One number is two less than a second number. If you take one-half of the first number and increase in by the second number, the result is at least 41. Find the least possible value for the second number.
HELP PLEASE!!!! I hate word problems!!!
1st#=x-2
2nd#=x
1/2(x-2)+x=41
x-2+2x=82
3x=84
x=28
No worries! I can help you solve this word problem step by step. Let's break it down:
Let's call the first number "x" and the second number "y".
We're given that one number (x) is two less than the second number (y), so we can write the equation: x = y - 2.
Now, let's focus on the second part of the problem. It says to take half of the first number (x/2) and increase it by the second number (y). The result should be at least 41. Mathematically, we can express this as:
(x/2) + y ≥ 41.
We want to find the least possible value for the second number (y). To do that, we need to find the minimum value that satisfies the equation (x/2) + y ≥ 41.
To simplify things, let's substitute x with y - 2 in the equation:
((y - 2)/2) + y ≥ 41.
Now, let's simplify this expression:
(y - 2)/2 + y ≥ 41.
To get rid of the fraction, we'll multiply through by 2:
(y - 2) + 2y ≥ 82.
Combine like terms:
3y - 2 ≥ 82.
Add 2 to both sides:
3y ≥ 84.
Divide both sides by 3:
y ≥ 28.
Therefore, the least possible value for the second number is 28.
So in summary, the second number must be at least 28 for the given conditions to be satisfied.