Question

There are three numbers. The first number is twice the second number. The third number is twice the first number. The sum of the three numbers is 112.

Answer 1

Answer:

x = 32 y = 16 z = 64

Step-by-step explanation:

x is the 1st number

y is the 2nd number

z is the 3rd number

The first number is twice the second number so

x = 2 y

The third number is twice the first number.

z = 2 x

Their sum is 112

x + y + z = 112

Plus in what we know:

x + y + z = 112

2 y + y + 2x = 112

3y + 2x = 112 Let's solve for y and subtract 2x from each side

3y = 112 - 2x  Divide both sides by 3

y = $\frac{112 - 2x}{3}$

Now plug our answer back in to solve for x.

x = 2y

x = 2 ( $\frac{112 - 2x}{3}$ )

x = (224 - 4x) / 3    Multiply each side by 3.

3x = 224 - 4x         Add 4x to each side

3x + 4x = 224

7x = 224              Divide each side by 7

7x / 7 = 224 / 7

x = 32

Now we can solve for z.

z = 2x

z = 2 ( 32 )

z = 64

Now we can solve for the numerical value of y.

x + y + z = 112

32 + y + 64 = 112

96 + y = 112       Subtract 96 from each side.

y = 112 - 96

y = 16

Answer 2

Answer:

1st number = 36

Second = 16

Third = 64

Step-by-step explanation:

We have 3 numbers

Let's call the second number g

Then the first would be , 2x g = 2g

Then the third would be twice the first = 2 x 2g = 4g

We add up all numbers and equate it to 112

2g + g +4g = 112

7g = 112

g = 112/7

g = 16

First number = 2g = 2x16 = 32

Second number = g = 16

Third number = 4g = 4x16 = 64

Therefore the numbers are 32, 16, 64.

I hope this helps!