Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 685: 49

Answer

$t=1$

Work Step by Step

$\frac{3}{t}-\frac{t^2-2t}{t^3} = \frac{4}{t^2}$ $t^3*(\frac{3}{t}-\frac{t^2-2t}{t^3} = \frac{4}{t^2})$ $\frac{3t^3}{t}-\frac{(t^2-2t)(t^3)}{t^3} = \frac{4t^3}{t^2}$ $3t^2 -(t^2-2t) = 4t$ $3t^2 -t^2+2t = 4t$ $2t^2+2t=4t$ $2t^2+2t-4t=4t-4t$ $2t^2-2t = 0$ $2t(t-1)=0$ $2t=0$ $t=0$ (we can't have zero in the denominator, so this answer is invalid) $t-1=0$ $t=1$ $t=1$ $\frac{3}{t}-\frac{t^2-2t}{t^3} = \frac{4}{t^2}$ $\frac{3}{1}-\frac{1^2-2*1}{1^3} = \frac{4}{1^2}$ $3-(1-2)/1 = 4$ $3-(-1)/1 = 4$ $3--1 = 4$ $4=4$
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