find the integers

Question

The product of two consecutive interns is 11 more than their sum

Answer 1

We can call the first integer x, an the second y. We can then write the following equations to solve this problem as:

y = x + 1

xy = x + y + 11

Substituting for y:

x (x+1) = x + (x+1) + 11

Simplifying each side:

x² + x = 2x + 12

Rearranging:

0 = x² - x +12

Factoring the polynomial gives

0 = (x - 4) (x + 3)

So x is either 4 or -3. If x is 4, then y is 5. If x is -3, then y is -2.

Answer 2

Hi Kisha, let x represent the first integer. Then the second integer will be x + 1.

Equation to solve:

x * (x + 1) = x + (x + 1) + 11

Or

x^2 + x = 2x + 12

Combine terms:

x^2 - x - 12 = 0

Solve by factoring:

x^2 - x - 12 = (x + 3) * (x - 4) = 0

Use the zero property rule:

x + 3 = 0

x = -3

And

x - 4 = 0

x = 4

There are two answers.

First answer:

x = -3, then x + 1 = -2

The two integers are -3 and -2.

Second answer:

x = 4, then x + 1 = 5

The two integers are 4 and 5.

Questions?