In fact this equation is factorable so the solution

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ours) so we need to work rates of each machine. Here are those computations. 1 Job = ( Work Rate of Machine A ) ´ ( t ) 1 Job = ( Work Rate of Machine B ) ´ ( t + 1) Þ Þ 1 t 1 Machine B = t +1 Machine A = Note that it’s okay that the work rates contain t. In fact they will need to so we can solve for it! Plugging into the word equation gives, © 2007 Paul Dawkins 109 http://tutorial.math.lamar.edu/terms.aspx College Algebra æ1ö æ1ö ç ÷ ( 2) + ç ÷ ( 2) = 1 èt ø è t +1 ø 2 2 + =1 t t +1 So, to solve we’ll first need to clear denominators and get the equation in standard form. 2ö æ2 ç+ ÷ ( t )( t + 1) = (1)( t )( t + 1) è t t +1ø 2 ( t + 1) + 2t = t 2 + t 4t + 2 = t 2 + t 0 = t 2 - 3t - 2 Using the quadratic formula gives, t= 3 ± 17 2 Converting to decimals gives, t= 3 + 17 = 3.5616 2 and t= 3 - 17 = -0.5616 2 Again, the negative doesn’t make any sense and so Machine A will work for 3.5616 hours to stuff a batch of envelopes by itself. Machine B will need 4.5616 hours to stuff a batch of envelopes by itself. Again, unlike the first example, note that the time for Machine B was NOT the second solution from...
View Full Document

Ask a homework question - tutors are online