Question30. Solve: $\frac{x+3}{x^{2}+5 x+6}=\frac{3}{2 x+4}-\frac{1}{x+3}$

Studdy Solution

STEP 1

What is this asking? We need to find the value of $x$ that makes this crazy equation with fractions and $x$ all over the place true! Watch out! We need to be careful about values of $x$ that would make the denominators zero, because dividing by zero is a big no-no in math!

STEP 2

1. Simplify the denominators
2. Combine the fractions
3. Solve for x
4. Check for extraneous solutions

STEP 3

Let's factor those denominators!
We can rewrite $x^2 + 5x + 6$ as $(x+2) \cdot (x+3)$ .
This helps us see what values of $x$ would make the denominator zero.

STEP 4

The equation then becomes: $\frac{x+3}{(x+2) \cdot (x+3)} = \frac{3}{2 \cdot (x+2)} - \frac{1}{x+3}$ Notice how $2x + 4$ can be rewritten as $2 \cdot (x+2)$ .
This makes things much clearer!

STEP 5

We can simplify the left side by dividing both the numerator and the denominator by $(x+3)$ , which gives us: $\frac{1}{x+2} = \frac{3}{2 \cdot (x+2)} - \frac{1}{x+3}$ Remember, we're assuming $x+3$ isn't zero, so $x$ can't be $-3$ .
We'll keep that in mind!

STEP 6

To combine the fractions on the right side, we need a common denominator.
Let's multiply the first fraction by $\frac{x+3}{x+3}$ and the second fraction by $\frac{2 \cdot (x+2)}{2 \cdot (x+2)}$ .
This gives us: $\frac{1}{x+2} = \frac{3 \cdot (x+3)}{2 \cdot (x+2) \cdot (x+3)} - \frac{2 \cdot (x+2)}{2 \cdot (x+2) \cdot (x+3)}$

STEP 7

Now we can subtract those fractions: $\frac{1}{x+2} = \frac{3x+9 - (2x+4)}{2 \cdot (x+2) \cdot (x+3)}$ $\frac{1}{x+2} = \frac{x+5}{2 \cdot (x+2) \cdot (x+3)}$

STEP 8

Let's multiply both sides of the equation by $2 \cdot (x+2) \cdot (x+3)$ to get rid of those pesky denominators!
We get: $2 \cdot (x+3) = x+5$ Remember, we're assuming $x$ is not $-2$ and $x$ is not $-3$ , because those values would make our denominators zero!

STEP 9

Distribute the $2$ on the left side: $2x+6 = x+5$

STEP 10

Subtract $x$ from both sides: $x+6 = 5$

STEP 11

Subtract $6$ from both sides to isolate $x$ : $x = -1$

STEP 12

Our potential solution is $x = -1$ .
Is this allowed?
Well, it doesn't make any of our original denominators zero, so we're good!

STEP 13

$x = -1$ is the solution!

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