The product of two numbers is
12. The sum of the larger number
and twice the smaller is 11. Find
the two numbers.
A. 4, 3
B. 8, 5/2
C. 3, 3/2
D. 3/2, -1
Let's assume the two numbers are x and y, with x being the larger number. We are given two equations:
1) xy = 12
2) x + 2y = 11
From equation 1, we can solve for y:
y = 12/x
Substituting this into equation 2, we get:
x + 2(12/x) = 11
x + 24/x = 11
Multiplying through by x to get rid of the fractions:
x^2 + 24 = 11x
x^2 - 11x + 24 = 0
Factoring the equation:
(x - 8)(x - 3) = 0
Setting each factor equal to zero and solving for x:
x - 8 = 0 --> x = 8
x - 3 = 0 --> x = 3
We have two possible values for x, so let's plug them back into equation 1 to find the corresponding values of y:
For x = 8:
y = 12/8 = 3/2
For x = 3:
y = 12/3 = 4
So the two possible pairs of numbers are (8, 3/2) and (3, 4).
Therefore, the correct answer is B. 8, 5/2.
Let's denote the two numbers as x and y, where x is the larger number. According to the given information, we have two equations:
1) x * y = 12
2) x + 2y = 11
We can solve these equations to find the values of x and y.
From equation 1), we can rewrite it as y = 12/x and substitute it into equation 2):
x + 2(12/x) = 11
Multiplying through by x to get rid of the fraction:
x^2 + 24 = 11x
Rearranging terms:
x^2 - 11x + 24 = 0
Factoring the quadratic equation:
(x - 8)(x - 3) = 0
Solving for x:
x - 8 = 0 --> x = 8
x - 3 = 0 --> x = 3
So, there are two possible solutions: x = 8 or x = 3.
Substituting these values back into equation 2), we can find the corresponding values of y:
For x = 8,
8 + 2y = 11
2y = 3
y = 3/2
For x = 3,
3 + 2y = 11
2y = 8
y = 4
Therefore, the possible pairs of numbers are (8, 3/2) and (3, 4).
The correct answer is B. 8, 5/2.