My 12th Problem

The first equation can be rewritten as follows:

$2f(x) + 3g(x) = 2x^2+2x+6$

Subtract this from the second equation, we get:

$-2g(x) = -2x - 8$

or: $-4g(x) = -4x -16$

Add this equation to the first equation, we get:

$f(x) + 3g(x) - 4g(x) = x^2+x+6-4x-16$

$f(x) - g(x) = x^2-3x-10$

We also have:

$f(x) = g(x)$

$\Leftrightarrow f(x) - g(x) = 0$

Which means:

$x^2 - 3x - 10 = 0$

$\Leftrightarrow (x+2)(x-5) = 0$

$\Leftrightarrow x = -2$ or $x = 5$

There is no need to solve for f(x) or g(x) explicitly. Realize that for this special 'x' call it x1, f(x1)=g(x1) = y. Now you can re-write the equations as 4y = x^2 +x + 6 and 6y = 2x^2 +4. And find the points of intersection between these two parabolas.

Equation System:

1ª) -2f(x) - 6g(x) = -2x^2 -2x - 12

2ª) 2f(x) +4g(x) = 2x^2 + 4

Result: -2g(x) = -2x -8 ---> g(x) = x + 4

substituting in the second equation:

2f(x) + 4(x+4) = 2x^2 + 4 ---> 2f(x) = 2x^2 - 4x -12 ---> f(x) = x^2 -2x -6

Equating both:

f(x) = g(x)

x^2 -2x -6 = x + 4 ---> x^2 -3x - 10 = 0 ---> (x + 2)(x - 5)

x = -2 or x = 5