Math Snap

STEP 1

What is this asking?
We need to find all the values of $x$ that make this equation true, and then add them together!
Watch out!
Remember to check if any solutions make the denominator zero, because that's a no-no!

STEP 2

1. Rewrite the equation
2. Solve for $x$
3. Check for invalid solutions
4. Calculate the sum

STEP 3

Let's make this equation easier to work with.
We can do this by getting rid of those pesky fractions!
Notice that $(x+1)$ and $(x-1)$ are in the denominators.
So, we can multiply both sides of the equation by $2 \cdot (x+1) \cdot (x-1)$ to get rid of the fractions.
Remember, we're multiplying every term by this!

STEP 4

$$2 \cdot (x+1) \cdot (x-1) \cdot \frac{x-1}{x+1} + 2 \cdot (x+1) \cdot (x-1) \cdot \frac{4}{x-1} = 2 \cdot (x+1) \cdot (x-1) \cdot \frac{5}{2}$$ Now, we can divide to one!

STEP 5

$$2 \cdot (x-1) \cdot (x-1) + 2 \cdot (x+1) \cdot 4 = (x+1) \cdot (x-1) \cdot 5$$ $$2(x-1)^2 + 8(x+1) = 5(x+1)(x-1)$$

STEP 6

Expand everything!
$$2(x^2 - 2x + 1) + 8x + 8 = 5(x^2 - 1)$$ $$2x^2 - 4x + 2 + 8x + 8 = 5x^2 - 5$$

STEP 7

Combine like terms and move everything to one side to get a nice quadratic equation:
$$2x^2 + 4x + 10 = 5x^2 - 5$$ $$0 = 3x^2 - 4x - 15$$

STEP 8

We can use the quadratic formula to solve for $x$!
Remember, for an equation of the form $ax^2 + bx + c = 0$, the solutions are given by:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our case, $a = \mathbf{3}$, $b = \mathbf{-4}$, and $c = \mathbf{-15}$.

STEP 9

Let's plug those values in:
$$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-15)}}{2 \cdot 3}$$ $$x = \frac{4 \pm \sqrt{16 + 180}}{6}$$$$x = \frac{4 \pm \sqrt{196}}{6}$$$$x = \frac{4 \pm 14}{6}$$

STEP 10

So, we have two possible solutions:
$$x = \frac{4 + 14}{6} = \frac{18}{6} = 3$$ $$x = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3}$$

STEP 11

Remember, we can't have $x$ values that make the denominator zero in the original equation!
Let's check:
• If $x = 3$, the denominators are $3+1=4$ and $3-1=2$.
These are fine!
• If $x = -\frac{5}{3}$, the denominators are $-\frac{5}{3} + 1 = -\frac{2}{3}$ and $-\frac{5}{3} - 1 = -\frac{8}{3}$.
These are also fine!

STEP 12

Now, we just add our solutions together:
$$3 + \left(-\frac{5}{3}\right) = 3 - \frac{5}{3} = \frac{9}{3} - \frac{5}{3} = \frac{4}{3}$$

The sum of the solutions is $\frac{4}{3}$.
So the answer is C!