I need help to prepare for the ACT's!!!!

[jessicaharvey31709]

Here is one problem that I cant seem to figure out how to solve. I have ADHD therefore its harder for me to concentrate. I do not want answers, I need someone to show me how its done.


Amy drove the 200 hundred miles to New Orleans at an average speed 10 mph faster than her usual average speed. If she completed the trip in 1 hour less than usual, what is her usual driving speed, in mph?

 

[BigGlenntheHeavy]

\(\displaystyle R \ \ \ \ X \ \ \ \ \ T \ = \ D\)

\(\displaystyle (x+10) \ \ \ X \ \ (y-1) \ \ = \ \ \ 200, \ latest \ going \ to \ New \ Orleans\)

\(\displaystyle x \ \ \ \ \ X \ \ \ y \ \ \ \ = \ 200, \ usual \ going \ to \ New \ Orleans\)

\(\displaystyle Therefore, \ (x+10)(y-1) \ = \ 200 \ = \ xy\)

\(\displaystyle (x+10)(y-1) \ = \ xy, \ xy-x+10y-10 \ = \ xy.\)

\(\displaystyle -x+10y \ = \ 10, \ x \ = \ 10y-10, \ hence \ 10y^2-10y \ = \ 200, \ subbing.\)

\(\displaystyle Hence \ 10(y^2-y-20) \ = \ 0 \ \implies \ 10(y-5)(y+4) \ = \ 0, \ \implies \ y \ = \ 5, \ hence \ x \ = \ 40.\)

\(\displaystyle I'll \ leave \ the \ check \ up \ to \ you.\)

 

[Denis]

Use ye olde formula: speed = distance divided by time.

u = usual speed, t = usual time

u = 200 / t [1]: usual case

u + 10 = 200 (t - 1) [2]: faster case

[1]: ut = 200
[2]: (u + 10)(t - 1) = 200

Are you ok with this; can you finish it off?
If not, ya'll come back now hear?

 

[Denis]

Geezzzz BigG, Jessica asked that we not do it, just show her how to start...