S., Salomeya

March 2003

Algebra 1-2

Monroy

Per. 7

 

Student Tutorial

 

            Mary works 6 hours for some days at the library, 3 hours for some days at the zoo, and a total of 33 hours. Jane works 4 hours for the same amount of days as Mary at the library, 1 hour for the same amount of days as Mary at the zoo, and a total of 15 hours. How many days did the girls work at the library and how many days did they work at the zoo?

 

Let x = days at the library, y = days at the zoo

 

6x + 3y = 33                                                    6(2) + 3y  = 33

4x + y = 15                                                      

12 + 3y = 33

                                                                        -12 = -12                                                        

6x + 3y = 33
· 3(4x + y = 15)                                              3y = 21                                               
                                                                        Έ3 = 3

6x + 3y = 33
12x + 3y = 45                                                  y = 7

6x + 3y = 33

-(12x + 3y = 45)                                             check: 4x + y = 15

                                                                        4(2) + (7) = 15

-6x = -12                                                         8 + 7 = 15

Έ -6 = Έ -6                                                      15 = 15

 

x = 2                                                                6x + 3y = 33

6(2) + 3(7) = 33

                                                                        12 + 21 = 33

33 = 33

 

            The first step in solving this problem would be to write an equation. Let x stand for the variable for the amount of days at the library and y stand for the number of days at the zoo. The first equation is Mary and the second is Jane. Mary’s equation is 6x + 3y = 33 because six is the number of hours she worked for “x” days and 3 is the number of hours she worked for “y” days. 33 is the total number of hours she worked. The same format is used for Jane’s equation.

            The next step is to decided how to solve my system. The most reasonable method would be elimination. You need to single out either the “x” or “y” variable from both equations. A simple way to do this would be to multiple the second equation by 3 since there are 3 “y”s in the first equation and one “y” in the second. After doing this, you will get the equations 6x + 3y = 33 and 12x + 3y = 45.

Now you can subtract the second equation from the first since the “y”s in both equations are equal. After doing this, you come up with the equation –6x = -12. Now your objective is to single out the “x”. To do this, you need to divide each side by –6 since there are –6 “x”s. Once this is done, you find that the “x” in both equations is 2.

            After this, you need to plug the “x” into one of the equations to find the “y”. Both equations show that, after plugging 2 into each “x” and solving the equation, “y” is 7. It is important to check your equation to make sure you have the right answer. This is a simple process. You need to plug in both the “x” and the “y” into either equation and see that both sides of the equation are equal.

As you can see, after checking the solution, the “x” is 2 and the “y” is 7. This means that Jane and Mary worked 2 hours at the library and 7 hours at the zoo.