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**Question 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 â‚¹ x and that of a pen to be â‚¹ y).Solution:** Let the cost of a notebook =

The cost of a pen =

According to the condition,

we have [Cost of a notebook] = 2 x [Cost of a pen]

i.e. [x] = 2 x [y]

or

x = 2y

or

x â€“ 2y = 0

Thus, the required linear equation is x â€“ 2y = 0.

**Question 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) 2x + 3y = ****(ii) x â€“ (y/5) â€“ 10 = 0 ****(iii) â€“2x + 3y = 6 ****(iv) x = 3y****(v) 2x = â€“5y ****(vi) 3x + 2 = 0 ****(vii) y â€“ 2 = 0 ****(viii) 5 = 2x****Solution:** (i) We have 2x + 3y =

âˆ´ (2)x + (3)y + (â€“) = 0

Comparing it with ax + bx + c = 0,

we have a = 2, b = 3 and c = â€“.

(ii) We have x â€“(y/5) â€“ 10 = 0

or

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

**Note:** Above equation can also be compared by:

Multiplying throughout by 5,

or 5x â€“ y â€“ 50 = 0

or 5(x) + (â€“1)y + (â€“50) = 0

Comparing with ax + by + c = 0,

we get a = 5, b = â€“1 and c = â€“50.

(iii) We have â€“2x + 3y = 6

â‡’ â€“2x + 3y â€“ 6 = 0

â‡’ (â€“2)x + (3)y + (â€“6) = 0

Comparing with ax + bx + c = 0,

we get a = â€“2, b = 3 and c = â€“6.

(iv) We have x = 3y

â‡’ x â€“ 3y = 0

â‡’ (1)x + (â€“3)y + 0 = 0

Comparing with ax + bx + c = 0,

we get a = 1, b = â€“3 and c = 0.

(v) We have 2x = â€“5y

â‡’ 2x + 5y = 0

â‡’ (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

â‡’ 3x + 2 + 0y = 0

â‡’ (3)x + (0)y + (2) = 0

Comparing with ax + by + c = 0,

we get a = 3, b = 0 and c = 2.

(vii) We have y â€“ 2 = 0

â‡’ (0)x + (1)y + (â€“2) = 0

Comparing with ax + by + c = 0,

we have a = 0, b = 1 and c = â€“2.

(viii) We have 5 = 2x

â‡’ 5 â€“ 2x = 0

â‡’ â€“2x + 0y + 5 = 0

â‡’ (â€“2)x + (0)y + (5) = 0

Comparing with ax + by + c = 0,

we get a = â€“2, b = 0 and c = 5.

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