Question

The sum of three number is 14. The sum of twice the first number, 4 times the second number, and 5 times the third number is 41. The difference between 7 times the first number and the second number is 51. Find the three
numbers.

Answer 1

Answer:

The numbers are 8 , 5 , 1

Step-by-step explanation:

Let the three numbers be x,y,z

Sum of three numbere is 14

⇒ x +y + z = 14 ---------------(I)

The sum of twice the first number, 4 times the second number, and 5 times the third number is 41.

⇒ 2x + 4y + 5z = 41 -----------(II)

The difference between 7 times the first number and the second number is 51.

⇒  7x -y = 51 ---------------(III)

Multiply equation (I) by (-5) and then add this wiht equation (II). Thus 'z' will be eliminated.

(I)*(-5)        -5x - 5y - 5z = -70

(II               2x + 4y + 5z = 41   {Now add}

                -3x   - y          = - 29  ------------(IV)

Multiply (III by (-1) and then add it with equation (IV)

(III)*(-1)        -7x + y = -51

(IV)             -3x - y = -29   {Now add}

                 -10x     = -80

                          x = -80 ÷ (-10)

x = 8

Plugin x = 8 in equation (III) and we obtain the value of y

7*8 - y = 51

56 - y = 51

     -y = 51 - 56

    -y = -5

   y = -5 ÷ (-1)

y = 5

Plugin x =8 and y =5 in equaiton (I)

8 + 5 + z = 14

    13 + z = 14

         z  = 14 - 13

 z = 1