Question

solve for x.\\(\frac{x + 7}{x + 3}=\frac{x - 6}{x - 7}+1\\)\
if there is more than one solution, separate them with commas.\
if there is no solution, click on
o solution\.\
x = \\(\square\\)

Explanation:

Step1: Find the common denominator

The denominators are \(x + 3\) and \(x - 7\), so the common denominator is \((x + 3)(x - 7)\). Multiply each term by \((x + 3)(x - 7)\) to eliminate the fractions:
\[
(x + 7)(x - 7)=(x - 6)(x + 3)+(x + 3)(x - 7)
\]

Step2: Expand each product

Expand the left side: \((x + 7)(x - 7)=x^{2}-49\) (using the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\))
Expand the right side: \((x - 6)(x + 3)=x^{2}+3x-6x - 18=x^{2}-3x - 18\) and \((x + 3)(x - 7)=x^{2}-7x+3x - 21=x^{2}-4x - 21\)
So the equation becomes:
\[
x^{2}-49=(x^{2}-3x - 18)+(x^{2}-4x - 21)
\]

Step3: Simplify the right side

Combine like terms on the right side:
\[
x^{2}-49=x^{2}-3x - 18+x^{2}-4x - 21
\]
\[
x^{2}-49=2x^{2}-7x - 39
\]

Step4: Move all terms to one side

Subtract \(x^{2}-49\) from both sides:
\[
0 = 2x^{2}-7x - 39-(x^{2}-49)
\]
\[
0 = 2x^{2}-7x - 39 - x^{2}+49
\]
\[
0=x^{2}-7x + 10
\]

Step5: Factor the quadratic equation

Factor \(x^{2}-7x + 10\):
We need two numbers that multiply to \(10\) and add to \(-7\). The numbers are \(-2\) and \(-5\). So:
\[
x^{2}-7x + 10=(x - 2)(x - 5)
\]

Step6: Solve for x

Set each factor equal to zero:
\(x - 2 = 0\) gives \(x = 2\)
\(x - 5 = 0\) gives \(x = 5\)

Step7: Check for extraneous solutions

We need to check if these solutions make the original denominators zero.
For \(x = 2\): \(x+3=5
eq0\), \(x - 7=-5
eq0\), so it is valid.
For \(x = 5\): \(x+3 = 8
eq0\), \(x - 7=-2
eq0\), so it is valid.

Answer:

\(2,5\)