what 2 numbers add to 3 but equal -18?

Answer 1

x+y =3

XY =-18

y = 18/x

x+18/x = 3 multiply both sides by x to eliminate the fraction

x^2 +18 =3x

x

x^2 -3x +18 = 0 factor and set each factor = 0, then solve for x

(x-6)(x+3) = 0

x =6 and -3 are the two numbers

6+(-3)= 3

6(-3) = -18

Answer 2

Let's crack into this problem, shall we?

So, you want to find two numbers that add to 3 and multiply to -18?

We can do this a number of different ways (pun intended), but the quickest way will be to take a system of equations approach.

METHOD #1: Quadratic Solutions (A.K.A. The "right" way to solve this problem)

Let's call your two numbers x and y.

From the question, you've given us two different conditions to work with, so we're going to turn these conditions into equations!

1. The first equation (x + y = 3) represents condition #1: two numbers that add to 3.
2. The second equation (x * y = -18) represents condition #2: two numbers whose multiple is -18.

Now, we just solve the system of equations through substitution and we're done!

1. Rewrite equation 1 to isolate y on the left side: y = 3 - x
2. Substitute the rewritten equation 1 into equation 2: x(3 - x) = -18
3. Write the new version of equation 2 as a quadratic by expanding x(3 - x) and adding 18 to both sides: 3x - x² + 18 = 0
4. Multiply both sides by -1 to make it prettier to work with, and rearrange the expression to put terms in order of their "power" (i.e., their "exponent"): x² - 3x - 18 = 0

This equation (x² - 3x - 18 = 0) factors beautifully!

1. x² - 3x - 18 = 0 (quadratic form)
2. (x - 6)(x + 3) = 0 (factored form)

And we've found our solution! The answers are x = 6 and x = -3.

Side note: Polynomial equations for 2+ unknowns, like (x - 6)(x + 3) = 0, where we know we want integer solutions, are called Diophantine Equations, and they're a pretty special class of equations.

Now, sure, we've done this one way, are there other ways to do it? Absolutely.

METHOD #2: The Sherlock Holmes approach, with some incredibly cool theory underneath it.

If you stop to listen to the math, it will tell you lots of useful information!

Start by "interviewing" the more "advanced" operation first (i.e., x * y = -18).

1. What does this tell us?
2. The product is negative, so one and ONLY one of the two values will be negative.
3. -18 is an even negative integer, so we're probably looking for a pair of integers, where one must be negative.

Time for a trick: There's a neat theorem called the Unique Factorization Theorem (one of the fundamental theorems of arithmetic) that can help here!

Every integer greater than 1 can be uniquely expressed as a product of prime numbers (up to the order of the factors).

Personally, I think that's pretty neat!

This means that 18 can be "written out" as a product of prime numbers:

1. 18 = 9 * 2, but we're not done, since 9 can also be factored.
2. 18 = (3 * 3) * 2, Tada! This is 18's unique prime factorization!

Let's review our notes and figure out what have we deduced so far.

1. We know the prime factors of 18 (2, 3, and 3)
2. We know we need to combine these factors to make two integers:
3. one integer will be the product of two factors
4. the other will just be a single factor
5. ... and one of the integers above must be negative.

Think of this like stacking building blocks – we're starting with three blocks (a '2' block and two '3' blocks), but we have to combine them into a two-stack and a one-stack, such that the sum of both stacks has to equal 3.

Let's investigate:

1. What if we start with '2' on its own and combine the two '3' blocks together to make 9?
2. -2 + (3 * 3) = 7, and 2 + (-3 * 3) = -7 Try again!
3. So '3' must be the solo block, and '2' * '3' must be the double block.
4. (2 * 3) + (-3) = (6 - 3) = 3 Well done.

Using prime factorization, we've found the answer to the original problem: the two values are -3 and 6.

METHOD #3: The pseudo-algorithmic way.

Let's try another method! This will be an algorithmic way to do this factorization method. This one goes out to my CS friends out there.

Now, we are making an assumption here.

This method assumes you know to find the prime factors of 18

From x*y = -18, we know to factor -18 down to prime number "building blocks". Write out the *positive* prime factors and assign them to letters for now:

1. A = 2;
2. B = 3;
3. C = 3;

From x * y = -18. we know either one or three of these blocks (A, B, C) must be negative, but from x+y = 3 we know that only one will be negative.

Let's systematically consider combinations of these factors, and their negatives, to find a pair that sums to 3. We'll represent x and y in terms of A, B, and C.

Possibility 1: x is a single positive or negative prime factor (i.e., -3, -2, 2, or 3)

1. Case 1.1: x = A = 2 => y = -9 (from xy = -18).
2. 2 + (-9) = -7 (Doesn't work)
3. Case 1.2: x = -A = -2 => y = 9
4. (-2) + 9 = 7 (Doesn't work)
5. Case 1.3: x = B = 3 => y = -6 (from xy = -18).
6. 3 + (-6) = -3 (Doesn't work)
7. Case 1.4: x = -B = -3 => y = 6
8. (-3) + 6 = 3 (WORKS!)

Since we found a solution, we can stop here.

The solution is x = -3 and y = 6.

In summary, we explored three different ways to solve this problem: using a standard quadratic approach, employing a deductive method based on prime factorization, and constructing a systematic, almost algorithmic, search.

All three methods lead us to the same answer, showcasing the interconnectedness of different mathematical concepts,

Hope that helps!