A positive integer is 29 more than 18 times another. Their product is 6354. Find the two integers.

Sir-Emo-Chappington  solved this with two variables.....here's a way to do it with one......

Let x be one integer   ...and let  18x + 29  be the second integer....so we have....

x (18x + 29) = 6354    simplify

18x^2 + 29x  = 6354     rearrange

18x^2 + 29x - 6354  =  0    this factors as

(18x + 353) (x - 18) = 0       the only integer produced here is when x = 18

And the second integer is 18(18)+ 29 =  353

I think this method is a little easier because we don't have to use two equations and a substitution to solve......but......SEC's method is perfectly valid, too !!!!

x = 29 + (18*y)

x * y = 6354

First we can insert either y or x into the other equation. For this I shall insert y's equivalent.

x * y = 6354

∴ y = ⁶³⁵⁴/_x

Insert that into the below equation like so:

x = 29 + (18*y)

∴ x = 29 + (18 * ⁶³⁵⁴/_x)

Now let's simplify and re-arrange:

x = 29 + (¹¹⁴³⁷²/_x)

Since we don't like that x being on the bottom of the fraction, let's multiply it by "x". of course everything else must be to keep this equation valid:

[x] * x = [29 + (¹¹⁴³⁷²/_x)] * x

x² = 29x + 114372

0 = -x² + 29x + 114372

We have now gotten ourselves a quadratic equation, so use the "quadratic forumula" to find all possible values of "x":

x = (-29 ± sqrt(29² - 4*-1*114372)) / (2 * -1)

x = (-29 ± sqrt(841 + 457488)) / -2

x = (-29 ± sqrt(458329)) / -2

x = (-29 ± 677) / -2

The two results are...

x = (-29 + 677) / -2 = -324

x = 353

Using these two values of x, let's find their associated values of y:

y = ⁶³⁵⁴/_x

y = ⁶³⁵⁴/_-324 = -19.61 _{[1 recurring]}

y = ⁶³⁵⁴/₃₅₃ = 18

So the two results are:

x = -324 and y = -19.61 _{[1 recurring]}

x = 353 and y = 18