the sum of 3 numbers is 49. the second number is twice the 1st, and the third is 1 less than the second. find the numbers
a+b+c=49
b=2a
c=b-1
a + 2a + (2a-1) = 49
5a - 1 = 49
5a = 50
a = 10
10,20,19
10,20,19
To solve this problem, we can use algebraic equations to represent the relationships between the numbers and their sum.
Let's assume the first number is "x". According to the problem, the second number is twice the first number, so it can be represented as "2x". Similarly, the third number is 1 less than the second number, so it can be represented as "2x - 1".
Now we know that the sum of the three numbers is 49. We can write an equation to represent this:
x + 2x + (2x - 1) = 49
Simplifying the equation:
x + 2x + 2x - 1 = 49
5x - 1 = 49
5x = 50
x = 10
So, the first number is 10. Now we can substitute this value back into our expressions for the second and third numbers:
Second number = 2x = 2 * 10 = 20
Third number = 2x - 1 = 2 * 10 - 1 = 20 - 1 = 19
Therefore, the numbers are 10, 20, and 19.