Find the speed of a jet given the time of travel back and forth

Question

The problem:

A jet flew from Tokyo to Bangkok, a distance of 4800km. On the return trip, the speed was decreased by 200 km/h. If the difference in the times of the flights was 2 hours, what was the jet's speed from Bangkok to Tokyo?

From the problem, we can see that the variables are:

Distance:

T to B = 4800km

B to T = 4800km

Speed:

T to B = $x$km/h

B to T = $x-200$km/h

Time:

T to B = $x$h

B to T = $x-2$h

The problem definitely needs to be solved using the distance formula, $d=s*t$. Rearranging it for time, we get $t=d/s$

What do I do from here? I know from solving similar previous problems that I have to factor it and get the x-intercepts (one of them will be the jet's speed) but how to I solve up to that point when there are two unknowns?

I would appreciate if you provide the full solution.

Thank you!

Answer 1

$R_1T_1 = 4800$

$R_2T_2 = 4800$

$R_1T_1 = (R_1-200)(T_1+2)$

$R_1T_1 = R_1T_1+2R_1-200T_1-400$

$2R_1-200\left(\frac{4800}{R_1}\right)-400=0$

$2R_1^2-400R_1-960000=0\implies R_1 = -600; +800 = +800$

$R_1 = 800; T_1 = 6$

$R_2 = 600; T_2 = 8$

Answer 2

Let $t_1, v_1$ denote time and speed from T to B and $t_2, v_2$ correspond to time and speed from B to T. So we can have the following relations

$$ v_2=v_1-200\\ t_2= t_1+2 $$

and

$$ 4800 = t_1 v_1 = t_2 v_2 \implies t_1 v_1 = t_2 v_2 \longrightarrow (1)$$

Substituting for $t_2, v_2$ in terms of $t_1,v_1$ in $(1)$ gives

$$ t_1 v_1= (t_1+2)(v_1-200) \implies v_1= 100 t_1 + 200 \longrightarrow (2). $$

substituting $(2)$ in the equation $t_1 v_1 = 4800 $ gives the quadratic equation in $t_1$

$$ t^2_1+2t_1-48 =0 \implies (t_1+8)(t_1-6)=0 \implies t_1 = 6 \implies v_1 =800 \implies v_2 =600. $$