Let x be the speed of the first car, then (x + 10) is the speed of the second car.
We can write the equation by the condition of the problem:
[tex] \frac{120}{x}- \frac{120}{x+10}=1 \\ 120(x+10)-120(x)=1(x)(x+10) \\ 120x+1200-120x=x^2+10x \\ x^2+10x-1200=0 \\ D=b^2-4ac=10^2-4*1*(-1200) = 100+4800 = 4900 \\ x_{1,2}= \frac{-bб \sqrt{D} }{2a} \\ \\ x_1= \frac{-10- \sqrt{4900} }{2}= \frac{-10-70}{2} = -\frac{80}{2}=-40 \ \ \ \O [/tex]
[tex]x_2= \frac{-10+ \sqrt{4900} }{2}= \frac{-10+70}{2} = \frac{60}{2}=30 [/tex] mph speed of the first car
Then 30+10=40 mph speed of the second car
Answer: 30mph and 40 mph.
I hope this helps.
