Answer
The only valid solution is $x = -\frac{3}{2}$, and the point of intersection is $(-\frac{3}{2}, \frac{20}{3})$.
Work Step by Step
We want to find when $f(x) = g(x)$, so we equate the right sides of the equations and solve for $x$:
$\dfrac{32x}{x - 3} - \dfrac{3}{x + 1} = \dfrac{2x + 18}{x^2 - 2x - 3}$
Make the expressions similar using their LCD:
$\dfrac{2x(x + 1)}{(x - 3)(x + 1)} - \dfrac{3(x - 3)}{(x + 1)(x - 3)} = \dfrac{2x + 18}{x^2 - 2x - 3}$
Combine the expressions on the left side::
$\dfrac{2x(x + 1) - 3(x - 3)}{(x - 3)(x + 1)} = \dfrac{2x + 18}{x^2 - 2x - 3}$
Distribute terms in the numerator:
$\dfrac{2x^2 + 2x - 3x + 9}{(x - 3)(x + 1)} = \dfrac{2x + 18}{x^2 - 2x - 3}$
Combine like terms:
$\dfrac{2x^2 - x + 9}{(x - 3)(x + 1)} = \dfrac{2x + 18}{x^2 - 2x - 3}$
Factor the denominator on the right side of the equation:
$\dfrac{2x^2 - x + 9}{(x - 3)(x + 1)} = \dfrac{2x + 18}{(x - 3)(x + 1)}$
Since the denominators on both sides of the equation are the same, we can multiply both sides by the denominator to get rid of the fractions:
$2x^2 - x + 9 = 2x + 18$
Move all terms to the left side of the equation:
$2x^2-x+9-2x-18=0\\
2x^2 - 3x - 9 = 0$
We can now factor by splitting the middle term.
We have a quadratic equation, which is in the form $ax^2 + bx + c = 0$. We need to find which factors multiplied together will equal $ac$ but when added together will equal $b$.
In this equation, $ac$ is $-18$ and $b$ is $-3$. The factors $-6$ and $3$ will work.
Let's rewrite the equation and split the middle term using these two factors:
$2x^2 - 6x + 3x - 9 = 0$
Group the first two terms and the last two terms:
$(2x^2 - 6x) + (3x - 9) = 0$
Factor out the greatest common monomial factor of each group:
$2x(x - 3) + 3(x - 3) = 0$
Factor out $(x-3)$:
$(2x + 3)(x - 3) = 0$
Use the Zero-Product Property by equating each factor equal to $0$, then solve each equation:
First factor:
$2x + 3 = 0$
$2x = -3$
$x = -\frac{3}{2}$
Second factor:
$x - 3 = 0$
$x = 3$
The potential solutions to this system of equations are $x = -\frac{3}{2}$ and $x = 3$.
We can now plug these values into either $f(x)$ or $g(x)$ to find the points of intersection of the two graphs. Let's use $g(x)$.
Let us plug in the first possible solution:
$g(x) = \frac{2(-\frac{3}{2}) + 18}{(-\frac{3}{2})^2 - 2(-\frac{3}{2}) - 3}$
Evaluate the exponent first:
$g(x) = \frac{2(-\frac{3}{2}) + 18}{\frac{9}{4} - 2(-\frac{3}{2}) - 3}$
Multiply to simplify:
$g(x) = \frac{-\frac{6}{2} + 18}{\frac{9}{4} + \frac{6}{2} - 3}$
Convert all terms to equivalent fractions so we can add:
$g(x) = \frac{-\frac{6}{2} + \frac{36}{2}}{\frac{9}{4} + \frac{12}{4} - \frac{12}{4}}$
Add the fractions to simplify:
$g(x) = \frac{\frac{30}{2}}{\frac{9}{4}}$
Multiply the numerator by the reciprocal of the denominator:
$g(x) = \frac{120}{18}$
Simplify the fraction:
$g(x) = \frac{20}{3}$
Let us plug in the second possible solution into $g(x)$:
$g(x) = \frac{2(3) + 18}{(3)^2 - 2(3) - 3}$
Evaluate the exponent first:
$g(x) = \frac{2(3) + 18}{9 - 2(3) - 3}$
Multiply to simplify:
$g(x) = \frac{6 + 18}{9 - 6 - 3}$
Add to simplify:
$g(x) = \frac{24}{0}$
This solution is not valid because the fraction is undefined.
Therefore, the only valid solution is $x = -\frac{3}{2}$, and the point of intersection is $\left(-\frac{3}{2}, \frac{20}{3}\right)$.
