Math, asked by gautamprajeinck, 10 months ago

Solve the equation 3x+4=5x+8 and represent the solution in (i) one variable and (ii) two variables

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Answered by Anonymous
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Answer:

This question is hard

Step-by-step explanation:

Answered by giriaishik123
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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 = 0or 5(x) + (–1)y + (–50) = 0Comparing 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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