Math, asked by deepika113366, 7 months ago

plz answer the ques

find a when (1 , - 1) is a solution of linear equation 2x - (2a + 5) Y = 5 .hence draw the graph .​

Answers

Answered by yuvraj0965
0

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.

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