the larger of two numbers is 1 more than twice the smaller.the sum of the numbers is 20 less than three times the larger.find the numbers

Look at my answer to Ulises a few lines down for hint.

i know i just need help setting the equation

Ok, call big one b

call small one s
b = 1 + 2 s
b + s = 3 b - 20

can u help me work it out please

from the first equation you know that you can use the quantity (1 + 2 s) for b

so do that
1 + 2 s + s = 3 (1 + 2 s) - 20
or
1 + 3 s = 3 + 6 s - 20
solve that for s, then go back and find b

b=13,s=6

Let's solve this problem step by step.

Let's assume the smaller number is 'x' and the larger number is 'y'.

We are given two pieces of information:

1) The larger number is 1 more than twice the smaller:
y = 2x + 1

2) The sum of the numbers is 20 less than three times the larger:
x + y = 3y - 20

Now, we can solve these two equations simultaneously to find the values of x and y.

Substituting the value of 'y' from equation (1) into equation (2), we get:
x + (2x + 1) = 3(2x + 1) - 20

Simplifying this equation:
x + 2x + 1 = 6x + 3 - 20
3x + 1 = 6x - 17

Bringing all the x terms to one side and constant terms to the other side:
3x - 6x = -17 - 1
-3x = -18
x = 6

Now that we have the value of x, we can substitute it back into equation (1) to find y:
y = 2x + 1
y = 2(6) + 1
y = 12 + 1
y = 13

So, the smaller number is 6 and the larger number is 13.