The sum of three numbers is 83. The third number is two times the second. The second number is seven less than the first. What are the numbers

Question

Please help break down solution 

Answer 1

The way to handle a problem like this is to rewrite the three numbers in terms of their relationship to one of the three numbers.  Let's call the numbers x, y, and z, where x is the first number, y is the second, and z is the third.

 

We know x + y + z = 83.

 

Since the third number is twice the second, we say that z = 2y.

Since the second number is seven less than the first, we say that y = x - 7.  We can rewrite this as x = y + 7.

 

Now all three can be written in terms of y.

 

x + y + z = 83

(y + 7) + y + 2y = 83

4y + 7 = 83

4y = 76

y = 19

 

The second number is 19.  The first is 19 + 7 or 26, and the third is 2 times 19 or 38.

 

A check reveals that 26 + 19 + 38 = 83.

Answer 2

Use substitution to solve this problem:

 

X + Y + Z = 83

-2Y + Z = 0

Y - X = -7

 

1)Express the second and third equations in terms of Y

 

-2Y + Z = 0 -> 2Y = Z 

 

Y - X = -7 -> X = Y + 7

 

2)Plug X and Z into first equation:

 

(Y + 7) + Y + 2Y = 83 -> 4Y + 7 = 83 -> Y = 19.

 

3)Plug Y back into equations 2 and 3.

 

-2*19 + Z = 0 -> Z = 38

 

19 - X = -7 -> X = 26.

 

4) The solution set is (26, 19, 38)

 

 

  Verification: 26 + 19 + 38 = 83

                    -2*(19) + 38 = 0

                     19 - 26 = -7.