Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10     1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
2. Serial order wise
3. Ex 3.5

Transcript

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 And 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 and Area increase by 67 square units, if length is increased by 3 units and breadth increased by 2 units (x – 5) (y + 3) = xy – 9 x (y + 3) – 5 (y + 3) = xy – 9 x (y + 3) – 5 (y + 3) = xy – 9 xy + 3x – 5y - 15 = xy – 9 xy + 3x – 5y - 15 - xy + 9 = 0 3x – 5y – 6 = 0 (x + 3) (y + 2) = xy + 67 x(y + 2) + 3(y + 2) = xy + 67 xy + 2x + 3y + 6 = xy + 67 xy + 2x + 3y + 6 - xy – 67 = 0 2x + 3y – 61 = 0 Hence, our equations are 3x – 5y – 6 = 0 …(1) 2x + 3y – 61 = 0 …(2) From (1) 3x – 5y – 6 = 0 3x = 6 + 5y x = (6 + 5𝑦)/3 Putting value of x in (2) 2x + 3y – 61 = 0 2((6 + 5𝑦)/3) + 3y – 61 = 0 Multiplying both sides by 3 3 × 2(((6+5𝑦))/3) + 3 × 3y – 3 × 61 = 0 2(6 + 5y) + 9y – 183 = 0 12 + 10y + 9y - 183 = 0 19y – 171 = 0 19y = 171 y = 171/19 y = 9 Putting y = 9 in equation (1) 3x – 5y – 6 = 0 3x – 5(9) – 6 = 0 3x – 45 – 6 = 0 3x – 51 = 0 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

Ex 3.5 