Need some major help with algebra?

TRAIN PROBLEM

So, we are trying to find an equation to describe the two trains locations. Imagine the track is a number line starting at 0, and ending at 128.

Train B starts at 0, and increases its location at 68 miles an hour. So, if we let 't' equal the amount of time..

Location = 0 + 68*t

Train A starts at mile 128, and decreases that mile marker at 60 miles an hour.

Location = 128 -60*t

Since were trying to find out when they cross, when their locations are the same, we say

68*t = 128 - 60*t

which solved is.. t = 16

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NUTS PROBLEM

We have two different equations we are given, An equation for the value of the nuts, and an equation for the total weight.

Price:

We will say that the number of lbs mixed nuts is x, and the number of lbs of trail mix is y. The resulting equation is...

2*x + .8*y = 1.2*(x+y) <--we need this (x+y) because the 1.2 is given in price per lb.

Weight:

This is pretty straight forward

x+y = 10

now we need to solve this system of equations

2*x + .8*y = 1.2*(x+y)

2*x + .8*y = 1.2*10 <-- From the second equation

2*x + .8*y = 12

2*x = 12 - .8y

x = 6 - .4y

Now, substitute this into the second equation

x+y=10

6 - .4y + y = 10

6 +.6y = 10

.6y = 4

y = 6.6667

now, re-use our new y value in the second equation to find x

x+y=10

x+6.6667=10

x = 3.33333

There we go, 3.33333 lb mixed nuts, 6.66667 lb trail mix

60 mph in one hour + 68 mph in the same hour but from the opposite direction = 128 mi traveled; so they meet/pass at the 1 hour mark.

Matthew was fine on the train until he messed up his math on the last equation. Always double ck your answers, and everybody else's !!

Parallel. Make the slope the identical and alternative within the two for x and the three for y then resolve. a. y=3x-three Perpendicular. The slope is the reverse reciprocal and plug within the (two,three) once more. b. y=-a million/three X + eleven/three I edited, considering the fact that did perpendicular with reverse and no longer reverse reciprocal.