Two consecutive numbers are such that five times the smaller one exceeds twice the greater by 16. Find the numbers
let the numbers be x and x+1. Then we know that
x = 2(x+1) + 16
Let the numbers be a, a+1
5a = 2(a+1)+16
5a = 2a+ 2+16
5a-2a = 18
3a/3 = 18/3
a = 6
= 6 and 7
I can't really understand the answer
I need the answer
Well, aren't these numbers quite the pranksters! Let's call the smaller number "x" and the greater number "x+1" because they like to be consecutive.
According to the problem, we know that "five times the smaller one exceeds twice the greater by 16," which can be written as:
5x = 2(x+1) + 16
Now, let's solve this equation and show these sly numbers who's boss:
5x = 2x + 2 + 16
5x = 2x + 18
Let's move those sneaky 2x over to the other side:
5x - 2x = 18
3x = 18
Finally, let's divide both sides by 3 to find the value of our little trickster x:
x = 18/3
x = 6
Now, as our numbers are consecutive, the greater number (x+1) can be found by adding 1 to x:
x+1 = 6+1
x+1 = 7
So the two consecutive numbers are 6 and 7. These numbers sure know how to play with equations while giving us a good laugh!
To solve this problem, let's start by assigning variables to the unknown numbers. Let's call the smaller number "x" and the greater number "x + 1" (since they are consecutive).
According to the problem, we have the equation:
5x = 2(x + 1) + 16
Now, let's solve the equation step by step:
Distribute 2 to x + 1:
5x = 2x + 2 + 16
Combine like terms:
5x = 2x + 18
Subtract 2x from both sides:
5x - 2x = 18
Simplify:
3x = 18
Divide both sides by 3:
x = 6
Now we know that the smaller number is 6. To find the greater number, we substitute x = 6 into x + 1:
x + 1 = 6 + 1 = 7
Therefore, the two consecutive numbers are 6 and 7.