Now, table for x – y – 1 = 0 or y = x – 1

Here, the lines intersecting at point B i.e. (3,2)

Hence, the unique solution is x = 3 and y = 2.

Question 18.

Solve the following system of linear equations graphically:

3x — 5y = 19, 3y — 7x + 1 = 0

Does the point (4, 9) lie on any of the lines? Write its equation.

Answer:

The given equation is 3x – 5y = 19

and 3y – 7x + 1 = 0 or 7x – 3y = 1

Now, let us find atleast two solutions of each of the above equations, as shown in the following tables.

Table for

Now, table for

From the graph, it is clear that lines represented by the equations 3x – 5y = 19 and 7x – 3y – 1 = 0 are intersecting at a point C i.e. ( – 2, – 5).

Yes, point (4,9) lie on 3y – 7x + 1 = 0.

Question 19.

Solve the following system of linear equations graphically: 2x — 3y = 1, 3x — 4y = 1 Does the point (3, 2) lie on any of the lines? Write its equation.

Answer:

The given equation is

2x – 3y = 1

and 3x – 4y = 1

Table for or

Now, table for or

Here, the lines intersecting at point F i.e. ( – 1, – 1)

Yes, point (3,2) lie on the line 3x – 4y = 1

Question 20.

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

2x + 3y = 7, (a + b) x + (2a – b) y = 3(a + b + 1)

Answer:

Given, pair of equations

2x + 3y = 7

and (a + b)x + (2a – b)y = 3(a + b + 1)

On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get

a1 = 2, b1 = 3 and c1 = – 7

and a2 = (a + b), b2 = (2a – b) and c2 = – (a + b + 1)

For infinitely many solutions,

Here,

On taking I and II terms, we get

⇒ 2(2a – b) = 3(a + b)

⇒ 4a – 2b = 3a + 3b

⇒ 4a – 3a – 3b – 2b = 0

⇒ a – 5b = 0 …(1)

On taking I and III terms, we get

⇒ 6(a + b + 1) = 7(a + b)

⇒ 6a + 6b + 6 = 7a + 7b

⇒ 6a – 7a + 6b – 7b = – 6

⇒ – a – b = – 6

⇒ a + b = 6 …(2)

Solving eqn (1) and (2), we get

⇒ b = 1

Now, substituting the value of b in eqn (2), we get

⇒ a + b = 6

⇒ a + 1 = 6

⇒ a = 5

Question 21.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x — 2y = – 3

2x + y = 4

Answer:

The given equation is

x – 2y = – 3

and 2x + y = 4

Table for

Now, table for

Here, the lines intersecting at point C i.e. (1,2)

The points which intersect the x axis are B ( – 3,0) and E (2,0)

Question 22.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

2x + 3y = 8

x — 2y = — 3

Answer:

The given equation is

2x + 3y = 8

and x – 2y = – 3

Table for

Now, table for

Here, the lines intersecting at point C i.e. (1,2)

The points which intersect at x axis are B (4,0) and E ( – 3,0).

Question 23.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x + 2y = 5

2x — 3y = — 4

Answer:

The given equation is

x + 2y = 5

and 2x – 3y = – 4

Table for

Now, table for

Here, the lines intersecting at point C i.e. (1,2)

The points which intersect the x axis are B (5,0) and E ( – 2,0)

Question 24.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x — y + 1 = 0

4x + 3y = 24

Answer:

The given equation is

x – y + 1 = 0

and

Table for

Now, table for

Here, the lines intersecting at point C i.e. (3,4)

The points which intersect the x axis are B ( – 1,0) and E (6,0)

Question 25.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x + 2y = 1

x— 2y = 7

Answer:

The given equation is

x + 2y = 1

and x – 2y = 7

Table for

Now, table for

Here, the lines intersecting at point B i.e. (4, – 1.5)

The points which intersect the x axis are B (1,0) and E (7,0)

Question 26.

