The larger of two numbers is ten more than twice the smaller number. The sum of the numbers is forty. Find each number.
To solve this problem, we can set up a system of equations based on the given information.
Let's assume that the smaller number is x, and the larger number is y.
The problem states that the larger number is ten more than twice the smaller number. Therefore, we can write the equation:
y = 2x + 10 (Equation 1)
The sum of the numbers is forty, so we can also write the equation:
x + y = 40 (Equation 2)
Now we have a system of two equations with two variables. We can solve this system by substituting Equation 1 into Equation 2.
Substituting y in Equation 2 with 2x + 10, we get:
x + (2x + 10) = 40
Simplifying,
3x + 10 = 40
Subtracting 10 from both sides,
3x = 30
Dividing both sides by 3,
x = 10
Now that we have found the value of x, we can substitute it back into Equation 1 to solve for y:
y = 2x + 10
y = 2(10) + 10
Simplifying,
y = 20 + 10
y = 30
Therefore, the smaller number is 10, and the larger number is 30.
x + 2x + 10 = 40
3x = 30
x = 10