Title Lecture_06.pptx ✅

No. 41 EXERCISE 3.4 Q.2(I)(III)

Q. Form the pair of linear equations in the following problems, and find their solutions by the elimination method : If we add 1 to the numerator and subtract 1 from denominator, fraction reduces to 1. It becomes if we only add 1 to the denominator. What is the fraction ? 1 2 Soln. Let the numerator of the fraction be x and denominator of a fraction be y (y ≠ 0) ∴ Original Fraction = x y (y ≠ 0) According to the first condition, x + 1 y – 1 = 1 x + 1 = y – 1 x – y = -2 ... (i) According to the second condition, y + 1 x 1 2 = 2x = y + 1 2x – y = 1 ... (ii) x = 3 x y = 3 5 Substituting x = 3 in (i), we get y = 5 ∴ Required fraction is 3 5 What do we have to find ? Fraction = Numerator Denominator Numerator + 1 Denominator - 1 = 1 What is the condition given to us Numerator Denominator + 1 = 1 2 What is the condition given to us Solve (i) and (ii) by Elimination Method Fraction = Numerator Denominator
The sum of the digits of a two–digit number is 9. Also, nine times number is twice the number obtained by reversing the order of the digits. Find the number. Soln. Let the digit in ten’s place be x and the digit in unit’s place be y ∴Original number = + 10 x y Number obtained by interchanging the digits = + 10 y x According to the first condition, 9 x + y = ... (i) According to the second condition, + (10 x y) 2 = 9 + (10 y x) + 90 x 9 y = + 20 y 2 x 88 x - 11 y = 0 8 x y = 0 ... (ii) Substituting x = 1 in (i), 1 x = ∴ Original Number = 10 x + y + 10 (1) 8 = = + 10 8 = 18 ∴ Required two digit number is 18. – What we need to find ? Which types of number ? Two digit Number i.e. 10, 11, 12… A two digit number consists of a digit in tens place and a digit in units place A two digit number = 10 x digit in tens place + 1 x digit in units place e.g. 25 = 10 x 2 + 1 x 5 Dividing throughout by 11 Solve (i) and (ii) by any of the four Methods 8 y =