RS Aggarwal Solutions Class 10 Ex 3C

Question 1: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

5x – y = 7, x – y + 1 = 0

Solution:

Given equations are 5x – y = 7 and x – y + 1 = 0

Graph of 5x – y = 7:

5x – y = 7 \( \Rightarrow y = 5x – 7 \)

Putting x = 0, we get y = -7

Putting x = 1, we get y = -2

Putting x = 2, we get y = 3

Hence, the table is,

x 0 1 2
y -7 -2 3

Plot the points A(0, 7), B(1, -2) and C(2, 3) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 5x – y = 7.

Graph of x – y + 1 = 0:

x – y + 1 = 0 \( \Rightarrow y =x+ 1 \)

Putting x = 0, we get y = 1

Putting x = 1, we get y = 2

Putting x = 2, we get y = 3

Hence, the table is,

x 0 1 2
y 1 2 3

Now, on the same graph paper as above plot the points P(0, 1) and Q(1, 2). The point C(2, 3) has already been plotted. Join PA.

https://lh4.googleusercontent.com/xK_dGWR7jX_HjegNF2LW2uNOMnJ2WV7RtZHcCqI0Wj0Y2TnbUpevz30pWUGpwFHllxgwqFPqub0YPuxCM-ZjdKvprpABJd6HNb_qZMsM9H-4LeOnsTwLa0_OlDBG-QwQuoE3aENn

Thus, line PA is the graph of 3x + 2y – 12 = 0. The two graph lines intersect at C(2, 3).

Hence, x = 2, y = 3 is the solution of the given system of equations.

Clearly, the vertices of \( \Delta APC \) formed by these lines and the y-axis are A(0, 7), P(0, 1) and C(2, 3).

Consider the triangle \( \Delta APC \) :

Height of the triangle = 2 units and base(AP) = 8 units

Area of the triangle \( \Delta APC \):

\( Area = \frac{1}{2} \times Base \times height = \frac{1}{2} \times 8 \times 2 \) sq. units

Area of \( \Delta APC \) = 8 sq. units

Question 2: On a graph paper, draw a horizontal line X’OX and a vertical line YoY’ as the x-axis and the y-axis respectively.

x – 2y = 2, 4x – 2y = 5

Solution:

Given equations are x – 2y = 2 and 4x – 2y = 5

Graph of x – 2y = 2:

x – 2y = 2 \( \Rightarrow y = \frac{x – 2}{2} \)

Putting x = 0, we get y = -1

Putting x = 1, we get y = -0.5

Putting x = 2, we get y = 0

Hence, the table is,

x 0 1 2
y -1 -0.5 0

Plot the points A(0, -1), B(2, 0) and C(1, -0.5) on the graph paper. Join AC and AB to get the graph line AB. Extend it both ways.

Thus, line AB is the graph of x – 2y = 2.

Graph of 4x – 2y = 5:

4x – 2y = 5 \( \Rightarrow y = \frac{4x – 5}{2} \)

Putting x = 0, we get y = -2.5

Putting x = 1, we get y = -0.5

Putting x = 2, we get y = -1.5

Hence, the table is,

x 0 1 2
y -2.5 -0.5 -1.5

Now, on the same graph paper as above plot the points P(0, -0.25) and Q(2, -1.5). The point C(1, -0.5) has already been plotted. Join PC and CQ to get PQ.

https://lh4.googleusercontent.com/PP2Edl-jVzsNyv88tIugNuni6Bxgz-fjgRgys86YZtWCjk-COr1mIoPjhQSYBHbA0HjqS7Y3aWu9QGZeGgSiokY0ENAobqUD4IjniPcV3v1wTfudbIFOpAuh_kYqJkDSVnyUx9cp

Thus, line PA is the graph of 4x – 2y = 5. The two graph lines intersect at (1, -0.5).

The given system of equations is consistent.

Question 3: On a graph paper, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and the y-axis respectively.

2x + 3y = 4, 4x + 6y = 12

Solution:

Given equations are 2x + 3y = 4 and 4x + 6y = 12

Graph of 2x + 3y = 4:

2x + 3y = 4 \( \Rightarrow y = \frac{-2x + 4}{3} \)

Putting x = 2, we get y = 2

Putting x = -1, we get y = 2

Putting x = -4, we get y = 4

Hence, the table is,

x 2 -1 -4
y 0 2 4

Plot the points A(2, 0), B(-1, 2) and C(-4, 4) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 2x + 3y = 4.

