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Ex 3.4, 2 - Form linear equation - If we add 1 to numerator - Elimination

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
  2. Serial order wise
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Ex 3.4, 2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction? Let numerator be x and denominator be y So, fraction is π‘₯/𝑦 Given that, if 1 is added to numerator and 1 is subtracted from the denominator, fraction becomes 1. (π‘π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ + 1)/(π·π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ βˆ’1)=1 (π‘₯ + 1)/(𝑦 βˆ’ 1)=1 (x + 1) = (y – 1) x – y = – 1 – 1 x – y = – 2 Also, if we add 1 to the denominator, fraction becomes 1/2. π‘π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ/(π·π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ + 1)=1/2 π‘₯/(𝑦 +1)=1/2 2x = y + 1 2x – y = 1 Hence, our equations are x – y = –2 …(1) 2x – y = 1 …(2) We use elimination method with equation (1) & (2) –x = –3 x = 3 Putting x = 3 in equation (1) x – y = –2 3 – y = –2 –y = – 2 – 3 –y = –5 y = 5 So, x = 3, y = 5 is the solution of our equation So, x = 3, y = 5 is the solution of our equation ∴ Numerator = x = 3 & Denominator = y = 5 Hence, original fraction = π‘π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ/π·π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ "= " π‘₯/𝑦=3/5 Ex 3.4,2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? Let present age of Nuri = x years & present age of Sonu = y years Five years ago , Nuri’s age = x – 5 years Sonu’s age = y – 5 years Given that Nuri was thrice as old as Sonu x – 5 = 3(y – 5) x – 5 = 3y – 15 x – 3y = – 10 Ten years later, Nuri’s age = x + 10 years Sonu’s age = y + 10 years Given that Nuri will be twice as old as sonu x + 10 = 2(y + 10) x + 10 = 2y + 2(10) x + 10 = 2y + 20 x – 2y = 20 – 10 x – 2y = 10 Hence, our equations are x – 3y = – 10 …(1) x – 2y = 10 …(2) Using elimination method with equation (1) & (2) –y = – 20 y = 20 Putting y = 20 in (1) x – 3y = –10 x – 3 (20) = –10 x – 60 = –10 x = –10 + 60 x = 50 So, the present age of Nuri = x = 50 years & the present age of Sonu = y = 20 years Ex 3.4 ,2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Number is of the form Let digit at units place = y and the digit at tens place = x Given that sum of digits of two digit number is 9 ∴ Digit at tens place + Digit at units place = 9 x + y = 9 It is given 9 times the number is twice the number obtained by reversing the digits 9 Γ— (Number) = 2 Γ— (Reversed number) 9 (10x + y) = 2 (10y + x) 90x + 9y = 20 y + 2x 90x – 2x + 9y – 20y = 0 88x – 11 y = 0 11(8x – y) = 0 (8x – y) = 0/11 (8x – y) = 0 8x – y = 0 Hence, our equations are x + y = 9 …(1) 8x – y = 0 …(2) We use elimination method with equation (1) & (2) 9x = 9 x = 9/9 ∴ x = 1 Putting x = 1 in (1) x + y = 9 1 + y = 9 y = 9 – 1 y = 8 So, x = 1, y = 8 is the solution of our equation ∴ Digit at units place = y = 8 & digit at tens place = x = 1 Therefore, Number = 10x + y = 10(1) + 8 = 10 + 8 = 18 So, the required number is 18 Ex 3.4 ,2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method : (iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received. Let the number of Rs. 50 notes = x And the number of Rs. 100 notes = y It is given that total notes is 25 So, (Number of Rs 50 notes) + (Number of Rs 100 notes) = 25 x + y = 25 Also, given that Total amount withdrawn = 2000 Money withdrawn with Rs 50 note + Money withdrawn with Rs 100 notes 50 Γ— (Number of Rs 50 notes ) + 100 Γ— (Number of Rs 100 notes ) 50 (x) + 100 (y) = 2000 50 (x + 2y) = 2000 x + 2y = 2000/50 x + 2y = 40 Hence, our equations are x + y = 25 …(1) x + 2y = 40 …(2) We use elimination method with equation (1) & (2) βˆ’ y = βˆ’ 15 y = 15 Putting y = 15 in equation (1) x + y = 25 x + 15 = 25 x = 25 – 15 x = 10 So, x = 10 & y = 15 are the solution of our equations Therefore number of Rs 50 notes = x = 10 And the number of Rs 100 notes = y = 15 Ex 3.4 ,2 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. Let the fixed charge for first three days = Rs x & Additional charge after three days = Rs y per day Given that Saritha paid Rs 27 for a book kept for seven days Fixed charge for first three days + (Number of additional days kept)Γ—(Additional charges per day) " x + (7 βˆ’ 3) y = 27" " x + 4y = 27" Also, "Susy paid Rs 21 for a book kept for five days" Fixed charge for first three days + (Number of additional days kept) Γ— (Additional charges per day) x + (5 – 3) y = 21 x + 2y = 21 We use elimination method with equation (1) & (2) We use elimination method with equation (1) & (2) 2y = 6 y = 6/2 y = 3 Putting y = 3 in equation (2) x + 2y = 21 x + 2(3) = 21 x + 6 = 21 x = 21 - 6 x = 15 Hence, x = 15, y = 3 is the solution of the equations Hence, x = 15, y = 3 is the solution of the equations Hence, Fixed charge for the first 3 days = x = Rs 15 Additional charges after 3 days = y = Rs 3 per day

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