one number is 8 more than twice another. their product is 8 more than twice their sum. find the numbers
N1=8+2*N2
N1*N2=8+2(N1+N2)
Let the smaller number be x; that means the other number must be 2x + 8.
Solve the equation
x(2x+8) = 2x + 8 + x + 8
which can also be written
2x^2 + 8x = 3x + 16
or
2x^2 + 5x -16 = 0
There seems to be no integer solution. The smaller number is
x = (-5 + sqrt 153)/4 = 1.84233
and the larger number is 2x + 1 = 11.68466
Their sum is 13.5267
and the product is 21.527
as required.
drwls, you made a slight mistake - you forgot a factor of 2 in the RHS of the equation.
If N1 = x and N2 = 2x + 8
then the equation is
x(2x+8) = 8 + 2*(2x + 8 + x)
Which simplifies to
x^2 - x - 12 = 0
which has two integer solutions.
Hope that helps!
To solve this problem, let's assume the first number as x and the second number as y.
According to the given information, one number is 8 more than twice another:
x = 2y + 8 ...equation 1
The product of the two numbers is 8 more than twice their sum:
xy = 2(x + y) + 8 ...equation 2
Now, let's solve these equations to find the values of x and y.
Substitute the value of x from equation 1 into equation 2:
(2y + 8)y = 2((2y + 8) + y) + 8
Simplify the equation:
2y^2 + 8y = 2(3y + 8) + 8
Expand and simplify further:
2y^2 + 8y = 6y + 16 + 8
2y^2 + 8y = 6y + 24
Now, bring all the terms to one side:
2y^2 + 8y - 6y - 24 = 0
2y^2 + 2y - 24 = 0
Divide the entire equation by 2 to simplify:
y^2 + y - 12 = 0
Now, we need to factorize the quadratic equation:
(y + 4)(y - 3) = 0
This gives us two possible values for y:
y + 4 = 0 or y - 3 = 0
Solving these equations, we find:
y = -4 or y = 3
Now, substitute these values back into equation 1 to find the corresponding values of x.
For y = -4:
x = 2(-4) + 8
x = -8 + 8
x = 0
So, one pair of numbers is x = 0 and y = -4.
For y = 3:
x = 2(3) + 8
x = 6 + 8
x = 14
So, another pair of numbers is x = 14 and y = 3.
Therefore, the two pairs of numbers that satisfy the given conditions are (0, -4) and (14, 3).