Question
solve for all values of x 3x/x-5 - 2/x+3 =x/x+3
Answers
GPT-4o mini
To solve the equation
\[
\frac{3x}{x-5} - \frac{2}{x+3} = \frac{x}{x+3},
\]
we will first eliminate the fractions by finding a common denominator. The common denominator for the fractions is \((x-5)(x+3)\).
We rewrite each term with this common denominator:
1. For \(\frac{3x}{x-5}\):
\[
\frac{3x}{x-5} = \frac{3x(x+3)}{(x-5)(x+3)}.
\]
2. For \(-\frac{2}{x+3}\):
\[
-\frac{2}{x+3} = -\frac{2(x-5)}{(x-5)(x+3)}.
\]
3. For \(\frac{x}{x+3}\):
\[
\frac{x}{x+3} = \frac{x(x-5)}{(x-5)(x+3)}.
\]
Now we can write the entire equation with a common denominator:
\[
\frac{3x(x+3) - 2(x-5)}{(x-5)(x+3)} = \frac{x(x-5)}{(x-5)(x+3)}.
\]
Since the denominators are the same, we set the numerators equal to each other:
\[
3x(x+3) - 2(x-5) = x(x-5).
\]
Now, we expand each term:
1. Expanding \(3x(x+3)\):
\[
3x^2 + 9x.
\]
2. Expanding \(-2(x-5)\):
\[
-2x + 10.
\]
3. Expanding \(x(x-5)\):
\[
x^2 - 5x.
\]
Putting these together gives us:
\[
3x^2 + 9x - 2x + 10 = x^2 - 5x.
\]
Now we combine like terms:
\[
3x^2 + 7x + 10 = x^2 - 5x.
\]
Next, we move all terms to one side of the equation:
\[
3x^2 + 7x + 10 - x^2 + 5x = 0.
\]
This simplifies to:
\[
2x^2 + 12x + 10 = 0.
\]
Next, we can divide the entire equation by 2:
\[
x^2 + 6x + 5 = 0.
\]
Now we can factor the quadratic:
\[
(x + 1)(x + 5) = 0.
\]
Setting each factor equal to zero gives us:
1. \(x + 1 = 0 \implies x = -1\)
2. \(x + 5 = 0 \implies x = -5\)
Therefore, the solutions are
\[
\boxed{-1} \quad \text{and} \quad \boxed{-5}.
\]
We should also check that these values do not make any denominators zero:
- For \(x = -1\), the denominator \(x - 5 = -1 - 5 = -6 \neq 0\) and \(x + 3 = -1 + 3 = 2 \neq 0\).
- For \(x = -5\), the denominator \(x - 5 = -5 - 5 = -10 \neq 0\) and \(x + 3 = -5 + 3 = -2 \neq 0\).
Both solutions are valid. Thus, the final answers are:
\[
\boxed{-1} \quad \text{and} \quad \boxed{-5}.
\]
\[
\frac{3x}{x-5} - \frac{2}{x+3} = \frac{x}{x+3},
\]
we will first eliminate the fractions by finding a common denominator. The common denominator for the fractions is \((x-5)(x+3)\).
We rewrite each term with this common denominator:
1. For \(\frac{3x}{x-5}\):
\[
\frac{3x}{x-5} = \frac{3x(x+3)}{(x-5)(x+3)}.
\]
2. For \(-\frac{2}{x+3}\):
\[
-\frac{2}{x+3} = -\frac{2(x-5)}{(x-5)(x+3)}.
\]
3. For \(\frac{x}{x+3}\):
\[
\frac{x}{x+3} = \frac{x(x-5)}{(x-5)(x+3)}.
\]
Now we can write the entire equation with a common denominator:
\[
\frac{3x(x+3) - 2(x-5)}{(x-5)(x+3)} = \frac{x(x-5)}{(x-5)(x+3)}.
\]
Since the denominators are the same, we set the numerators equal to each other:
\[
3x(x+3) - 2(x-5) = x(x-5).
\]
Now, we expand each term:
1. Expanding \(3x(x+3)\):
\[
3x^2 + 9x.
\]
2. Expanding \(-2(x-5)\):
\[
-2x + 10.
\]
3. Expanding \(x(x-5)\):
\[
x^2 - 5x.
\]
Putting these together gives us:
\[
3x^2 + 9x - 2x + 10 = x^2 - 5x.
\]
Now we combine like terms:
\[
3x^2 + 7x + 10 = x^2 - 5x.
\]
Next, we move all terms to one side of the equation:
\[
3x^2 + 7x + 10 - x^2 + 5x = 0.
\]
This simplifies to:
\[
2x^2 + 12x + 10 = 0.
\]
Next, we can divide the entire equation by 2:
\[
x^2 + 6x + 5 = 0.
\]
Now we can factor the quadratic:
\[
(x + 1)(x + 5) = 0.
\]
Setting each factor equal to zero gives us:
1. \(x + 1 = 0 \implies x = -1\)
2. \(x + 5 = 0 \implies x = -5\)
Therefore, the solutions are
\[
\boxed{-1} \quad \text{and} \quad \boxed{-5}.
\]
We should also check that these values do not make any denominators zero:
- For \(x = -1\), the denominator \(x - 5 = -1 - 5 = -6 \neq 0\) and \(x + 3 = -1 + 3 = 2 \neq 0\).
- For \(x = -5\), the denominator \(x - 5 = -5 - 5 = -10 \neq 0\) and \(x + 3 = -5 + 3 = -2 \neq 0\).
Both solutions are valid. Thus, the final answers are:
\[
\boxed{-1} \quad \text{and} \quad \boxed{-5}.
\]