Mr burr the head custodian is working on putting new numbers on 42 of the 81 classroom doors. He found two numbers for the 23rd door he was working on in the 6th grade hallway whose sum is 35 and their product is 306. What are the two numbers?
Eq1: X+Y = 35.
Eq2: XY.
X + Y = 35
Y = 35-X
In Eq2, replace Y with 35-X
X(35-X) = 306
-X^2 + 35x = 306
-X^2 + 35x - 306 = 0.
X^2 _ 35 + 306 = 0
(x-17)(x-18) = 0.
X = 17, and 18.
Correction:
Eq2: XY = 306.
To find the two numbers for the 23rd door, let's consider the sum and the product of the two numbers.
Let's assume the two numbers are x and y.
According to the problem, the sum of the two numbers is 35:
x + y = 35 ...[Equation 1]
The product of the two numbers is 306:
xy = 306 ...[Equation 2]
To find the values of x and y, we can solve this system of equations.
One way to do this is by substitution:
From Equation 1, we can express y in terms of x:
y = 35 - x
Substituting this value of y into Equation 2:
x(35 - x) = 306
Expanding the equation:
35x - x^2 = 306
Rearranging the equation:
x^2 - 35x + 306 = 0
Now, we have a quadratic equation. We can solve it using factoring, completing the square, or the quadratic formula. In this case, it can be easily factored:
(x - 17)(x - 18) = 0
This means that either (x - 17) = 0 or (x - 18) = 0.
If we solve for x, we get two possible values: x = 17 or x = 18.
Now, substituting these values back into Equation 1:
For x = 17: 17 + y = 35 => y = 35 - 17 => y = 18
For x = 18: 18 + y = 35 => y = 35 - 18 => y = 17
Therefore, the two numbers for the 23rd door are 17 and 18.