The sum of three times a first number n twice second number is 8. If the second number is subtravted from twice the first number, the result is 3, find the numbers.
3a + 2b = 8
2a - b = 3
7a = 14
a = 2
b = 1
Tell whether the first number is a multiple of the second number
5. 22,2
6. 29,3
7. 25,5
8. 40,8
To solve this problem, we can set up a system of two equations based on the given information.
Let's first define our variables:
- Let n be the first number.
- Let m be the second number.
According to the problem statement, we have two conditions:
1) "The sum of three times a first number n and twice the second number is 8."
This can be written as: 3n + 2m = 8.
2) "If the second number is subtracted from twice the first number, the result is 3."
This can be written as: 2n - m = 3.
Now we have a system of equations:
3n + 2m = 8
2n - m = 3
To solve this system, we can use the substitution method or the elimination method. Let's use the elimination method:
Multiply the second equation by 2 so that we can eliminate the 'm' variable:
4n - 2m = 6 (equation obtained by multiplying the second equation by 2)
Now, we can eliminate 'm' by subtracting the modified second equation from the first equation:
(3n + 2m) - (4n - 2m) = 8 - 6
3n + 2m - 4n + 2m = 2
-n + 4m = 2
To solve for 'n', isolate 'n' in terms of 'm' in the equation -n + 4m = 2:
-n = -4m + 2
n = 4m - 2
Now, substitute our expression for 'n' into any of the original equations. Let's use the first equation:
3n + 2m = 8
3(4m - 2) + 2m = 8
12m - 6 + 2m = 8
14m - 6 = 8
14m = 8 + 6
14m = 14
m = 14/14
m = 1
Now, substitute the value of 'm' back into the expression we found for 'n':
n = 4m - 2
n = 4(1) - 2
n = 4 - 2
n = 2
So, the first number (n) is 2 and the second number (m) is 1.