Solve the following system of equations graphically. Also find the points where the lines intersect x – axis.

x + 2y = 1

x — 2y = —7

Answer:

The given equation is

x + 2y = 1

and x – 2y = – 7

Table for

Now, table for

Here, the lines intersecting at point B i.e. ( – 3,2)

The points which intersect the x axis are B (1,0) and E ( – 7,0)

Question 27.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x — y = 4

3y — x = 3

Answer:

The given equation is

2x – y = 4

and 3y – x = 3

Table for 2x – y = 4 or y = 2x – 4

Now, table for

Here, the lines intersecting at point C, i.e. (3,2)

The point which intersects at y axis are A (0, – 4) and D (0,1)

Question 28.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x + 3y — 12 = 0

2x — y — 4 = 0

Answer:

The given equation is

2x + 3y = 12

and 2x – y – 4 = 0

Table for

Now, table for

Here, the lines intersecting at point C, i.e. (3,2)

The points which intersects at y axis is A (0,4) and D (0, – 4)

Question 29.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x — y — 5 = 0

x — y — 3 = 0

Answer:

The given equation is

2x – y = 5

and x – y = 3

Table for 2x – y = 5 or y = 2x – 5

Now, table for x – y = 3 or y = x – 3

Here, the lines intersecting at point C, i.e. (2, – 1)

The point which intersects at y axis are A (0, – 5) and D (0, – 3)

Question 30.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

2x — y — 4 = 0

x + y + 1 = 0

Answer:

The given equation is

2x – y = 4

and x + y + 1 = 0

Table for 2x – y = 4 or y = 2x – 4

Now, table for x + y + 1 = 0 or y = – (x + 1)

Here, the lines intersecting at point C, i.e. (1, – 2)

The point which intersects at y axis are A (0, – 4) and D (0, – 1)

Question 31.

Solve the following system of equations graphically. Also find the points where the lines meet the y – axis.

3x + y — 5 = 0

2x — y — 5 = 0

Answer:

The given equation is

3x + y = 5

and 2x – y = 5

Table for 3x + y = 5 or y = 5 – 3x

Now, table for 2x – y = 5 or y = 2x + 5

Here, the lines intersecting at point C i.e. (2, – 1)

The point which is intersect at y axis are A (0,5) and D (0, – 5)

Question 32.

Solve the following system of linear equations graphically.

3x + 2y — 4 = 0

2x — 3y — 7 = 0

Shade the region bounded by the lines and the x – axis.

Answer:

The given equation is

3x + 2y = 4

and 2x – 3y = 7

Table for

Now, table for

Here, the lines intersecting at a point C i.e. (2, – 1).

Question 33.

Solve the following system of linear equations graphically.

3x — 2y – 1 = 0

2x — 3y + 6 = 0

Shade the region bounded by the lines and the x – axis.

Answer:

The given equation is

3x – 2y = 1

and 2x – 3y = – 6

Table for

Now, table for

Here, the lines intersecting at a point C, i.e. (3,4).

Question 34.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and x – axis; also find the area of the shaded region.

2x + y = 6

2x — y = 0

Answer:

The given equation is

2x + y = 6

and 2x – y = 0

Table for 2x + y = 6 or y = 2x – 6

Now, table for 2x – y = 0 or y = 2x

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(),O(0,0) and B(3,0).

Area =

Question 35.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and x – axis; also find the area of the shaded region.

2x + 3y = —5

3x — 2y = 12

Answer:

The given equation is

2x + 3y = – 5

and 3x – 2y = 12

Table for

Now, table for

Here, the lines are intersecting at a point C ().

The coordinates of the vertices of are C(),B(0) and D(4,0)

()

Question 36.

Solve the following pair of linear equations graphically and shade the region bounded by these lines and x – axis; also find the area of the shaded region.

4x — 3y + 4 = 0

4x + 3y — 20 = 0

Answer:

The given equation is

4x – 3y = – 4

and 4x + 3y = 20

Table for

Now, table for