Question
solve for all values of x
x/x+2 - 5=- 2/x+2
x/x+2 - 5=- 2/x+2
Answers
Answered by
GPT-4o mini
To solve the equation
\[
\frac{x}{x+2} - 5 = -\frac{2}{x+2},
\]
let's first combine the terms on the left-hand side. To get started, we can first eliminate the fraction by isolating the fractions on one side:
\[
\frac{x}{x+2} + \frac{2}{x+2} = 5.
\]
Now we can combine the left side:
\[
\frac{x + 2}{x+2} = 5.
\]
This simplifies to:
\[
\frac{x + 2}{x+2} = 5.
\]
Now, since \(x + 2 \neq 0\), we can multiply both sides by \(x + 2\):
\[
x + 2 = 5(x + 2).
\]
Now we expand the right side:
\[
x + 2 = 5x + 10.
\]
Next, let's move all terms with \(x\) to one side and constant terms to the other side:
\[
2 - 10 = 5x - x,
\]
which simplifies to:
\[
-8 = 4x.
\]
Now, divide both sides by 4:
\[
x = -2.
\]
Now, we need to verify that this solution does not make any denominator zero in the original equation.
Substituting \(x = -2\) into the original denominators \(x + 2\):
\[
-2 + 2 = 0.
\]
Since this value makes the denominator zero, \(x = -2\) is not a valid solution for the equation.
Thus, we conclude that there are **no valid solutions** for the given equation.
\[
\frac{x}{x+2} - 5 = -\frac{2}{x+2},
\]
let's first combine the terms on the left-hand side. To get started, we can first eliminate the fraction by isolating the fractions on one side:
\[
\frac{x}{x+2} + \frac{2}{x+2} = 5.
\]
Now we can combine the left side:
\[
\frac{x + 2}{x+2} = 5.
\]
This simplifies to:
\[
\frac{x + 2}{x+2} = 5.
\]
Now, since \(x + 2 \neq 0\), we can multiply both sides by \(x + 2\):
\[
x + 2 = 5(x + 2).
\]
Now we expand the right side:
\[
x + 2 = 5x + 10.
\]
Next, let's move all terms with \(x\) to one side and constant terms to the other side:
\[
2 - 10 = 5x - x,
\]
which simplifies to:
\[
-8 = 4x.
\]
Now, divide both sides by 4:
\[
x = -2.
\]
Now, we need to verify that this solution does not make any denominator zero in the original equation.
Substituting \(x = -2\) into the original denominators \(x + 2\):
\[
-2 + 2 = 0.
\]
Since this value makes the denominator zero, \(x = -2\) is not a valid solution for the equation.
Thus, we conclude that there are **no valid solutions** for the given equation.
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