Ex 3.5, 4 (v) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables (Term 1)

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Ex 3.5,4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle. Let Length of rectangle be x units & Breadth of rectangle be y units Hence, Area = Length × Breadth Area = xy Given that, Area gets reduced by 9 square units, If length is reduced by 5 units and breadth increased by 3 units So, New Area = New Length × New Breadth Old Area − 9 = (Length − 5) × (Breadth + 3) xy – 9 = (x – 5) (y + 3) xy – 9 = x (y + 3) – 5 (y + 3) xy – 9 = xy + 3x – 5y − 15 0 = xy + 3x – 5y − 15 − xy + 9 3x – 5y – 6 = 0 3x − 5y = 6 Also, Area increases by 67 square units, If length is increased by 3 units and breadth increased by 2 units So, New Area = New Length × New Breadth Old Area + 67 = (Length + 3) × (Breadth + 2) xy + 67 = (x + 3) (y + 2) xy + 67 = x(y + 2) + 3(y + 2) xy + 67 = xy + 2x + 3y + 6 0 = xy + 2x + 3y + 6 − xy − 67 2x + 3y − 61 = 0 2x + 3y = 61 Hence, our equations are 3x – 5y = 6 …(1) 2x + 3y = 61 …(2) From (1) 3x – 5y – 6 = 0 3x = 6 + 5y x = (𝟔 + 𝟓𝒚)/𝟑 Putting value of x in (2) 2x + 3y = 61 2 ((6 + 5𝑦)/3) + 3y = 61 Multiplying both sides by 3 3 × 2 (((6 + 5𝑦))/3) + 3 × 3y = 3 × 61 2(6 + 5y) + 9y = 183 12 + 10y + 9y = 183 19y = 183 − 12 19y = 171 y = 171/19 y = 9 Putting y = 9 in equation (1) 3x – 5y = 6 3x – 5(9) = 6 3x – 45 = 6 3x = 6 + 45 3x = 51 x = 51/3 x = 17 Therefore x = 17, y = 9 is the solution Hence, Length of rectangle = x = 17 units Breadth of rectangle = y = 9 units