IDNLearn.com: Where curiosity meets clarity and questions find their answers. Whether it's a simple query or a complex problem, our experts have the answers you need.
Sagot :
Answer:
Mr. Pinter's class has 42 students, Mrs. Rupert's class has 21 students, and Mrs. Althouse's class has 43 students.
Step-by-step explanation:
This question is solved using a system of equations.
I am going to say that:
Mr. Pinter's class has x students.
Mrs. Rupert's class has y students.
Mrs. Althouse's class has z students.
Mr. Pinter's class has twice as many students as Mrs. Rupert's class.
This means that:
[tex]x = 2y[/tex]
Mrs. Althouse's class has 20 less than three times Mrs. Rupert's class.
This means that:
[tex]z = 3y - 20[/tex]
Together they have 106 students.
This means that:
[tex]x + y + z = 106[/tex]
We have x and z has a function of y, so:
[tex]2y + y + 3y - 20 = 106[/tex]
[tex]6y = 126[/tex]
[tex]y = \frac{126}{6}[/tex]
[tex]y = 21[/tex]
And:
[tex]x = 2y = 2(21) = 42[/tex]
[tex]z = 3y - 20 = 3(21) - 20 = 63 - 20 = 43[/tex]
Mr. Pinter's class has 42 students, Mrs. Rupert's class has 21 students, and Mrs. Althouse's class has 43 students.
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
