How many students are there in Class A?

eddy2017 · Jun 14, 2021

Hi, dear tutors: i am having difficulties in continuing to solve this ratio problem.
There are 75 students in a classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A?

Step 1
i added class A and B and equaled them to 75. Did the same with the classes A and B

A+B=75

A+C =61

I proceeded to set up the ratio given in the problem.
Ratio of # students in B and C

B/C= 5/3
now, i am at sea about what to do next.
thanks,
eddy

lev888 · Jun 14, 2021

eddy2017 said:
Hi, dear tutors: i am having difficulties in continuing to solve this ratio problem.
There are 75 students in a classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A?

Step 1
i added class A and B and equaled them to 75. Did the same with the classes A and B

A+B=75

A+C =61

I proceeded to set up the ratio given in the problem.
Ratio of # students in B and C

B/C= 5/3
now, i am at sea about what to do next.
thanks,
eddy

You have a system of equations. Have you solved one before? What methods are you familiar with? Heard of substitution?

Deleted member 4993 · Jun 14, 2021

eddy2017 said:
Hi, dear tutors: i am having difficulties in continuing to solve this ratio problem.
There are 75 students in a classes A and B altogether. There are 61 students in classes A and C altogether. The ratio of the number of students in Class B to Class C is 5:3. How many students are there in Class A?

Step 1
i added class A and B and equaled them to 75. Did the same with the classes A and B

A+B=75

A+C =61

I proceeded to set up the ratio given in the problem.
Ratio of # students in B and C

B/C= 5/3
now, i am at sea about what to do next.
thanks,
eddy

You have correctly found 3 equations for 3 unknowns.

A+B=75 .................................................... (1)

A+C =61................................................... (2)

B/C= 5/3.................................................. (3)

Using eqns. (1) and (2) can you eliminate 'A' from those two equations (think subtraction)

eddy2017 · Jun 14, 2021

lev888 said:
You have a system of equations. Have you solved one before? What methods are you familiar with? Heard of substitution?

yes, i have heard of system of equations and substitution. i have watched a couple of videos on youtube about that. let me give a try and i will send my work. the system here would be the first two equations, right?.
these ?.
A+B=75 .................................................... (1)

A+C =61................................................... (2)

thanks to you both!

eddy2017 · Jun 14, 2021

lev888 said:
You have a system of equations. Have you solved one before? What methods are you familiar with? Heard of substitution?

I can do a system of equation by substitution but with a variable with numbers:
like, 2x+y=11
7x-4y=-14
this is pretty easy.
but here I have two equations that have only letters/variables/ no numbers except after the equal sign. can't do it without a hint. i tried but get stuck, like,
A+B=75
A+C =61
2A + BC=136
it is quite different from the system i showed you above.

pka · Jun 14, 2021

eddy2017 said:
I can do a system of equation by substitution but with a variable with numbers:
like, 2x+y=11 & 7x-4y=-14
this is pretty easy.
A+B=75
A+C =61

Subtract and get \(B-C=14\)
Then \(\dfrac{B}{C}=\dfrac{5}{3}\) gives us \(5C=3B\) which gives \(B-\dfrac{3}{5}B=14\)
Can you solve for \(B~?\)

eddy2017 · Jun 14, 2021

pka said:
No,no, no. you subtract and get \(B-C=14\)
Then \(\dfrac{B}{C}=\dfrac{5}{3}\) gives us \(5C=3B\) which gives \(B-\dfrac{3}{5}B=14\)
Can you solve for \(B~?\)

Question # 2
5C=3B
C= which gives......
Did not get it.
can you go step by step there. i do not see why B-3/5 B

pka · Jun 14, 2021

Have you had eight grade mathematics as well as algebra one?
From \(\dfrac{B}{C}=\dfrac{5}{3}\) gives us \(5C=3B\) which gives \(C=\dfrac{3}{5}B\)
So \(B-C=B-\dfrac{3}{5}B=14\)
Thus \(5B-3B=70\) that is what any middle schooler is expected be able to do.

eddy2017 · Jun 14, 2021

pka said:
Have you had eight grade mathematics as well as algebra one?
From \(\dfrac{B}{C}=\dfrac{5}{3}\) gives us \(5C=3B\) which gives \(C=\dfrac{3}{5}B\)
So \(B-C=B-\dfrac{3}{5}B=14\)
Thus \(5B-3B=70\) that is what any middle schooler is expected be able to do.

