A fraction becomes 1 4 when 2 is subtracted from its numerator and 3 is added to its denominator and on adding 6 to numerator and multiplying denominator by 3 it becomes 2 3 . Find the fraction.

Let the numerator = x and the denominator = y So, the fraction = x y According to the question, Condition 1 : x - 2 y + 3 = 1 4 ⇒ 4 x - 2 = y + 3 ⇒ 4 x - 8 = y + 3 ⇒ 4 x - y = 11 ⇒ y = 4 x - 11 . . . . . i Condition 2 : x + 6 3 y = 2 3 ⇒ 3 x + 18 = 6 y ⇒ 3 x + 18 = 6 4 x - 11 ⇒ 3 x + 18 = 24 x - 66 ⇒ 21 x = 84 ⇒ x = 4 From equation i , y = 4 × 4 - 11 = 5 So, the numerator is 4 and the denominator is 5 . Hence, the fraction is 4 5 .

If in △ A B C , ∠ C = 2 ∠ B = ∠ A + ∠ B + 20 , then find all the three angles of the triangle.

Given ∠ C = 2 ∠ B = ∠ A + ∠ B + 20 ⇒ 2 ∠ B = ∠ A + ∠ B + 20 ⇒ ∠ B = ∠ A + 20 . . . . . i We know that ∠ A + ∠ B + ∠ C = 180 ° (Sum of the interior angles of a triangle = 180 ° ) ∠ A + ∠ B + 2 ∠ B = 180 ° (From the given statement) ⇒ ∠ A + 3 ∠ B = 180 ° ⇒ ∠ A + 3 ∠ A + 20 = 180 ° ⇒ ∠ A + 3 ∠ A + 60 = 180 ° ⇒ 4 ∠ A = 120 ° ⇒ ∠ A = 30 ° From equation i , ∠ B = 50 ° And, ∠ C = 2 ∠ B = 2 × 50 ° = 100 ° ∴ ∠ A = 30 ° , ∠ B = 50 ° , ∠ C = 100 ° .

Solve the following system of equation by substitution method: x + y = 7 3 x - 2 y = 11

The given system of equation is x + y = 7 . . . . . . . . . . . . . . . . i 3 x - 2 y = 11 . . . . . . . . . . . . . . i i From ( i ) y = 7 - x . . . . . . . . . . . i i i Substituting y = 7 - x in ( i i ) , we get 3 x - 2 7 - x = 11 ⇒ 3 x - 14 + 2 x = 11 ⇒ 5 x = 25 ⇒ x = 5 ∴ From i i i , y = 7 - 5 ⇒ y = 2 Hence, the required solution is x = 5 and y = 2 .

The cost of 5 oranges and 3 apples is R s . 35 and the cost of 2 oranges and 4 apples is R s . 28 . Let us find the cost of an orange and an apple.

Let the cost of an orange = R s x And the cost of an apple = R s y According to the question, 5 x + 3 y = 35 . . . . . . i 2 x + 4 y = 28 . . . . . . i i Step 1 : Multiply equation i by 2 and equation i i by 5 to make the coefficients of x equal. Then we get the equations as : 10 x + 6 y = 70 . . . . . . i i i 10 x + 20 y = 140 . . . . . . . i v Step 2 : Subtracting equation i i i from equation i v , we get 14 y = 70 ⇒ y = 70 14 = 5 Step 3 : Putting the value of y in equation i , we get 5 x + 3 × 5 = 35 ⇒ 5 x = 35 - 15 ⇒ 5 x = 20 ⇒ x = 4 Hence, the cost of an orange is R s . 4 and the cost of an apple is R s . 5 .

Use the elimination method to find all possible solutions of the following pair of linear equations : 2 x + 3 y = 8 4 x + 6 y = 7

Given a pair of linear equations : 2 x + 3 y = 8 . . . . . . i 4 x + 6 y = 7 . . . . . . i i Step 1 : Multiply equation i by 2 and equation i i by 1 to make the coefficients of x equal. Then we get the equations as : 4 x + 6 y = 16 . . . . . . i i i 4 x + 6 y = 7 . . . . . . . i v Step 2 : Subtracting equation i v from equation i i i , we get 4 x - 4 x + 6 y - 6 y = 16 - 7 ⇒ 0 = 9 , which is a false statement. Therefore, the pair of equations has no solution.

The sum of two numbers is 7 . If the sum of these numbers is 7 times of its difference, then find the numbers.

Let the numbers are x and y . One of them must be greater than or equal to the other. Let us assume that x is greater than or equal to y . The sum of the two numbers is 7 . Thus, we have x + y = 7 . . . . i The sum of the two numbers is seven times of its difference. Thus, we have x + y = 7 ( x - y ) ⇒ 7 = 7 x - y ⇒ x - y = 1 . . . . . ii Adding equations i and ii , 2 x = 8 ⇒ x = 4 From equation i , 4 + y = 7 ⇒ y = 7 - 4 ⇒ y = 3 Hence, x = 4 and y = 3 .