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3 Answers

3v +  c = 22  ---> c = 22 - 3v
2v + 4c = 28 ---> 2v + 4(22 - 3v) = 28
                          2v + 88 - 12v = 28
                        -10v + 88 = 28
                                - 88  - 88
                         - 10v = - 60
                               v = 6 students

Comments

This problem can be solved by using the rules of system of equations. You have two situations: 1) when 3 vans and  1 car can transport 22 student; 2) when 2 vans and 4 cars can transport 28 students. Let's name students in the van V; students in the car - C.  Now think this way: in first case 3 vans can take V students (3V) and 1 car - C students (1C).Let's write the equation for this case: 3V+ 1C = 22

Think the same way about the second case: 2 vans can take V students (2V), 4 cars - 4C students. Total number of students in the second case - 28. Let's write the equation for this case: 2V + 4C = 28. 

Now you have two equations of the system, write them one under another and solve by SUBSTITUTION method.

3V+ 1C = 22             1c = -3V + 22   Substitute 1C by its equivalent (-3V + 22) in the green system:                                                                                                                                            2V + 4C = 28.          2v + 4(-3V +22) = 28
                                 2v  - 12v +88= 28
 Combine  Vs together:
                               -10v         + 88 = 28
                                                - 88 - 88
                                - 10v = - 60
                                   v = 6 students

 

Hi Yahnaijah, I don't know where this word problem is from (there are some standardized tests for which you can't assume any starting information), but, for this one I did.  I started out by using my real world knowledge that vans are bigger than cars.  Then, I started plugging in reasonable numbers for van capacity (number of people who can fit into one van).

    Since, I know that 3 doesn't go into 22 evenly, I want to subtract a whole number from 22 to find numbers that will allow 3 to go into it evenly. I started by assuming that only 1 person could fit into a car (this might be weird in real life, but not in a math problem), leaving 21  --->  so each van fits 7 students:

  3V + 1C = 22   --->  3(7) + 1(1) -------->  21 + 1 = 22

but this doesn't work for the second equation: 2V + 4C = 28  ----> 2(6) + 4(1) -----> 12 + 4 = 16 (not 28, which is what we need)

Since the sum is too small, I can deduce that the addends (numbers being added together) are too small.  The only way I can change them is to change the values for the variables V (Vans) and C (Cars).  When one value goes up, the other goes down, because there is only a finite (limited) number of students I can have in total.  

Let's go back to the first equation.  Again, since, I know that 3 doesn't go into 22 evenly, I want to subtract a whole number from 22 to find numbers that will allow 3 to go into it evenly.  The next number down from 21 that is a multiple of 3 is 18.  3 goes into 18, 6 times, so that would mean that 6 students would fit into each Van, V=6.  ---->  That leaves a remainder of 4, which means that 4 students would fit into each Car, C=4.  These numbers seem reasonable in a real world setting, but let's see if the math works out when we plug the numbers into the second equation:

2V + 4C = 28 --------> 2(6) + 4(4) ----------->12 + 16 = 28  ------->  Yay!

 

But, remember to answer the correct question.  The question only asks about Vans and not Cars.  V=6, so 6 students can fit in each Van.

 

Comments

Great, thanks!  I  know it looks wordy, but it wouldn't see so wordy if I were saying it in person.  ;)  Glad I could help.