Graph of 4x + 6y = 12:

4x + 6y = 12 \( \Rightarrow y = \frac{-4x + 12}{6} \)

Putting x = 3, we get y = 0

Putting x = 0, we get y = 2

Putting x = 6, we get y = -2

Hence, the table is,

x 3 0 6
y 0 2 -2

Now, on the same graph paper as above plot the points P(3, 0) and Q(0, 2) and R(6, -2). Join PQ and PR to get QR.

https://lh4.googleusercontent.com/e-DZcu0dnqolu3Sb4eSgFkvUtqovXigFcJs7lBpa8M9JVPBObIvHr0AB8hpQeUMYeszDOAFhWDMZV7PdpFJ7A2kPA3zNzDBf7dpjKnLXw9o-d5K49BHNaBTQGicuH8_vYX1wA6N7

Thus, line QR is the graph of 4x + 6y = 12.

It is not clear from the graph that two graph lines are parallel and do not intersect when produced.

Hence, the given system of equation is consistent.

Question 4: On a graph paper, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and the y-axis respectively.

2y – x = 9, 4y – 2x = 20

Solution:

Given equations are 2y – x = 9 and 4y – 2x = 20

Graph of 2y – x = 9:

2y – x = 9 \( \Rightarrow y = \frac{x + 9}{2} \)

Putting x = 1, we get y = 5

Putting x = -1, we get y = 4

Putting x = -3, we get y = 3

Hence, the table is,

x 1 -1 -3
y 5 4 3

Plot the points A(1, 5), B(-1, 4) and C(-3, 3) on the graph paper. Join AB and BC to get the graph line AC. Extend it both ways.

Thus, line AC is the graph of 2y – x = 9.

Graph of 4y – 2x = 20:

4y – 2x = 20 \( \Rightarrow y = \frac{2x + 20}{4} \)

Putting x = 0, we get y = 5

Putting x = 2, we get y = 6

Putting x = -2, we get y = 4

Hence, the table is,

x 0 2 -2
y 5 6 4

Now, on the same graph paper as above plot the points P(0, 5) and Q(2, 6) and R(-2, 4). Join PQ and PR to get QR.

https://lh4.googleusercontent.com/L9zCguBKhv6V4vbF8Dk15mhcFcynMGZc6YwBcAQEofWeKCvl4PIx8z6HUiDaKYJ9iaDCtkFWO4gE3aOmP8truwIG-vQUKuh0tpThZPOhRcVWXkrcV0ilpdT8hXE_X9mASsY6fZ84

Thus, line QR is the graph of 4y – 2x = 20.

It is not clear from the graph that two graph lines are parallel and do not intersect when produced.

Hence, the given system of equation is consistent.

Question 5: On a graph paper, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and the y-axis respectively.

3x – y = 5, 6x – 2y = 10

Solution:

Given equations are 3x – y = 5 and 6x – 2y = 10

Graph of 3x – y = 5:

3x – y = 5 \( \Rightarrow y = 3x – 5 \)

Putting x = 1, we get y = -2

Putting x = 0, we get y = -5

Putting x = 2, we get y = 1

Hence, the table is:

x 1 0 2
y -2 -5 1

Plot the points A(1, -2), B(0, -5) and C(2, 1) on the graph paper. Join AB and AC to get the graph line BC. Extend it both ways.

Thus, line BC is the graph of 3x – y = 5.

Graph of 6x – 2y = 10:

6x – 2y = 10 \( \Rightarrow y = \frac{6x – 10}{2} \)

Putting x = 0, we get y = -5

Putting x = 1, we get y = -2

Putting x = 2, we get y = 1

Hence, the table is:

x 0 1 2
y -5 -2 1

These points are same as obtained above:

https://lh3.googleusercontent.com/cnEIpd2bxqCIee1lJGar29P4Mg8SvXdecsroFcOp9lf-7yzNhWG7Gj9S7m311oSDQyHehLoscGAyfCbwLxSuy7RU0IoVv9oDWaIk9gqPOkvden796AFfEiQCJAFOpw_JsD3vj4lh

From the graph, it is clear that these two lines coincide. Both equations represent same graph. Hence, these lines have infinitely many solutions.

