Three more than twice the smaller number is equal to the bigger number. If the sum of the numbers is 33, what is the smaller number?
Let's represent the smaller number as 'x'.
The bigger number can be represented as 2x + 3, since it is three more than twice the smaller number.
We are given that the sum of the numbers is 33, so we can write the equation:
x + 2x + 3 = 33
Combining like terms, we get:
3x + 3 = 33
Subtracting 3 from both sides:
3x = 30
Dividing both sides by 3:
x = 10
Therefore, the smaller number is 10.
Let's use algebra to solve this problem.
Let's assume the smaller number is "x" and the bigger number is "y".
According to the problem, three more than twice the smaller number is equal to the bigger number, so we can write this as an equation:
2x + 3 = y
We also know that the sum of the numbers is 33, so we can write a second equation:
x + y = 33
Now, we have a system of two equations with two variables:
2x + 3 = y
x + y = 33
To solve this system, we can use the substitution method.
Rearranging the second equation, we get:
y = 33 - x
Now, we substitute this value of "y" in the first equation:
2x + 3 = 33 - x
Let's solve for "x":
Add "x" to both sides:
2x + x + 3 = 33
Combine like terms:
3x + 3 = 33
Subtract 3 from both sides:
3x = 30
Divide both sides by 3:
x = 10
Therefore, the smaller number is 10.
2s + 3 = b
s + b = 33 ... b = 33 - s
2s + 3 = 33 - s
3s = 30