Qudadratic - Distance / Partial Solution

[Guest]

A and B traveled to meet each other. A left C at the same time B left D. They traveled the direct route from C to D. Upon arriving A traveled 18 more miles than B. A could have made B's time in 15 3/4 days, but B could have made A's time in 28 days. Calculate distance between C and D.


C M D
---------------------------------------------------------------------------
A ---------------------------------> <-----------------------------------B

Let X = CM = distance A traveled.

Let X - 18 = MD = distance B traveled.

X - 18 / 15 3/4 = distance A travels in 1 day.

x / 28 = distance B travels in 1 day.

Note : travel times are equal.


How do I solve ?

[Guest]

Distance B travels from D is 47.3 miles
Distance A travels from C is 47.3 + 18 = 65.3 miles

Distance C to D is 65.3 + 47.3 = 112.6 miles.

[Guest]

Please post your solution. Thanks.

[Guest]

A travels P + 18 miles
B travels P miles

Distance between C and D = (2xP + 18)

(P + 18)/R = P/Q = Time taken to meet.

P/R = 15 3/4 = 63/4 = the days that A said he could have travelled the distance B travelled

(P + 18)/Q = 28 = the days that B said he could have travelled the distance that A travelled

This gives 3 simultaneous equations
Solve for P to get 47.3 miles that B travelled, then work out distance between C and D = (2xP + 18)

[Guest]

Sorry, I omitted the lines " Let P = Distance etc"

Let P = distance B travels from D
Let Q = Speed of B
Let R = Speed of A

A travels P + 18 miles
B travels P miles

Distance between C and D = (2xP + 18)


(P + 18)/R = P/Q = Time taken to meet.

P/R = 15 3/4 = 63/4 = the days that A said he could have travelled the distance B travelled

(P + 18)/Q = 28 = the days that B said he could have travelled the distance that A travelled

This gives 3 simultaneous equations
Solve for P, then work out distance between C and D = (2xP + 18)

[Guest]

I can't solve them.

[Guest]

Solve by isolating the unknowns and substituting them in one of the other equations

(P + 18)/Q = 28 gives us .... Q = (P + 18)/28 OR.... 1/Q = 28/(P + 18)

P/R = 63/4 gives us 63R = 4P OR.... R = 4P/63 OR.... 1/R = 63/4P

Substitute the above into..... (P + 18)/R = P/Q to leave an equation expressed only in terms of P

Then solve for P .......

As you can see the equations are very simple, but if you cannot do this level of simultaneous manipulation you would be better to practice on a few of very simple 2 unknowns equations first.

[Guest]

"Solve by isolating the unknowns and substituting them in one of the other equations

(P + 18)/Q = 28 gives us .... Q = (P + 18)/28 OR.... 1/Q = 28/(P + 18)

P/R = 63/4 gives us 63R = 4P OR.... R = 4P/63 OR.... 1/R = 63/4P

Substitute the above into..... (P + 18)/R = P/Q to leave an equation expressed only in terms of P

Then solve for P ......."


I don't understand.

[Guest]

The first equation is rearranged to give "Q = something" and also I did "1/Q = something"

(P + 18)/Q = 28 gives us .... Q = (P + 18)/28 OR.... 1/Q = 28/(P + 18)

Can you see that Q is isolated on its own to equal something.

It turns out Q is expressed as an equation involving P only on the RHS.

So the RHS can be substituted into the 3rd equation for everywhere you see a Q.

And do the same for the other equation involving R. Substitute it in everywhere you see a R.

Example......

If A = 20
and B = 30

and C = A + B

We can substitute the values for A and B in the equation for C .......

Then ...... C = 20 + 30
So C = 50

This is the same type of thing you have got to do for this problem but it involves more unknowns.
You need to practice some simple equations like the example given before tackling the main problem in this post.