Mr. Kohl has a breaker containing n milliliters of solution to distribute to the students in his chemistry class. If he gives each student 3 milliliters of solution, he will have 5 milliliters left over. In order to give each student 4 milliliters of solution, he will need an additional 21 milliliters. How many students are in the class? *

There were 26 students in his class and the teacher had 83 ml of the solution.

Step-by-step explanation:

Mr. Kohl has a "x" amount of solution, if he divides it by the number of students "n" he'll give each student 3 milliliters  and have a left over of 5 milliliters. If the amount of solution Mr. Kohl had was "x + 21" then he'd be able to give each student 4 milliliters of the solution. From these informations we have:

x = 3*n + 5

(x + 21)/n = 4

x + 21 = 4*n

x = 4*n - 21

Now that we have two equations and two variables we can solve the system of equations, as seen bellow:

3*n + 5 = 4*n - 21

3*n - 4*n = -21 - 5

-n = -26

n = 26

x = 4*26 - 21 = 83 ml

There were 26 students in his class and the teacher had 83 ml of the solution.

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.