Question

Solve for $x$:

$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\ne 2,4$

Answer 1

Step 1

Express the given equation in standard quadratic form:

Multiply both sides by $3\left(x-2\right)\left(x-4\right)$

$\begin{array}{rcl}& \Rightarrow & \left(\frac{x-1}{x-2}+\frac{x-3}{x-4}\right)3\left(x-2\right)\left(x-4\right)=\frac{10}{3}\times 3\left(x-2\right)\left(x-4\right)\\ & \Rightarrow & 3\left(x-1\right)\left(x-4\right)+3\left(x-3\right)\left(x-2\right)=10\left(x-2\right)\left(x-4\right)\end{array}$

Expand the above equation:

$\begin{array}{rcl}& \Rightarrow & 3\left(x^2-5x+4\right)+3\left(x^2-5x+6\right)=10\left(x^2-6x+8\right)\\ & \Rightarrow & 3x^2-15x+12+3x^2-15x+18=10x^2-60x+80\\ & \Rightarrow & 6x^2-30x+30=10x^2-60x+80\\ & \Rightarrow & 6x^2-10x^2-30x+60x+30-80=0\\ & \Rightarrow & -4x^2+30x-50=0\\ & \Rightarrow & 4x^2-30x+50=0\\ & \Rightarrow & 2x^2-15x+25=0\end{array}$

Step 2:

Solve the above quadratic equation:

Use factorization method to solve the equation.

$\begin{array}{rcl}& \Rightarrow & 2x^2-10x-5x+25=0\\ & \Rightarrow & 2x\left(x-5\right)-5\left(x-5\right)=0\\ & \Rightarrow & \left(2x-5\right)\left(x-5\right)=0\\ & \Rightarrow & 2x-5=0\mathrm{or}x-5=0\\ & \Rightarrow & x=\frac{5}{2}\mathrm{or}x=5\end{array}$

So, $x=\frac{5}{2}\mathrm{and}x=5$ are the roots of given equation.

Final answer:

Hence, the required solution is $x=\frac{5}{2}\mathrm{and}x=5$.