I'm sorry pka. 5C=3B
and i do nit know why we are doing subtraction. answer my doubts. don't make me waste my time.
i lost you at the proportion. I am not a middle schooler. i am studying algebra and a lot of things math on my own. but i have my doubts. i come to you for help. so i hope if you see your student is not following go step by step. be patient. what kind of help are you giving here. for the gifted?. for the math whizzes?. c'mon for god's sake!

eddy2017 · Jun 14, 2021

i can go to another site where they give you the answers if you want to. i do not like it that way. i want to reason and know why something is like it is. if you volunteer do it with love and kindness and patient. that is the only volunteering i know, my man. you know what happens. it is not about being a volunteer is about being one at heart. you're not a teacher. a teacher should have patience. it is night and i am here at my desktop working and following this problem and you come up with crap like that. stop it!

eddy2017 · Jun 14, 2021

stop making denigrating comments on the knowledge or lack thereof of your students!. it is not nice and do not reflect the civility and culture that people like you should have at all times.

lev888 · Jun 14, 2021

eddy2017 said:
yes, i have heard of system of equations and substitution. i have watched a couple of videos on youtube about that. let me give a try and i will send my work. the system here would be the first two equations, right?.
these ?.
A+B=75 .................................................... (1)

A+C =61................................................... (2)

thanks to you both!

Ok, assuming you now see that there are 3 equations, let's use the substitution method.
The idea is to take one equation, let's say x+2y=3, isolate one variable: x=3-2y, then substitute "3-2y" for x in all _other_ equations. What do we get? We get a system with 1 fewer variable and 1 fewer equation. Repeat until we get a single equation with single variable. Solve it - we now know the value of one variable. We plug this value into one of the previous equations with 2 variables - this makes it a one variable equation. Solve it and repeat the process until we find all variables. Does this agree with videos you watched? Can you start solving your system?

eddy2017 · Jun 14, 2021

lev888 said:
Ok, assuming you now see that there are 3 equations, let's use the substitution method.
The idea is to take one equation, let's say x+2y=3, isolate one variable: x=3-2y, then substitute "3-2y" for x in all _other_ equations. What do we get? We get a system with 1 fewer variable and 1 fewer equation. Repeat until we get a single equation with single variable. Solve it - we now know the value of one variable. We plug this value into one of the previous equations with 2 variables - this makes it a one variable equation. Solve it and repeat the process until we find all variables. Does this agree with videos you watched? Can you start solving your system?

thank you lev, thank you. i was about to give up. thanks. give me time and i will get back with work done and doubts. thanks.
i will reply. it is just late now for me, but i will work on it tomorrow and give you my work.
yes, i think the videos i have been watching deal witth what you are talking about.
this is one i was studying and found it easy. just pasted it here for you to see. you can delete it. don't know if it ios according to forum rules.

skeeter · Jun 14, 2021

Venn diagrams are a great visual aid …

Steven G · Jun 14, 2021

eddy2017 said:
I'm sorry pka. 5C=3B
and i do nit know why we are doing subtraction. answer my doubts. don't make me waste my time.
i lost you at the proportion. I am not a middle schooler. i am studying algebra and a lot of things math on my own. but i have my doubts. i come to you for help. so i hope if you see your student is not following go step by step. be patient. what kind of help are you giving here. for the gifted?. for the math whizzes?. c'mon for god's sake!

Excuse me, but you are expected to think, which is all pka is asking of you.

Cubist · Jun 15, 2021

You can subtract (and also add) equations like this (LHS = left hand side, RHS = right hand side)...

LHS1 = RHS1
LHS2 = RHS2

Subtracting gives...

LHS1 - LHS2 = RHS1 - RHS2

In your case you have two equations:-

A + B = 75
A + C = 61

When you subtract it can be easier think in the following layout (don't write this in an exam)...

Code:

  LHS1     =     RHS1
- LHS2         - RHS2

  (A + B)  =    75
- (A + C)     - 61

  (    B)  =    75
- (    C)     - 61

     B - C =    14

Or doing it on single lines...

(LHS1) - (LHS2) = (RHS1) - (RHS2)

(A + B) - (A + C) = (75) - (61)

A + B - A - C = 14

B - C = 14

pka is direct, and he has a great knowledge of the order that maths should be taught in the educational system. He was just pointing out what stage this is normally taught. I'm sure that he didn't intend any insult. You might have missed the lesson, or the teacher skipped it, or whatever. There's no need to be upset because you can learn it now. As you can see above it's quite easy! Very soon you'll just do these steps in your head.

eddy2017 · Jun 15, 2021

lev888 said:
Ok, assuming you now see that there are 3 equations, let's use the substitution method.
The idea is to take one equation, let's say x+2y=3, isolate one variable: x=3-2y, then substitute "3-2y" for x in all _other_ equations. What do we get? We get a system with 1 fewer variable and 1 fewer equation. Repeat until we get a single equation with single variable. Solve it - we now know the value of one variable. We plug this value into one of the previous equations with 2 variables - this makes it a one variable equation. Solve it and repeat the process until we find all variables. Does this agree with videos you watched? Can you start solving your system?

