# Thread: Solving for x & y with the formula..... x+(y*1.5)=1000

1. ## Solving for x & y with the formula..... x+(y*1.5)=1000

I've looked everywhere to try and solve this formula and I know it must be simple for others, so I'm hoping someone here can help me. I'm not even sure if it's algebra or some other kind of math and there may be more than one answer.

I'm looking for a formula that will tell me every combination for what X and y have to represent for the formula 1000=x+(y*1.5) to be correct

For example, I know right off the bat that the formula would be correct if x=250 and y=500 but how can I determine with a formula what combinations for X & Y in the above formula will equal 1000 exactly? If there is more than one answer, would an "if" formula in Excel be better (but I don't know how to create that formual)

If there is an excel formula for this, that would be even better.

Thank you all!

2. ## Re: Solving for x & y with the formula..... x+(y*1.5)=1000

Hi, GrumpyGuy!

Your given equation is $\displaystyle x+1.5y=1000$, but you can rearrange it in terms of $\displaystyle y$ as $\displaystyle y=\tfrac{2}{3}(1000-x)$.

Using function notation, we have $\displaystyle f(x)=\tfrac{2}{3}(1000-x)$.

The domain of $\displaystyle f$ (all possible values of $\displaystyle x$) is all real numbers, i.e., anything! The range (all possible values of $\displaystyle f(x)$) of $\displaystyle f$ is also all real numbers, because you can set $\displaystyle f(x)$ equal to any real number and solve for a real-valued $\displaystyle x$. For the most part (given no restrictions or boundaries), the domain and range of a simple linear function is all real numbers.

The theme is that your equation (or function) has infinite real-valued solutions.

I have enclosed an Excel file with sample x-values in the A column and $\displaystyle y$ values in the B column. I manually entered some $\displaystyle x$ values and used the Excel function =(2/3)*(1000-x) to compute the corresponding $\displaystyle f(x)$ values (or $\displaystyle y$ values).

Good luck!
-Andy

3. ## Re: Solving for x & y with the formula..... x+(y*1.5)=1000

Keeping x and y positive integers, there are 333 solutions.
n: x,y
1: 1,666
2: 4,664
3: 7,662
.....
84: 250,500
....
331: 991,6
332: 994,4
333: 997,2

y = (2000 - 2x) / 3

Silly problem unless some restrictions are placed...

4. ## Re: Solving for x & y with the formula..... x+(y*1.5)=1000

Thanks DenisB. I posted a reply to Abender that clarifies what I was trying to do. Once it's posted, I think you'll have a better idea. Yes there restrictions that would consist of more exacting values multiplied against Y and the number chosen instead of 1000 that I think would generate less cmbinations for X and Y to make the formula correct

5. ## Re: Solving for x & y with the formula..... x+(y*1.5)=1000

Originally Posted by GrumpyGuy
....that I think would generate less combinations for X and Y to make the formula correct
That statement makes no sense:
a formula is a formula is a formula...

SIMPLE: to get a unique solution,
you need 2 equations.