Question

1. If 2 2 A x xy y    7 5 9 , 2 2 B x xy y     4 5 and 2 2 C y x xy    4 3 6 then show that A+B+C=0. 2. a) Subtract the sum of 5 4 6 x y z   and    8 2 x y z from then sum of 12 3 x y z   and    3 5 8 . x y z
OR b) Simplify: 5 4 7 3 2 4 3 3 2 x y x z y z x y z         

Answer 1

Step-by-step explanation:

Given,

2

x+y−8

=

3

x+2y−14

=

4

3x−y

By taking first two. we get,

2

x+y−8

Substitute y=6 in (i).

x−6=−4

∴x=2

∴x=2,y=6

Answer 2

Answer:

EXERCISE: 4.1

1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

(Take the cost of a notebook to be Rs. x and that of a pen to be Rs. y).

Sol: Let the cost of a notebook = Rs x

The cost of a pen = y

According to the condition, we have

[Cost of a notebook] = 2 × [Cost of a pen]

i.e. [x] = 2 × [Y]

or x = 2y

or x – 2y = 0

Thus, the required linear equation is × – 2y = 0.

2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) (ii) (iii) –2x + 3y = 6 (iv) x = 3y

(v) 2x = –5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x

Sol: (i) We have

Comparing it with ax + bx + c = 0, we have a = 2, b = 3 and

(ii) We have

Comparing with ax + bx + c = 0, we get

Note: Above equation can also be compared by:

Multiplying throughout by 5,

or 5x – y – 50 = 0

or 5(x) + (–1)y + (–50) = 0

Comparing with ax + by + c = 0, we get a = 5, b = –1 and c = –50.

(iii) We have –2x + 3y = 6

⇒ –2x + 3y – 6 = 0

⇒ (–2)x + (3)y + (–6) = 0

Comparing with ax + bx + c = 0, we get a = –2, b = 3 and c = –6.

(iv) We have x = 3y

x – 3y = 0

(1)x + (–3)y + 0 = 0

Comparing with ax + bx + c = 0, we get a = 1, b = –3 and c = 0.

(v) We have 2x = –5y

⇒ 2x + 5y =0

⇒ (2)x + (5)y + 0 = 0

Comparing with ax + by + c = 0, we get a = 2, b = 5 and c = 0.

(vi) We have 3x + 2 = 0

⇒ 3x + 2 + 0y = 0

⇒ (3)x + (10)y + (2) = 0

Comparing with ax + by + c = 0, we get a = 3, b = 0 and c = 2.

(vii) We have y – 2 = 0

⇒ (0)x + (1)y + (–2) = 0

Comparing with ax + by + c = 0, we have a = 0, b = 1 and c = –2.

(viii) We have 5 = 2x

⇒ 5 – 2x = 0

⇒ –2x + 0y + 5 = 0

⇒ (–2)x + (0)y + (5) = 0

Comparing with ax + by + c = 0, we get a = –2, b = 0 and c = 5.