Question 6: On a graph paper, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and the y-axis respectively.

2x + y = 6, 6x + 3y = 18

Solution:

Given equations are 2x + y = 6 and 6x + 3y = 18

Graph of 2x + y = 6:

2x + y = 6 \( \Rightarrow y = -2x + 6 \)

Putting x = 3, we get y = 0

Putting x = 1, we get y = 4

Putting x = 2, we get y = 2

Hence, the table is,

x 3 1 2
y 0 4 2

Plot the points A(3, 0), B(1, 4) and C(2, 2) on the graph paper. Join AC and BC to get the graph line of equation 2x + y = 6.

Graph of 6x + 3y = 18:

6x + 3y = 18 \( \Rightarrow y = \frac{-6x + 18}{3} \)

Putting x = 0, we get y = -5

Putting x = 1, we get y = -2

Putting x = 2, we get y = 1

Hence, the table is:

x 3 1 2
y 0 4 2

These points are same as obtained above:

https://lh5.googleusercontent.com/v1y67If15LvahdaymXLoutjSot-39f4qyg1pKt6_ispYY1cKF0YRxTaZGKawdGMT8L6p2SlsC8AE3sgoeGeh9eWjKPVhQj3y47EJMXj_PLcSVRT16mXidf6blyoFGVs5CjJFHvMR

Thus, we find that the two line graphs coincide. Hence, the given system of equations has infinitely many solutions.

Solve for x and y

Question 7: x + y = 8, 2x – 3y = 1

Solution:

The given equations are.

x + y = 8 …..(1)

2x – 3y = 1 …..(2)

Multiplying (1) by 3 and (2) by 1, we get,

3x + 3y = 24 …..(3)

2x – 3y = 1 …..(4)

Adding (3) and (4), we get

\( 5x \, = \, 25 \Rightarrow x \, = \, \frac{25}{5} \Rightarrow x \, = \, 5 \)

Substituting x = 5 in (1), we get

5 + y = 8 \( \Rightarrow \) y = 8 – 5 = 3

∴ x = 5 and y = 3

Question 8: x + y = 3, 4x – 3y = 26

Solution:

The given equations are.

x + y = 3 …..(1)

4x – 3y = 26 …..(2)

Multiplying (1) by 3 and (2) by 1, we get,

3x + 3y = 9 …..(3)

4x – 3y = 26 …..(4)

Adding (3) and (4), we get

7x = 35 \( \Rightarrow \) x = 5

Substituting x = 5 in (1), we get

5 + y = 3 \( \Rightarrow \) y = 3 – 5 = -2

∴ x = 5 and y = -2

Question 9: x- y = 3, 3x – 2y

Solution:

The given equations are.

x- y = 3 …..(1)

3x – 2y = 10 …..(2)

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

x – y = 3 …..(3)

3x – 2y = 10 …..(4)

Subtracting (3) from (4), we get

x = 4

Substituting x = 4 in (1), we get

4 – y = 3 \( \Rightarrow \) y = 4 – 3 = 1

∴ x = 4 and y = 3

Question 10: 2x + 3y = 0, 3x + 4y = 5

Solution:

The given equations are.

2x + 3y = 0 …..(1)

3x + 4y = 5 …..(2)

Multiplying (1) by 4 and (2) by 3, we get,

8x + 12y = 0 …..(3)

9x + 12y = 15 …..(4)

Subtracting (3) from (4), we get

x = 15

Substituting x = 15 in (1), we get

(2)(15) + 3y = 0 \( \Rightarrow \) 3y = 0 – 30

3y = -30 or y = -10

∴ x = 15 and y = -10

Question 11: 2x – 3y = 13, 7x – 2y = 20

Solution:

The given equations are.

2x – 3y = 13 …..(1)

7x – 2y = 20 …..(2)

Multiplying (1) by 2 and (2) by 3, we get,

4x – 6y = 26 …..(3)

21x – 6y = 60 …..(4)

Subtracting (3) from (4), we get

17x = 34 \( \Rightarrow \) x = 2

Substituting x = 2 in (1), we get

(2)(2) – 3y = 13 \( \Rightarrow \) 4 – 3y = 13

-3y = 13 – 4 or -3y = 9 or y = -3

∴ x = 2 and y = -3


Practise This Question

If white ray falls on the system in the give way then how many colours will we see after all dispersion has happened?