Solve the following problem. Be sure to show all steps (V.E.S.T.) and work in order to receive full credit.

The sum of three numbers is 26. The second number is twice the first and the third number is 6 more than the second. Find the numbers.

Respuesta :

Answer:

The numbers are 4, 8 and 14

Step-by-step explanation:

Required

Determine the three numbers

We start by declaring the variables

Let the three numbers be [tex]x,\ y\ and\ z\ respectively[/tex]

Next, we write the relationships between the numbers

[tex]y = 2 * x[/tex] --- Second = Twice the first

[tex]z = 6 + y[/tex] --- Third = 6 more than second

[tex]x + y + z = 26[/tex] --- Sum of the three numbers

Then, we solve for each number

[tex]y = 2 * x[/tex]

[tex]z = 6 + y[/tex]

[tex]x + y + z = 26[/tex]

Substitute 6 + y for x in [tex]x + y + z = 26[/tex]

[tex]x + y + 6 + y =26[/tex]

Collect Like Terms

[tex]x + y + y = 26 - 6[/tex]

[tex]x + 2y = 20[/tex]

Substitute 2 * x for y in [tex]x + 2y = 20[/tex]

[tex]x + 2(2*x) = 20[/tex]

[tex]x + 4*x = 20[/tex]

[tex]x + 4x = 20[/tex]

[tex]5x = 20[/tex]

Solve for x

[tex]x = \frac{20}{5}[/tex]

[tex]x = 4[/tex]

Recall that:

[tex]y = 2 * x[/tex]

[tex]y = 2 * 4[/tex]

[tex]y = 8[/tex]

Recall that:

[tex]z = 6 + y[/tex]

[tex]z = 6 + 8[/tex]

[tex]z = 14[/tex]

Hence, the numbers are 4, 8 and 14

To test the result:

[tex]x + y + z = 26[/tex]

[tex]4 + 8 + 14 = 26[/tex]

[tex]26= 26[/tex]