The equation to solve is,
[tex]\frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{(x-1)(x-2)}[/tex]
Let's use algebra to simplify the equation and solve for 'x'. The steps are shown below:
[tex]\begin{gathered} \frac{3}{x-2}=\frac{1}{x-1}+\frac{7}{(x-1)(x-2)} \\ \frac{3}{x-2}=\frac{1(x-2)+7}{(x-1)(x-2)} \\ \frac{3}{x-2}=\frac{x-2+7}{(x-1)(x-2)} \\ \frac{3}{x-2}=\frac{x+5}{(x-1)(x-2)} \\ \text{Cross Multiplication >>>>} \\ 3(x-1)(x-2)=(x-2)(x+5) \\ 3\lbrack x^2-3x+2\rbrack=x^2+3x-10 \\ 3x^2-9x+6=x^2+3x-10 \\ 3x^2-9x+6-x^2-3x+10=0 \\ 2x^2-12x+16=0 \\ x^2-6x+8=0 \\ (x-2)(x-4)=0 \\ x=2,4 \end{gathered}[/tex]
The solution is x = 2 and x = 4.
Looking back at the original question, we can see that we cannot put x = 2 into the equation. It will make the denominator equal to 0. So, we disregard this value of x.
Thus, the only solution is 'x = 4'.
AnswerSolving for
x gives us
x = 4The value of
x cannot equal 2
