Chapter 3: Pair of Linear Equations in Two Variables

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 triangles ACE and BDE, respectively.

Let the area of triangle formed with x-axis = T₁

T₁ = Area of △ACE = ½ × AC × PE

T₁ = ½ × 4 × 4 = 8

And area of triangle formed with y-axis = T₂

T₁ = Area of △BDE = ½ × BD × QE

T₁ = ½ × 4 × 1 = 2

T₁:T₂ = 8:2 = 4:1

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)

Table for line y = x,

Table for line x = 3y,

Table for line x + y = 8

Plotting the points A (1, 1), B(2,2), C (3, 1), D (6, 2), we get the straight lines AB and CD.

Similarly, plotting the points P (0, 8), Q(4, 4) and R(8, 0), we get the straight line PQR.

AB and CD intersect the line PR on Q and D, respectively.

So, △OQD is formed by these lines. Hence, the vertices of the △OQD 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.

Table for line 2x – y – 4 = 0

Plotting the graph, we get,

From the graph, we get,

AB = OB-OA = 5-3 = 2

AD = 2

BC = 6

Thus, quadrilateral ABCD is a trapezium, then,

Area of quadrilateral ABCD = ½ × (distance between parallel lines) = ½ × (AB) × (AD + BC)

= 8 sq units

Let the cost of a pen and a pencil box be Rs x and Rs y, respectively.

According to the question,

4x + 4y = 100

Or x + y = 25 …(i)

3x = y + 15

Or 3x-y = 15 …(ii)

On adding Equation (i) and (ii), we get,

4x = 40

So, x = 10

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

y = 25-10 = 15

Hence, the cost of a pen = Rs. 10

The cost of a pencil box = Rs. 15

3x – y = 2 …(i)

2x -3y = 2 …(ii)

x + 2y = 8 …(iii)

Let the equations of the line (i), (ii) and (iii) represent the side of a ∆ABC.

On solving (i) and (ii),

We get,

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

7x = 7

x = 1

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

3×1-y = 3

y = 0

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

On solving lines (ii) and (iii),

We get,

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

7y = 14

y = 2

Substituting y=2 in Eq. (iii), we get

x + 2 × 2 = 8

x + 4 = 8

x = 4

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

On solving lines (iii) and (i),

We get,

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

7x = 14

x = 2

Substituting x=2 in Eq. (i), we get

3 × 2 – y = 3

y = 3

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

Hence, the vertices of the ∆ABC formed by the given lines are as follows,

A (2, 3), B (1, 0) and C (4, 2).

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

Now, she has taken time to travel 2 km by rickshaw, t₁ = (2/x) hr

Speed = distance/ time

She has taken time to travel the remaining distance i.e., (14 – 2) = 12km

By bus t₂ = (12/y) hr

By the first condition,

t₁ + t₂ = ½ = (2/x) + (12/y) … (i)

Now, she has taken time to travel 4 km by rickshaw, t₃ = (4/x) hr

and she has taken time to travel the remaining distance i.e., (14 – 4) = 10km, by bus = t₄ = (10/y) hr

By the second condition,

t₃ + t₄ = ½ + 9/60 = ½ + 3/20

(4/x) + (10/y) = (13/20) …(ii)

Let (1/x) = u and (1/y) = v

Then equations (i) and (ii) become

2u + 12v = ½ …(iii)

4u + 10v = 13/20…(iv)

(4u + 24v) – (4u + 10v) = 1–13/20

14v = 7/20

v = 1/40

Substituting the value of v in Eq. (iii),

2u + 12(1/40) = ½

2u = 2/10

u = 1/10

x = 1/u = 10km/hr

y = 1/v = 40km/hr

Hence, the speed of rickshaw = 10 km/h

And the speed of the bus = 40 km/h.

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