Question
Solve for $x$ :
$
\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} ; x \neq 3,5
$
$
\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} ; x \neq 3,5
$
✓
Answer
Consider the equation:
$\begin{array}{l}\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} \\\Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3} \\
\Rightarrow \frac{x^2-5 x-2 x+10+x^2-3 x-4 x+12}{x^2-3 x-5 x+15}=\frac{10}{3}\end{array}$
$\begin{array}{l}\Rightarrow \frac{2 x^2-14 x+22}{x^2-8 x+15}=\frac{10}{3} \\\Rightarrow 3\left(2 x^2-14 x+22\right)=10\left(x^2-8 x+15\right) \\\Rightarrow 6 x^2-42 x+66=10 x^2-80 x+150 \\\Rightarrow 10 x^2+150-6 x^2-66-80 x+42 x=0 \\4 x^2-38 x+84=0 \\2\left(2 x^2-19 x+42\right)=0 \\2 x^2-19 x+42=0\end{array}$
Splitting the middle term,
$\begin{array}{l}2 x^2-12 x-7 x+42=0 \\2 x(x-6)-7(x-6)=0 \$x-6)(2 x-7)=0 \\x-6=0 \quad \text { or } 2 x-7=0 \\x=6 \quad \text { or } \quad x=\frac{7}{2}\end{array}$
$\begin{array}{l}\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} \\\Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3} \\
\Rightarrow \frac{x^2-5 x-2 x+10+x^2-3 x-4 x+12}{x^2-3 x-5 x+15}=\frac{10}{3}\end{array}$
$\begin{array}{l}\Rightarrow \frac{2 x^2-14 x+22}{x^2-8 x+15}=\frac{10}{3} \\\Rightarrow 3\left(2 x^2-14 x+22\right)=10\left(x^2-8 x+15\right) \\\Rightarrow 6 x^2-42 x+66=10 x^2-80 x+150 \\\Rightarrow 10 x^2+150-6 x^2-66-80 x+42 x=0 \\4 x^2-38 x+84=0 \\2\left(2 x^2-19 x+42\right)=0 \\2 x^2-19 x+42=0\end{array}$
Splitting the middle term,
$\begin{array}{l}2 x^2-12 x-7 x+42=0 \\2 x(x-6)-7(x-6)=0 \$x-6)(2 x-7)=0 \\x-6=0 \quad \text { or } 2 x-7=0 \\x=6 \quad \text { or } \quad x=\frac{7}{2}\end{array}$
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