<p>i'm trying to help my brother w/ his algebra hw and i can't figure out this problem. </p>
<p>t/(t-5) = 1/(t+5) - 17/(25 - t^2)</p>
<p>solve for t</p>
<p>You multiply the denominator and numerator for the first fraction by t+5, multiply the denominator and numerator for the second fraction by t-5, then subtract the first fraction and solve when equal to zero.</p>
<p>The (25-t^2) can be broken down into (5-t)(5+t) = -(t-5)(t+5)
Multiply each term by this to eliminate fractions, simplify and solve the quadratic.</p>
<p>Edit: Or what Zachsta said.</p>
<p>this is simple.</p>
<p>ill do it step by step for you</p>
<p>t/(t-5) = 1/(t+5) + 17/(t^2 - 25)
t/(t-5) = 1/(t+5) + 17/(t + 5)(t - 5)</p>
<p>multiply t/(t-5) by (t+5)/(t+5). multiply 1/(t+5) by (t-5)/(t-5)</p>
<p>now you have a common denominator of (t+5)(t-5)</p>
<p>so youll get t^2 + 4t + 5 = 17</p>
<p>t can be either -6 or 2</p>
<p>thanks guys i just forgot to make (25 - t^2) common with the other denominators</p>
<p>Has anyone heard of heard of a TI-89? ;)</p>