Question 1080325
let x = number of cars going eastbound
let y = number of cars going westbound.


on day 1 you have 3 times as many cars going eastbound as on day 2.
on day 1 you have 1/2 as many cars going westbound as on day 2.


draw a table similar to one shown below:


<pre>


                        east             west             total

day 1                    3x               y                390
day 2                     x              2y                430

total                    4x              3y                820

</pre>


since 3x is 3 times x, your first requirements is satisfied because 3 times as many cars are going east on day 1 as on day 2.


since y is 1/2 of 2y, your second requirement is satisfied because 1/2 as many cars are going west on day 1 as on day 2.


you have 2 equations that need to be solved simultaneously.


they are:


3x + y = 390
x + 2y = 430


multiply both sides of the first equation by 2 and leave the second equation as is to get:


6x + 2y = 780
x + 2y = 430


subtract the second equation from the first to get:


5x = 350


solve for x to get x = 530 / 5 = 70


in the first original equation, solve for y as follows:


start with 3x + y = 390
replace x with 70 to get 3*70 + y = 390
simplify to get 210 + y = 390
subtract 210 from both sides to get y = 390 - 210
simplify to get y = 180


you have:


x = 70
y = 180


substitute in both original equation to get:


3x + y = 390 becomes 3*70 + 180 = 390 which becomes 210 + 180 = 390 which becomes 390 = 390 which is true.


x + 2y = 430 becomes 70 + 2*180 = 430 which becomes 70 + 360 = 430 which becomes 430 = 430 which is true.


the solutions look good.


the question was how many cars were eastbound total and how many cars were westbound total.


3x + x = 4x = 4*70 = 280
y + 2y = 180 + 360 = 540


total cars both ways for both days = 820
total of 280 were eastbound
total of 540 were westbound