Question 154399
Remember:
Speed of the 2 planes in {{{miles/hour}}}
{{{Plane1=P[1]= x(mi/hr)}}}
{{{Plane2=P[2]=2x(mi/hr)}}},  twice the speed of {{{P[1]}}}
As they left the airport, the 2 plane made a combined distance of {{{2700miles}}} in {{{3hours}}}. So,
{{{P[1]+P[2]=2700miles/3hours}}} -------------------------> working eqn
{{{x+2x=cross(2700)900/cross(3)}}}
{{{3x=900(mi/hr)}}}
{{{cross(3)x/cross(3)=cross(900)300/cross(3)}}}
{{{x=300(mi/hr)}}} ----------------------------> speed of {{{P[1]}}}
Also, {{{P[2]=2*300=600(mi/hr)}}}, speed of {{{P[2]}}}
In doubt? Go back working eqn:
{{{300+2(300)=2700/3}}}
{{{300+600=900}}}
{{{900(mi/hr)=900(mi/hr)}}}
Thank you,
Jojo