We know that two jets leave Harrisburg at the same, time, one flying east, and another flying west.
We will denote the speed of the second jet by x (in km/h). Thus, the speed of the first jet is x+20. Remembering that:
[tex]v=\frac{d}{t}[/tex]
where v is speed, d is distance and t is time, we know that for the first jet:
[tex]x+20=\frac{d_1}{4}\Rightarrow4x+80=d_1[/tex]
Where d₁ represents the distance of the first jet from the starting point. For the second jet:
[tex]x=\frac{d_2}{4}\Rightarrow4x=d_2[/tex]
Where d₂ represents the distance of the second jet from the starting point.
We also know that:
[tex]d_1+d_2=6000[/tex]
As:
Thus, we have that:
[tex]\begin{gathered} (4x+80)+(4x)=6000 \\ \text{And solving for x, we get:} \\ 8x+80=6000 \\ 8x=5920 \\ x=\frac{5920}{8}=740 \end{gathered}[/tex]
This means that the second jet has a speed of 740km/h, and the first jet has a speed of 760km/h (20km/h greater than the second one).
