Question 336523
Rate*Time=Distance
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{{{R1*T1=D1}}}
{{{R2*T2=D2}}}
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{{{D1=110
{{{D2=80}}}
{{{T1+T2=4}}}
{{{R1=R2+20}}}
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Substituting,
{{{(R2+20)*T1=110}}}
{{{T1=110/(R2+20)}}}
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{{{R2*T2=80}}}
{{{T2=80/R2}}}
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And finally,
{{{T1+T2=4}}}
{{{110/(R2+20)+80/R2=4}}}
{{{110+80(R2+20)=4*R2*(R2+20)}}}
{{{110R2+80R2+1600=4R2^2+80R2}}}
{{{4R^2-110R2-1600=0}}}
Use the quadratic formula,
{{{R2 = (110 +- sqrt( (110)^2-4*4*(-1600) ))/(2*4) }}}
{{{R2 = (110 +- sqrt(12100+25600 ))/(8) }}}
{{{R2 = (110 +- sqrt(37700))/(8) }}}
{{{R2 = (110 +- 10sqrt(377))/(8) }}}
Only the positive result makes sense for this problem.
{{{R2=(110+10sqrt(377))/8}}}
So then 
{{{R1=(110+10sqrt(377))/8+20}}}
{{{R2=(270+10sqrt(377))/8}}} or approximately,
{{{highlight(R2=58.02)}}}mph