pleas help....

Question

the sum of three numbers is 26. twice the first minus the second is 2 less than the third. The third is the second minus three times the first. Find the numbers.

Answer 1

Let's call the three numbers a, b, and c.

Now we can turn the information we are given into equations.

The sum of the three numbers is 26:

a + b + c = 26

Twice the first (2 times a) minus the second (2 times a minus b) is 2 less than the third:

2a - b = c - 2

The third is the second minus three times the first:

c = b - 3a

Counting what we have here, we now have three equations and three variables: enough to solve the whole system of equations.

The third equation gives us c directly, so we can start there and substitute into the second equation:

2a - b = (b - 3a) - 2

2a + 3a = b + b - 2

5a = 2b - 2

Let's get one of these variables on its own so we can continue with the substitution:

5a + 2 = 2b

b = (5a + 2) / 2

Now we have c in terms of a and b, and b in terms of just a. So let's use the first equation and substitute to find out what a is:

a + b + c = 26

a + (5a + 2) / 2 + (b - 3a) = 26

a + (5/2)a + 1 + (5a + 2) / 2 - 3a = 26

7/2a + 1 + 5/2a + 1 - 3a = 26

12/2a + 2 - 3a = 26

6a - 3a = 26 - 2

3a = 24

a = 8

At last, we have solved for one of the variables. Now, plug this into the equation for b to find b:

b = (5a + 2) / 2 = (5(8) + 2) / 2 = (40 + 2) / 2 = 42 / 2 = 21

Now we have a and b. Time to find c!

a + b + c = 26

(8) + (21) + c = 26

29 + c = 26

c = 26 - 29

c = -3

So our values for a, b, and c are 8, 21, and -3.