One integer is 3 less than 4 times another. Their product is 27. Find the integers.

another ---- x

one integer ---- 4x-3

x(4x-3) = 27
4x^2 - 3x - 27 = 0

(x-3)(4x+9) = 0
x = 3 or x = -9/4
but x was to be an integer
so
x =3

3 and 9

To find the integers, let's assume one of the integers as x and the other integer as y.

Based on the given information:
1. "One integer is 3 less than 4 times another" translates to x = 4y - 3.
2. "Their product is 27" translates to x * y = 27.

Now, we have a system of two equations:

Equation 1: x = 4y - 3
Equation 2: x * y = 27

To solve the system, we can substitute the value of x from equation 1 into equation 2:

(4y - 3) * y = 27

Now, we can simplify and solve for y:

4y^2 - 3y = 27
4y^2 - 3y - 27 = 0

This is a quadratic equation, so we can solve it by factoring or using the quadratic formula. Since factoring may not be straightforward, let's use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 4, b = -3, and c = -27. Substituting these values:

y = (-(-3) ± √((-3)^2 - 4*4*(-27))) / (2*4)
y = (3 ± √(9 + 432)) / 8
y = (3 ± √441) / 8
y = (3 ± 21) / 8

Now, we have two possible values for y:

y1 = (3 + 21) / 8 = 24 / 8 = 3
y2 = (3 - 21) / 8 = -18 / 8 = -9/4

Now, substitute each value of y back into equation 1 to find the corresponding values of x:

For y1 = 3:
x = 4y - 3
x = 4 * 3 - 3
x = 12 - 3
x = 9

For y2 = -9/4:
x = 4y - 3
x = 4 * (-9/4) - 3
x = -9 - 3
x = -12

Therefore, the two integers are 9 and -12.