One integer is 3 times another. The product is 4 more than their sum. Find the two integers.

Question

One integer is 3 times another. The product is 4 more than their sum. Find the two integers.

Answer 1

Here's what we know:

b = 3a (one integer is 3 times the other)

a * b = a + b + 4 (the product is 4 more than the sum)

 

We have a value for one of our variables, so let's plug it in:

a(3a) = a + (3a) + 4

3a²  = 4a + 4

3a² - 4a - 4 = 0

 

I'm going to use the quadratic formula because it's not going to factor out evenly:

(-b ± √(b² - 4ac)) / 2a

 

(-(-4) ± √((-4)² - 4 * 3 * -4)) / (2 * 3)

Take care of the signs and multiplication issues:

(4 ± √(16 + 48)) / 6

Add together the root numbers:

(4 ± √64) / 6

find the root:

(4 ± 8) / 6

Now to look at the plus/minus part:

12 / 6 or -4 / 6

2 or -²/₃

 

Since the question asked us for an integer, we're going to use 2.

a = 2

 

b = 2 * 3

b = 6

 

Check

2 * 6 = 2 + 6 + 4

12 = 12