Given the equation below
[tex]\frac{x+1}{x-3}-\frac{1}{x+4}=\frac{28}{x^2+x-12}[/tex]
To find the value(s) of x, we find the factors of the denominators
[tex]\begin{gathered} \frac{x+1}{x-3}-\frac{1}{x+4}=\frac{28}{x^2+x-12} \\ \text{Where} \\ x^2+x-12=x^2+4x-3x-12 \\ =x(x+4)-3(x+4)=(x-3)(x+4) \\ x^2+x-12=(x-3)(x+4) \\ \text{Substitute (x-3)(x+4) for x}^2+x-12 \\ \frac{x+1}{x-3}-\frac{1}{x+4}=\frac{28}{(x-3)(x+4)} \end{gathered}[/tex]
Since they have a common denominator i.e both the right and left handside, we solve for x by eliminating the denominators
[tex]\begin{gathered} \frac{x+1}{x-3}-\frac{1}{x+4}=\frac{28}{(x-3)(x+4)} \\ \frac{(x+1)(x+4)-(x-3)}{(x-3)(x+4)}=\frac{28}{(x-3)(x+4)} \\ \text{Eliminate the deominators} \\ (x+1)(x+4)-(x-3)=28_{} \end{gathered}[/tex]
Expand the brackets
[tex]\begin{gathered} (x+1)(x+4)-(x-3)=28_{} \\ x(x+4)+1(x+4)-x+3=28 \\ x^2+4x+x+4-x+3=28 \\ \text{Collect like terms} \\ x^2+4x+x-x+4+3-28=0 \\ x^2+4x-21=0 \\ x^2+7x-3x-21=0 \\ \text{Factorize the equation} \\ x(x+7)-3(x+7)=0 \\ (x-3)(x+7)=0 \\ x-3=0 \\ x=3 \\ x+7=0 \\ x=-7 \\ x=3\text{ or -7} \end{gathered}[/tex]
Given that the values of x are 3 or -7
To find which of the solution satisfies the given equation,
Find the undefined points of the equation, i.e
[tex]\begin{gathered} (x-3)(x+4)=0 \\ x=3\text{ or -4} \end{gathered}[/tex]
Since the equation is undefined for x = 3,
Hence, the answer is x = -7
