Question
The following rational equation has denominators that contain variables. For this equation, a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. b. Keeping the restrictions in mind, solve the equation.
StartFraction 2 x Over x plus 3 EndFraction equals 7 minus StartFraction 6 Over x plus 3 EndFraction
Question content area bottom
Part 1
a. What are the value or values of the variable that makes the denominators zero?
xequals
enter your response here
StartFraction 2 x Over x plus 3 EndFraction equals 7 minus StartFraction 6 Over x plus 3 EndFraction
Question content area bottom
Part 1
a. What are the value or values of the variable that makes the denominators zero?
xequals
enter your response here
Answers
GPT-4o mini
To find the values of \( x \) that make the denominator zero in the equation
\[
\frac{2x}{x + 3} = 7 - \frac{6}{x + 3}
\]
we first identify the denominators, which are \( x + 3 \) in both cases.
### Part 1: Finding Restrictions
Set the denominator equal to zero:
\[
x + 3 = 0
\]
Solving for \( x \):
\[
x = -3
\]
Thus, the value that makes the denominator zero is
\[
x = -3.
\]
### Part 2: Solve the Equation
Now we keep this restriction in mind and solve the equation, excluding \( x = -3 \) from our solution set.
Starting with the original equation:
\[
\frac{2x}{x + 3} = 7 - \frac{6}{x + 3}
\]
To eliminate the denominators, multiply both sides by \( x + 3 \) (noting that \( x \neq -3 \)):
\[
2x = (7 - \frac{6}{x + 3}) (x + 3)
\]
This expands into:
\[
2x = 7(x + 3) - 6
\]
Distributing on the right side:
\[
2x = 7x + 21 - 6
\]
Simplify:
\[
2x = 7x + 15
\]
Now, we bring the terms involving \( x \) to one side:
\[
2x - 7x = 15
\]
\[
-5x = 15
\]
Dividing both sides by -5:
\[
x = -3
\]
### Conclusion:
Thus, the value of \( x \) that causes the denominator to be zero is:
\[
\boxed{-3}
\]
Since \( x = -3 \) is not in the domain of the original equation (it makes the denominator zero), there are actually **no valid solutions** to the equation that satisfy all conditions.
\[
\frac{2x}{x + 3} = 7 - \frac{6}{x + 3}
\]
we first identify the denominators, which are \( x + 3 \) in both cases.
### Part 1: Finding Restrictions
Set the denominator equal to zero:
\[
x + 3 = 0
\]
Solving for \( x \):
\[
x = -3
\]
Thus, the value that makes the denominator zero is
\[
x = -3.
\]
### Part 2: Solve the Equation
Now we keep this restriction in mind and solve the equation, excluding \( x = -3 \) from our solution set.
Starting with the original equation:
\[
\frac{2x}{x + 3} = 7 - \frac{6}{x + 3}
\]
To eliminate the denominators, multiply both sides by \( x + 3 \) (noting that \( x \neq -3 \)):
\[
2x = (7 - \frac{6}{x + 3}) (x + 3)
\]
This expands into:
\[
2x = 7(x + 3) - 6
\]
Distributing on the right side:
\[
2x = 7x + 21 - 6
\]
Simplify:
\[
2x = 7x + 15
\]
Now, we bring the terms involving \( x \) to one side:
\[
2x - 7x = 15
\]
\[
-5x = 15
\]
Dividing both sides by -5:
\[
x = -3
\]
### Conclusion:
Thus, the value of \( x \) that causes the denominator to be zero is:
\[
\boxed{-3}
\]
Since \( x = -3 \) is not in the domain of the original equation (it makes the denominator zero), there are actually **no valid solutions** to the equation that satisfy all conditions.