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.