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?

1) {x,y,z}={4,-5,3}
2) 200 km

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.

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

Hello, njdevils1833!

Another approach . . .

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.

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a train leaves toronto for montreal at the same time as another train leaves montreal for toronto

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