I will work on the substitution method later. I can't now. Thanks to skeeter for the interesting thing with venn diagram.
Apologies to pka. I just blew up. I take this very seriously. If im in doubt i ask questions. Adress them,pls, or provide a tutorial. It is not a question of thinking. A lot of people do a lot of thinking but sometimes that does not help. If i ask a question please explain it as clearly as possible. Im not here for homework. Thanks. I'll be back with my work.

eddy2017 · Jun 15, 2021

lev888 said:
Ok, assuming you now see that there are 3 equations, let's use the substitution method.
The idea is to take one equation, let's say x+2y=3, isolate one variable: x=3-2y, then substitute "3-2y" for x in all _other_ equations. What do we get? We get a system with 1 fewer variable and 1 fewer equation. Repeat until we get a single equation with single variable. Solve it - we now know the value of one variable. We plug this value into one of the previous equations with 2 variables - this makes it a one variable equation. Solve it and repeat the process until we find all variables. Does this agree with videos you watched? Can you start solving your system?

okay, this is what i have been able to do:
i set up the two equations. i am going to do a system of equations for the first two. don't know to set up B/C=5/3

A+B= 75
A-C=61 ( Now i see that i have to subtract)

i'ma solve this one first
A+B=74 solving for A ( I sub B from both sides)
A+B-B=75-B
A=75-B
Now i will use A to solve the other
A-C=61
solving for C,
A-A-C=61-A
-C=61-A ( Now i will divide both sides of the equation by -1, to get rid of the negative sign)
C= 61÷ -1 =-61 -A÷-1 =A
C = -61+A

Now, i plug the earlier result, A=75-B, in for an plug in the value of C for wherever C is

let me continue to solve for C now
C=-61+A
-61+ 75-B
14-B

i am in doubt here. 14-B is still C, is it the value of C now?>

eddy2017 · Jun 15, 2021

let me rectify. i think i know i made a mistake here. because i have the value of A and i did not plug it in
Now i will use A to solve the other
A-C=61
solving for C,
A-C=61
75-B-C=61 ADD B TO BOTH SIDES
-B -C+75+B= 61+B
-C+75 = B+ 61 I WILL ADD -75 TO BOTH SIDES

-C +75 -75=B+61 -75
-C=B-14 I WILL DIVIDE BOTH SIDES BY -1 TO GET TID OF THE ENGATICE SIGN ON THE VARIABLE

-C/-1= B-14/-1

C= -B + 14
this is the value of C
SO RECTIFIYING
A=75-B
C=-B+14

now, the value of B

A+B= 75 PLUGGIN' IN VALUES
75-B+B=75
and here is where I am in doubt cos B=0 here.

Deleted member 4993 · Jun 15, 2021

eddy2017 said:
okay, this is what i have been able to do:
i set up the two equations. i am going to do a system of equations for the first two. don't know to set up B/C=5/3

A+B= 75
A-C=61 ( Now i see that i have to subtract)

i'ma solve this one first
A+B=74 solving for A ( I sub B from both sides)
A+B-B=75-B
A=75-B
Now i will use A to solve the other
A-C=61
solving for C,
A-A-C=61-A
-C=61-A ( Now i will divide both sides of the equation by -1, to get rid of the negative sign)
C= 61÷ -1 =-61 -A÷-1 =A
C = -61+A

Now, i plug the earlier result, A=75-B, in for an plug in the value of C for wherever C is

let me continue to solve for C now
C=-61+A
-61+ 75-B
14-B

i am in doubt here. 14-B is still C, is it the value of C now?>

There are no other tutor present at the forum right now - I'll get in the foray just to be expedient.

You have 3 "linear" equations - with 3 unknowns.

The general rule of

"solving for 'n' unknowns" from 'n' linear equations is - we systematically reduce the equations, such that we are left with "one equation and 1 unknown".

In this case we plan to reduce the equations - first two equations and two unknowns.

There are several methods (which you may learn later) to reduce the number of equations and variables.

In this case simple observation tells me that if we subtract (2) from (1), we will "eliminate" A - and we get:

B - C = 14 .................................................... (4)

B/C = 5/3.......... → ...... B = 5/3 * C ....... (5)

Now we use (5) in (4) and we get one equation and one unknown

5/3 * C - C = 14

C = 21 (skipped some steps)...................(6)

using (6) in (2) we get .......................................................... A = 40 answer

check

B/C = 5/3 → B = 5/3 * 21 = 35

A + B = 40 + 35 = 75 .........................................checks (1)

A + C = 40 + 21....................................................checks (2)

B/C = 35/21 = 5/3 ............................................. checks (3)

How many students are there in Class A?

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