Question
LaTeX: \frac{5}{x^2+4x+3}+\frac{2}{x^2+x-6}=\frac{3}{x^2-x-2}
Answers
GPT-4o mini
To solve the equation
\[
\frac{5}{x^2+4x+3}+\frac{2}{x^2+x-6}=\frac{3}{x^2-x-2},
\]
we first need to factor the quadratics in the denominators.
1. Factor \( x^2 + 4x + 3 \):
\[
x^2 + 4x + 3 = (x + 1)(x + 3).
\]
2. Factor \( x^2 + x - 6 \):
\[
x^2 + x - 6 = (x + 3)(x - 2).
\]
3. Factor \( x^2 - x - 2 \):
\[
x^2 - x - 2 = (x - 2)(x + 1).
\]
Now we substitute these factorizations back into the original equation:
\[
\frac{5}{(x + 1)(x + 3)} + \frac{2}{(x + 3)(x - 2)} = \frac{3}{(x - 2)(x + 1)}.
\]
Next, we find a common denominator for all fractions. The common denominator is
\[
(x + 1)(x + 3)(x - 2).
\]
Now we rewrite each term with the common denominator:
1. For the first term:
\[
\frac{5}{(x + 1)(x + 3)} \cdot \frac{(x - 2)}{(x - 2)} = \frac{5(x - 2)}{(x + 1)(x + 3)(x - 2)}.
\]
2. For the second term:
\[
\frac{2}{(x + 3)(x - 2)} \cdot \frac{(x + 1)}{(x + 1)} = \frac{2(x + 1)}{(x + 1)(x + 3)(x - 2)}.
\]
3. For the right side:
\[
\frac{3}{(x - 2)(x + 1)} \cdot \frac{(x + 3)}{(x + 3)} = \frac{3(x + 3)}{(x - 2)(x + 1)(x + 3)}.
\]
The equation now looks like:
\[
\frac{5(x - 2) + 2(x + 1)}{(x + 1)(x + 3)(x - 2)} = \frac{3(x + 3)}{(x + 1)(x + 3)(x - 2)}.
\]
Since the denominators are the same, we can set the numerators equal to each other:
\[
5(x - 2) + 2(x + 1) = 3(x + 3).
\]
Expanding both sides, we have:
Left side:
\[
5x - 10 + 2x + 2 = 7x - 8.
\]
Right side:
\[
3x + 9.
\]
Now, we equate the two sides:
\[
7x - 8 = 3x + 9.
\]
Subtract \(3x\) from both sides:
\[
4x - 8 = 9.
\]
Add \(8\) to both sides:
\[
4x = 17.
\]
Dividing both sides by \(4\), we find:
\[
x = \frac{17}{4}.
\]
Finally, we should check the solution to ensure it does not make any denominator zero:
- \(x + 1 = \frac{17}{4} + 1 = \frac{21}{4}\) (not zero)
- \(x + 3 = \frac{17}{4} + 3 = \frac{29}{4}\) (not zero)
- \(x - 2 = \frac{17}{4} - 2 = \frac{9}{4}\) (not zero)
Since none of the denominators are zero, the solution is valid. Therefore, the final solution is:
\[
\boxed{\frac{17}{4}}.
\]
\[
\frac{5}{x^2+4x+3}+\frac{2}{x^2+x-6}=\frac{3}{x^2-x-2},
\]
we first need to factor the quadratics in the denominators.
1. Factor \( x^2 + 4x + 3 \):
\[
x^2 + 4x + 3 = (x + 1)(x + 3).
\]
2. Factor \( x^2 + x - 6 \):
\[
x^2 + x - 6 = (x + 3)(x - 2).
\]
3. Factor \( x^2 - x - 2 \):
\[
x^2 - x - 2 = (x - 2)(x + 1).
\]
Now we substitute these factorizations back into the original equation:
\[
\frac{5}{(x + 1)(x + 3)} + \frac{2}{(x + 3)(x - 2)} = \frac{3}{(x - 2)(x + 1)}.
\]
Next, we find a common denominator for all fractions. The common denominator is
\[
(x + 1)(x + 3)(x - 2).
\]
Now we rewrite each term with the common denominator:
1. For the first term:
\[
\frac{5}{(x + 1)(x + 3)} \cdot \frac{(x - 2)}{(x - 2)} = \frac{5(x - 2)}{(x + 1)(x + 3)(x - 2)}.
\]
2. For the second term:
\[
\frac{2}{(x + 3)(x - 2)} \cdot \frac{(x + 1)}{(x + 1)} = \frac{2(x + 1)}{(x + 1)(x + 3)(x - 2)}.
\]
3. For the right side:
\[
\frac{3}{(x - 2)(x + 1)} \cdot \frac{(x + 3)}{(x + 3)} = \frac{3(x + 3)}{(x - 2)(x + 1)(x + 3)}.
\]
The equation now looks like:
\[
\frac{5(x - 2) + 2(x + 1)}{(x + 1)(x + 3)(x - 2)} = \frac{3(x + 3)}{(x + 1)(x + 3)(x - 2)}.
\]
Since the denominators are the same, we can set the numerators equal to each other:
\[
5(x - 2) + 2(x + 1) = 3(x + 3).
\]
Expanding both sides, we have:
Left side:
\[
5x - 10 + 2x + 2 = 7x - 8.
\]
Right side:
\[
3x + 9.
\]
Now, we equate the two sides:
\[
7x - 8 = 3x + 9.
\]
Subtract \(3x\) from both sides:
\[
4x - 8 = 9.
\]
Add \(8\) to both sides:
\[
4x = 17.
\]
Dividing both sides by \(4\), we find:
\[
x = \frac{17}{4}.
\]
Finally, we should check the solution to ensure it does not make any denominator zero:
- \(x + 1 = \frac{17}{4} + 1 = \frac{21}{4}\) (not zero)
- \(x + 3 = \frac{17}{4} + 3 = \frac{29}{4}\) (not zero)
- \(x - 2 = \frac{17}{4} - 2 = \frac{9}{4}\) (not zero)
Since none of the denominators are zero, the solution is valid. Therefore, the final solution is:
\[
\boxed{\frac{17}{4}}.
\]