Question

There are two factors of -36 such that one factor is 11 less than half of the other factor. Choose all the pairs of these factors:

-2 and 18

-6 and 6

3 and -12

4 and -9

Answer 1

3 and -12

Answer 2

Can set up a system of equations...

xy = -36
x = (y/2) - 11

So we substitute that into the first equation

\[\large (\frac{y}{2} - 11)y = -36\]
\[\large \frac{y^2}{2} - 11y = -36\]
\[\large y^2 - 22y = -72\]
\[\large y^2 - 22y + 121 = 49\]
\[\large (y - 11)^2 = 49\]
\[\large y - 11 = \sqrt{49}\]
This breaks into 2 equations
\[\large y - 11 = 7\]
and
\[\large y - 11 = -7\]

Solving we get
\[\large y = 18\]
\[\large y = 4\]

if those are 2 possible 'y' values...we can us them both to solve for 2 different 'x' values too

xy = -36
x(18) = -36
x = -2

and

xy = -36
x(4) = -36
x = -9

So we have 2 pairs...

-2 and 18

and

-9 and 4