Ok as odd as this sound i am stuck on this math problem...?

Question

This is the question:

The sum of three numbers is 137. The second number is 4 more thantwo times the first number. the Third number is 5 less than three times the first number. Find the three numbers.

Thanks

Answer 1

1st number --> x
2nd number --> 2x + 4
3rd number --> 3x - 5

Adding:
x + 2x + 4 + 3x - 5 = 137
6x - 1 = 137
6x = 138
x = 23

Now just substitute x to find the other numbers.
2nd number --> 2x + 4 = 2(23) + 4 = 46 + 4 = 50
3rd number --> 3x - 5 = 3(23) - 5 = 69 - 5 = 64

The numbers are 23, 50, 64

Double-check:
23 + 50 + 64 = 137

Answer 2

lol....funny intro. I like it! :)

Three numbers right?

You don't know any of them....so call then "a", "b" and "c".

They add up to 137....so: a + b + c = 137
--------------------------

The second number is 4 more than two times the first number.

So: b = 2a + 4
--------------------------

Third number is 5 less than three times the first number.

So: c = 3a - 5
++++++++++++++

So you have 3 equations and 3 unknowns.

a + b + c = 137
b = 2a + 4
c = 3a - 5

Sub the last two equations into the first:

==> a + (2a + 4) + (3a - 5) = 137
==> a + 2a + 3a + 4 - 5 = 137
==> 6a = 138
==> a = 23

Ok...so you know that "a" = 23.

Now, plug this into the other two equations, do the arithmetic and you will then know "b" and "c"....you CAN do this part.

I hope this helps you.

Answer 3

a + b + c = 137-------(1)

b = 2a + 4

c = 3a-5

substitute b and c in eqn(1)

a + 2a + 4 + 3a-5 = 137

6a - 1 = 137

6a = 138: a = 138/6

a = 23

b = 46+ 4 = 50

c = 69 - 5 = 64

Answer 4

the first number is 23
the second is 2x23 + 4 = 50
the last is 3x23 - 5 = 64

23 + 50 + 64 = 137

Answer 5

let x-first number, y=second number, z=third number
y=4+2x
z=3x-5
so x+y+z=137
therefore (4+2x)+(3x-5)+x=137
6x-1=137
6x=138
x=23
y=50
z=64