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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.

1 Answer

3 votes

Answer:

The numbers are 8 , 5 , 1

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

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