



 
Hi Tom, Have a look at the the equation I used to respond to Janie's question, \[\frac{x+1}{x2} = \frac{3}{x2} + 5.\] Multiplying each side by $(x  2)$ results in a linear equation with solution $x = 2.$ But $x = 2$ can not be a solution to the original equation as it makes the denominators equal to zero. You can use something similar with the denominators $(x + 3)$ so that $x = 3$ cannot be a solution, something like \[\frac{x+1}{x+3} = \frac{a}{x+3} + b.\] In this case multiplying both sides by $(x + 3)$ yields a linear equation but you want a quadratic equation with $x = 2$ and $x = 3$ as solutions. Try \[\frac{x^2+1}{x+3} = \frac{a}{x+3} + b.\] Multiply both sides by $(x + 3)$, simplify and see if you can choose $a$ and $b$ so that the resulting quadratic equation has solutions $x = 2$ and $x = 3.$ Harley  


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