Question 387257
A car traveled 3276 mi at a certain speed.  If the speed had been 13 mph faster, the trip could have been made in 6 hr less time.  Find the speed.
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let original speed be x mph
increased speed = x+13 mph
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time with original speed - time qith increased speed = 6
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t=d/r
3276/x - 3276/(x+13)=6
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LCD = x(x+13)
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(3276(x+13)-3276x)/x(x+13)=6
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multiply by x(x+13)
3276x+42588-3276x=6x(x+13)
42588=6x^2+78x
6x^2+78x-42588=0
/6
x^2+13x-7098=0
x^2+91x-78x-7098=0
x(x+91)-78(x+91)=0
(x+91)(x-78)=0
{{{(highlight(x= 78))}}} mph the speed
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