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.