The taxi fare in the city is as follows: for the first kilometre the fare is rupees 50 and for subsequent distance it is rupees 15 per kilometre. Taking the distance covered as x kilometre and Total fare as Rupees y. Write a linear equation for this information. plz help fast in 5 min plz plz on copy
Answers
Answer:
The taxi fare in the city is as follows: for the first kilometre the fare is rupees 50 and for subsequent distance it is rupees 15 per kilometre. Taking the distance covered as x kilometre and Total fare as Rupees y. Write a linear equation for this information. plz help fast in 5 min plz plz on copy
Answer: 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 = 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.
EXERCISE: 4.2
1. Which one of the following options is true, and why? y = 3x + 5 has
(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutionsSol: Option (iii) is true because a linear equation has an infinitely many solutions.
2. Write four solutions for each of the following equations:
(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4ySol: (i) 2x + y = 7 When x = 0, 2(0) + y = 7 ⇒ 0 + y = 7 ⇒ y =7 ∴ Solution is (0, 7). When x = 1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5 ∴ Solution is (1, 5). When x = 2, 2(2) + y = 7 ⇒ y = 7 – 4 ⇒ y = 3 ∴ Solution is (2, 3). When x = 3, 2(3) + y = 7 ⇒ y = 7 – 6 ⇒ y = 1 ∴ Solution is (3, 1).(ii) πx + y = 9 When x = 0 π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9 ∴ Solution is (0, 9). When × = 1, π(1) + y = 9 ⇒ y = 9 – π ∴ Solution is {1, (9 – π)} When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π ∴ Solution is {2, (9 – 2π)} When × = –1, π(–1) + y = 9 ⇒ – π + y = 9 ⇒ y = 9 + π ∴ Solution is {–1, (9 + π)}(iii) x = 4y When x = 0, 4y = 0 ⇒ y = 0 ∴ Solution is (0, 0). When x = 1, 4y = 1 ⇒ y = 0 ∴ Solution is (0, 0) When x = 4, 4y = 4 ⇒
3. Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) (v) (1, 1)Sol: (i) (0, 2) means x = 0 and y = 2 Putting x = 0 and y = 2 in x – 2y = 4, we have L.H.S. = 0 – 2(2) = –4 But R.H.S. = 4 L.H.S. ≠ R.H.S. ∴ x = 0, y = 0 is not a solution.(ii) (2, 0) means x = 2 and y = 0 ∴ Putting x = 2 and y = 0 in x – 2y = 4, we get L.H.S. = 2 – 2(0) = 2 – 0 = 2 But R.H.S. = 4 L.H.S. ≠ R.H.S. ∴ (2, 0) is not a solution.(iii) (4, 0) means x = 4 and y = 0 Putting x = 4 and y = 0 in x – 2y = 4, we get L.H.S. = 4 – 2(0) = 4 – 0 = 4 But R.H.S. = 4 L.H.S. ≠ R.H.S. ∴ (4, 0) is a solution.(v) (1, 1) means x = 1 and y = 1 Putting x = 1 and y = 1 in x – 2y = 4, we get L.H.S. = 1 – 2(1) = 1 – 2 = –1 But R.H.S. = 4 ⇒ L.H.S ≠ R.H.S. ∴ (1, 1) is not a solution.