One number is 6 more than another. If the sum of the smaller number and 3 times the larger number is 34, find the two numbers.
what am i suppose to do
i am reading as this but i know i am wrong
34=(x+6)+y*3
n + 3(n+6)=34
n is the smaller. the sum of the smaller and three times the larger is 34.
solve for n first.
The sum of 2 numbers is 56. 11 mire than twice the smaller number. Whatz the number
To solve the problem, let's start by assigning variables. Let's call the smaller number "n" and the larger number "n + 6" since it is stated that one number is 6 more than the other.
We are given that the sum of the smaller number and 3 times the larger number is 34. Therefore, we can set up the following equation:
n + 3(n + 6) = 34
Simplifying this equation, we can distribute the 3:
n + 3n + 18 = 34
Combining like terms:
4n + 18 = 34
Now, let's isolate the variable by subtracting 18 from both sides of the equation:
4n = 34 - 18
4n = 16
Next, divide both sides of the equation by 4 to solve for n:
n = 16 / 4
n = 4
So, the smaller number "n" is 4.
To find the larger number, we need to substitute the value of n back into one of the given equations.
We can use the statement that one number is 6 more than the other:
n + 6 = 4 + 6 = 10
Therefore, the larger number is 10.
The two numbers are 4 and 10.