## Draw the graph of the following equations. Read two more solutions from the graph in each case. Also, find the coordinates of the points whe

Question

Draw the graph of the following equations. Read two more solutions from the graph in each case. Also, find the coordinates of the points where the line intersects the two axes:- (a) x – 3y = 6 (b) 2x – 5y = 10 (c) 3x + 4y = 12. Please draw with graph

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2021-09-07T03:28:38+00:00
2021-09-07T03:28:38+00:00 2 Answers
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Answer:## Answer: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 ……….

Answer: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.