Question

The first of the three numbers exceeds twice the second number by 4, while the third number is twice the first. If the sum of the three numbers is 54, find the numbers.

Answer 1

let the second number be x

1st number =2x+4

third number=2(2x+4)=4x+8

sum=54

2x+4+x+4x+8=54

7x+12=54

7x=42

x=6

1st number =2x+4

2*6+4=16

2nd number =6

third number =2(2x+4)

2*16=32

Answer 2

Answer:

Let the three numbers be x y and z

According to question . x=2y+4 ..(1)

Also.... z=2x ..(2)

and. x+y +z =54..(3)

Comparing equation 1 and 2 we get

z= 4y+8

putting value of x and z in equation (3)

2y+4+y+4y+8=54

y=6

putting value of y in eq.(1)

x=16

and in eq(2) z =2x=32

Hence the value of x.y and z is 16 ,6 and 32 respectively.

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