Question:
Solution:
Let the following equation:
![\sqrt[]{12-x}=\text{ x}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B12-x%7D%3D%5Ctext%7B%20x%7D)
this is equivalent to:
![(\sqrt[]{12-x})^2=x^2](https://tex.z-dn.net/?f=%28%5Csqrt%5B%5D%7B12-x%7D%29%5E2%3Dx%5E2)
this is equivalent to:
this is equivalent to:
thus, we can conclude that
x= 3.
A) I would make the positive integer x and then form an equation.
x + 30 = x^2 - 12
x + 42 = x^2
0 = x^2 - x - 42 this can be factorised
(x - 7) ( x + 6) Therefore x = 7 or x = -6
Since the question asks for a positive integer the answer is 7.
B) two positive numbers x and y.
X - y = 3
x^2 + y^2 = 117
Use these simultaneous equations to figure out each number.
Rearrange the first equation
x = y + 3
Then substitute it into the second equation.
(y+3)^2 + y^2 = 117
y^2 + 6y + 9 + y^2 = 117
2y^2 + 6y - 108 = 0
then factorise this.
(2y - 12) (y + 9)
This means that y = 6 or y = -9 but it’s 6 because that’s the only positive number.
Use y to find x
x = y + 3
x = 6 + 3
x = 9
So the answers are x = 9 and y = 6.
The answer to your question is $1978
The answer is .1391
Hope this helps!
(If your confused, I just put this in a calculator)
