The present age of a woman is 3 years more than three times the age of her daughter. Three years later, the woman’s age will be 10 years more than twice the age of her daughter. Find their present ages.

Let the present age of daughter be x . The present age of woman = 3 x + 3 After three years, Age of daughter = 3 + x Age of woman = 3 + 3 + 3 x = 6 + 3 x ATQ, 6 + 3 x = 10 + 2 3 + x 6 + 3 x = 10 + 6 + 2 x 3 x - 2 x = 16 - 6 x = 1 0 Present age of daughter = 10 years. Present age of woman = 3 + 3 × 10 = 3 + 30 = 33 years.

Solve graphically the system of equations 2 x - 3 y - 17 = 0 , 4 x - y - 13 = 0 and shade the region between the lines and the x - axis.

Given: 2 x - 3 y - 17 = 0 4 x - y - 13 = 0 Table of value of " x " & " y " for the equation 2 x - 3 y - 17 = 0 or y = 2 x - 17 3 x 1 4 7 y - 5 - 3 - 1 Again table the value of " x " & " y " for the equation 4 x - y - 13 = 0 or x = y + 13 4 x 4 3 5 y 3 - 1 7 The graph for the equation 2 x - 3 y - 17 = 0 in the line " P Q " and the graph for the equation 4 x - y - 13 = 0 in the line " U V " We also observe that the required shaded region is A B C .

A train covered a certain distance at a uniform speed. If the train had been 10 km / hr faster, it would take 2 hours less than the scheduled time. And if the train were slower by 10 km / hr , it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Let the actual speed of the train be x km / hr and the actual time taken y hours. Distance = Speed × Time Distance covered = x y km __________ 1 If the speed is increased by ( 10 km / hr ) , then time of journey is reduced by 2 hours. When speed is ( x + 10 ) km / hr , time of journey is y - 2 hours . ∴ Distance covered = x + 10 ( y - 2 ) ⇒ x y = x + 10 y - 2 ⇒ x y = x y + 10 y - 2 x - 20 ⇒ - 2 x + 10 y - 20 = 0 ⇒ - 2 x + 3 y - 12 = 0 _________ 2 When the speed is reduced by ( 10 km / hr ) , then the time of journey is increased by 3 hours . When speed is ( x - 10 ) km / hr , time of journey is ( y + 3 ) . ∴ Distance covered = x - 10 ( y + 3 ) ⇒ x y = x - 10 y + 3 ⇒ x y = x y - 10 y + 3 x - 30 ⇒ 0 = - 10 y + 3 x - 30 ⇒ 3 x - 10 y - 30 = 0 _______________ 3 Thus, we obtain the following system of equations: - x + 5 y - 10 = 0 and 3 x - 10 y - 30 = 0 By using cross multiplication, we have, x 5 × - 30 - - 10 × - 10 = - y - 1 × - 30 - ( 3 × - 10 ) = 1 - 1 × - 10 - ( 3 × 5 ) ⇒ x - 150 - 100 = - y 30 + 30 = 1 10 - 15 ⇒ x - 250 = - y 60 = 1 - 5 ⇒ x = - 250 - 5 and y = - 60 - 5 ⇒ x = 50 and y = 12 Distance = x y km = 50 × 12 = 600 km Hence, the length of the journey is 600 km .

Solve the following pair of equation by reducing them to a pair of linear equations: 1 3 x + y + 1 3 x - y = 3 4 1 2 3 x + y - 1 2 3 x - y = - 1 8

Given, 1 3 x + y + 1 3 x - y = 3 4 ________ 1 And 1 2 3 x + y - 1 2 3 x - y = - 1 8 ⇒ 1 3 x + y - 1 3 x - y = - 1 4 ________ 2 Let a = 1 3 x + y and b = 1 3 x - y . Equation 1 ⇒ a + b = 3 4 ⇒ 4 a + 4 b = 3 _______ 3 Equation 2 ⇒ a - b = - 1 4 ⇒ 4 a - 4 b = - 1 ______ 4 Equation 3 + Equation 4 , we get, 4 a + 4 b + 4 a - 4 b = 3 - 1 ⇒ 8 a = 2 ⇒ a = 2 8 ⇒ a = 1 4 Substituting a = 1 4 in equation 3 , we get, 4 × 1 4 + 4 b = 3 ⇒ 1 + 4 b = 3 ⇒ b = 2 4 ⇒ b = 1 2 Therefore, 1 3 x + y = 1 4 ⇒ 3 x + y = 4 _________ 5 And 1 3 x - y = 1 2 ⇒ 3 x - y = 2 __________ 6 Equation 5 + equation 6 , we get, ⇒ 3 x + y + 3 x - y = 2 + 4 ⇒ 6 x = 6 ⇒ x = 1 Substituting x = 1 in equation 5 , we get, 3 × 1 - y = 4 ⇒ y = - 4 + 3 ⇒ y = - 1 Therefore, x = 1 and y = - 1

Solve the system of equations 2 x + 3 y = 17 and 3 x - 2 y = 6 using cross-multiplication method.

Given, 2 x + 3 y = 17 ⇒ 2 x + 3 y - 17 = 0 _____ 1 And 3 x - 2 y = 6 ⇒ 3 x - 2 y - 6 = 0 ______ 2 For system of linear equations a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 , the solution is given by x b 1 c 2 - b 2 c 1 = y c 1 a 2 - c 2 a 2 = 1 a 1 b 2 - a 2 b 1 . Here, a 1 = 2 , b 1 = 3 , c 1 = - 17 , a 2 = 3 , b 2 = - 2 and c 2 = - 6 ∴ x 3 - 6 - - 2 - 17 = y - 17 3 - - 6 2 = 1 2 - 2 - 3 3 ⇒ x - 18 - 34 = y - 51 + 12 = 1 - 4 - 9 ⇒ x - 52 = y - 39 = 1 - 13 ⇒ x - 52 = 1 - 13 and y - 39 = 1 - 13 ⇒ x = 4 and y = 3 Therefore, the required solution is 4 , 3 .