Question
$9- 46$ The given equation is either linear or equivalent to a linear equation. Solve the equation.$$\frac{1}{3-t}+\frac{4}{3+t}+\frac{16}{9-t^{2}}=0$$
Step 1
So, we multiply both sides of the equation by $9-t^2$ to get rid of the denominators. This gives us $$ (9-t^2)\left(\frac{1}{3-t}\right)+(9-t^2)\left(\frac{4}{3+t}\right)+(9-t^2)\left(\frac{16}{9-t^{2}}\right)=0 $$ Show more…
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