• Sep 15th 2012, 06:36 PM
njdevils1833
Hi, i was just wondering if anyone could help me out with these two math questions from my homework.

question 1.) solve the system of equations
x+3y-z=-14
7x+6y+z=1
4x-2y-5z=11

question 2.)a tain leaves toronto for montreal at the same time as another train leaves montreal for toronto. the cities are 500km apart. The trains pass each other 2 hours later. the train leaving montreal is travelling 50km/h faster than the one leaving from toronto. at what distance away from toronto do the trains pass each other?
• Sep 15th 2012, 07:19 PM
MaxJasper
1) {x,y,z}={4,-5,3}
2) 200 km
• Sep 15th 2012, 07:41 PM
MarkFL
1.) One method we may use is elimination. I would begin by taking the first two equations:

\$\displaystyle x+3y-z=-14\$

\$\displaystyle 7x+6y+z=1\$

Notice if we add the equations together, we will eliminate the z:

\$\displaystyle 8x+9y=-13\$

Next, we could take -5 times the first equation and add it to the third to also eliminate z: Then you will have a 2X2 system in x and y, which you may solve, then you will be able to find z from any of you original equations.

2.) Let train 1 be the train that departed Toronto, and train 2 be the train that departed Montreal. We will subscript all variables accordingly. We are being asked to find \$\displaystyle d_1\$

After traveling for 2 hours, the trains meet, so we know the sum of their traveled distances is 500 km:

\$\displaystyle t_1=t_2=2\$

\$\displaystyle d_1+d_2=500\$

We are also told:

\$\displaystyle v_2=v_1+50\$

The relationship between distance, speed and time is \$\displaystyle d=vt\$, so we may state:

\$\displaystyle v_1t_1+v_2t_2=500\$

Using the above information, you can find the quantity in question.
• Sep 16th 2012, 06:09 AM
Wilmer
Quote:

Originally Posted by njdevils1833
a tain leaves toronto for montreal at the same time as another train leaves montreal for toronto. the cities are 500km apart. The trains pass each other 2 hours later. the train leaving montreal is travelling 50km/h faster than the one leaving from toronto. at what distance away from toronto do the trains pass each other?

Toronto(@v mph).........x miles............>[2 hours]<..............500-x miles.............(@v+50)Montreal
Use speed = distance / time ; solve for x
• Sep 16th 2012, 06:58 AM
Soroban
Hello, njdevils1833!

Another approach . . .

Quote:

2) A train leaves Toronto for Montreal at the same time as another train leaves Montreal for Toronto.
The cities are 500km apart. .The trains pass each other 2 hours later.
The train leaving Montreal is travelling 50km/hr faster than the one leaving from Toronto.
At what distance away from Toronto do the trains pass each other?

Let \$\displaystyle v\$ = speed of the train leaving Toronto.
Then \$\displaystyle v+50\$ = speed of the train leaving Montreal.

Code:

```       :  2v    :    2(v+50)    :       *-------->*<--------------*       T        X              M       : - - - - - 500 - - - - - :```
In 2 hours, they meet at \$\displaystyle X.\$
The train from Toronto has traveled \$\displaystyle 2v\$ km.
The train from Montreal has traveled \$\displaystyle 2(v+50)\$ km.

Together, they covered 500 km: .\$\displaystyle 2v + 2(v+50) \:=\:500\$
. . Hence: .\$\displaystyle v \,=\,100\$ km/hr.

Therefore, \$\displaystyle X\$ is: \$\displaystyle 2v = 200\$ km from Toronto.