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 1 Over x minus 3 EndFraction minus StartFraction 3 Over x plus 7 EndFraction equals StartFraction 10 Over x squared plus 4 x minus 21 EndFraction
Question content area bottom
Part 1
a. What is/are the value or values of the variable that make(s) the denominators zero?
xequals
enter your response here
StartFraction 1 Over x minus 3 EndFraction minus StartFraction 3 Over x plus 7 EndFraction equals StartFraction 10 Over x squared plus 4 x minus 21 EndFraction
Question content area bottom
Part 1
a. What is/are the value or values of the variable that make(s) the denominators zero?
xequals
enter your response here
Answers
GPT-4o mini
To determine the values of \( x \) that make the denominators zero in the equation
\[
\frac{1}{x - 3} - \frac{3}{x + 7} = \frac{10}{x^2 + 4x - 21},
\]
we need to find the denominators first:
1. The first denominator is \( x - 3 \), which is equal to zero when \( x = 3 \).
2. The second denominator is \( x + 7 \), which is equal to zero when \( x = -7 \).
3. The third denominator is \( x^2 + 4x - 21 \). We can find the values of \( x \) that make this denominator zero by solving \( x^2 + 4x - 21 = 0 \).
To solve \( x^2 + 4x - 21 = 0 \), we can factor the quadratic:
\[
x^2 + 4x - 21 = (x + 7)(x - 3) = 0.
\]
Setting each factor to zero gives us:
- \( x + 7 = 0 \) which leads to \( x = -7 \)
- \( x - 3 = 0 \) which leads to \( x = 3 \)
Thus, the values that make the denominators zero are:
1. \( x = 3 \)
2. \( x = -7 \)
### Part 1
a. The values of the variable that make the denominators zero are:
\[
x = 3, \, x = -7
\]
### Part 2
Now, let's solve the equation keeping the restrictions in mind:
The equation to solve is:
\[
\frac{1}{x - 3} - \frac{3}{x + 7} = \frac{10}{x^2 + 4x - 21}.
\]
Since we already identified that \( x^2 + 4x - 21 = (x - 3)(x + 7) \), we can rewrite the equation as:
\[
\frac{1}{x - 3} - \frac{3}{x + 7} = \frac{10}{(x - 3)(x + 7)}.
\]
Next, we will combine the fractions on the left-hand side:
To combine these fractions, we need a common denominator, which is \( (x - 3)(x + 7) \):
\[
\frac{(x + 7) - 3(x - 3)}{(x - 3)(x + 7)} = \frac{10}{(x - 3)(x + 7)}.
\]
Now we simplify the numerator on the left-hand side:
\[
(x + 7) - 3(x - 3) = x + 7 - 3x + 9 = -2x + 16.
\]
So now we have:
\[
\frac{-2x + 16}{(x - 3)(x + 7)} = \frac{10}{(x - 3)(x + 7)}.
\]
Since the denominators are equal (provided \( x \neq 3 \) and \( x \neq -7 \)), we set the numerators equal to each other:
\[
-2x + 16 = 10.
\]
Next, we solve for \( x \):
\[
-2x + 16 = 10 \\
-2x = 10 - 16 \\
-2x = -6 \\
x = 3.
\]
However, since \( x = 3 \) is one of our restrictions (it makes the denominator zero), it is not a solution.
Since we have already established our restrictions of \( x \neq 3 \) and \( x \neq -7 \), and we found that the only possible solution leads back to one of the restrictions, we conclude that there are no valid solutions.
### Final result
**No valid solutions exist for the equation according to the restrictions on the variables.**
\[
\frac{1}{x - 3} - \frac{3}{x + 7} = \frac{10}{x^2 + 4x - 21},
\]
we need to find the denominators first:
1. The first denominator is \( x - 3 \), which is equal to zero when \( x = 3 \).
2. The second denominator is \( x + 7 \), which is equal to zero when \( x = -7 \).
3. The third denominator is \( x^2 + 4x - 21 \). We can find the values of \( x \) that make this denominator zero by solving \( x^2 + 4x - 21 = 0 \).
To solve \( x^2 + 4x - 21 = 0 \), we can factor the quadratic:
\[
x^2 + 4x - 21 = (x + 7)(x - 3) = 0.
\]
Setting each factor to zero gives us:
- \( x + 7 = 0 \) which leads to \( x = -7 \)
- \( x - 3 = 0 \) which leads to \( x = 3 \)
Thus, the values that make the denominators zero are:
1. \( x = 3 \)
2. \( x = -7 \)
### Part 1
a. The values of the variable that make the denominators zero are:
\[
x = 3, \, x = -7
\]
### Part 2
Now, let's solve the equation keeping the restrictions in mind:
The equation to solve is:
\[
\frac{1}{x - 3} - \frac{3}{x + 7} = \frac{10}{x^2 + 4x - 21}.
\]
Since we already identified that \( x^2 + 4x - 21 = (x - 3)(x + 7) \), we can rewrite the equation as:
\[
\frac{1}{x - 3} - \frac{3}{x + 7} = \frac{10}{(x - 3)(x + 7)}.
\]
Next, we will combine the fractions on the left-hand side:
To combine these fractions, we need a common denominator, which is \( (x - 3)(x + 7) \):
\[
\frac{(x + 7) - 3(x - 3)}{(x - 3)(x + 7)} = \frac{10}{(x - 3)(x + 7)}.
\]
Now we simplify the numerator on the left-hand side:
\[
(x + 7) - 3(x - 3) = x + 7 - 3x + 9 = -2x + 16.
\]
So now we have:
\[
\frac{-2x + 16}{(x - 3)(x + 7)} = \frac{10}{(x - 3)(x + 7)}.
\]
Since the denominators are equal (provided \( x \neq 3 \) and \( x \neq -7 \)), we set the numerators equal to each other:
\[
-2x + 16 = 10.
\]
Next, we solve for \( x \):
\[
-2x + 16 = 10 \\
-2x = 10 - 16 \\
-2x = -6 \\
x = 3.
\]
However, since \( x = 3 \) is one of our restrictions (it makes the denominator zero), it is not a solution.
Since we have already established our restrictions of \( x \neq 3 \) and \( x \neq -7 \), and we found that the only possible solution leads back to one of the restrictions, we conclude that there are no valid solutions.
### Final result
**No valid solutions exist for the equation according to the restrictions on the variables.**