It can not be done
The multiples of 40 are...
1 and 40, which add up to 41
2 and 20, which add up to 22
4 and 10, which add up to 14
and
5 and 8, which add up to 13
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First we have to set up our two equations:
X + Y = 10 XY = 40
We solve the first equation for Y: X + Y = 10 => X + Y - X = 10 - X => Y = 10 - X
Substitute the Y value in the first equation into the second equation:
XY = 40 => X(10 - X) = 40 => 10X - X^2 = 40
Rearrange the terms to produce a quadratic: -X^2 + 10X = 40 => -X^2 + 10X - 40 = 40 - 40 => -X^2 + 10X - 40 = 0
Flip the signs: -X^2 + 10X - 40 = 0 => +X^2 - 10X + 40 = 0 => X^2 - 10X + 40 = 0
Here, coefficient of X^2 is 1, X is -10 and the constant is 40. So, a = 1, b = -10, c = 40.
Substitute these values into the quadratic formula to find X: X = [-b ± √(b^2 - 4ac)]/2a => X = [-(-10) ± √((-10)^2 - 4(1)(40)]/2(1) => X = [10 ± √(100 - 160)]/2 => X = [10 ± √(-60)]/2 => X = [10 ± √(-1 * 2 * 2 * 3 * 5)]/2 => X = [10 ± √(-1) * √2 * √2 * √3 * √5]/2 => X = [10 ± i * 2 * √3 * √5]/2 => X = [10 ± 2√(3*5)i]/2] => X = (10 ± 2√15i)/2 => X = 10/2 ± 2√15i/2 => X = 5 ± √15i X = 5 + √15i or X = 5 - √15i
Substitute these into the first equation to find their respective Y values.
For X = 5 + √15i: Y = 10 - X => Y = 10 - (5 + √15i) => Y = 10 - 5 - √15i => Y = 5 - √15i
For X = 5 - √15i; Y = 10 - X => Y = 10 - (5 - √15i) => Y = 10 - 5 + √15i => Y = 5 - √15i
We see that the X in equation two equals the Y in equation 1 and vice versa. So these are the only two values.
So, the two numbers are 5 - √15i and 5 + √15i.
It was quite a famous question, being featured on the Area Magne as an unsolvable question because of the methods at the time used to solve equations not being equipped for complex, even negative numbers. Including famous methods such as Cardano's Method, Geometric methods, etc. Nowadays it would be quite an easy question involving very basic use of the imaginary number(i = √-1), and a knowledge of how to solve linear systems of equations(a fancy way of saying a bunch of related equations). But it was enough to stump math geniuses of that time. Now that I think about it, a lot of high school math today would stump mathematicians of old. But, we still owe it to them.
Hope this helps :)
The two numbers that add to make 10 and multiply to make 40 are 5 and 8. This is because 5 + 8 = 13 and 5 * 8 = 40. These numbers satisfy both conditions simultaneously.
Oh, dude, that's like basic math 101. The two numbers are 8 and 2. When you add them, you get 10, and when you multiply them, you get 40. It's like the easiest riddle ever.
Well, honey, the two numbers you're looking for are 5 and 8. They add up to 10 and when you multiply them, you get 40. It's as simple as that!