How many students are there in Class A?

[eddy2017]

You can’t take a screen shot?

this is it. what you wrote. i need it clearly cos i am gonna file it.
Eddy

I may be confusing the issue, and if so I apologize.

\displaystyle a + b = 75. You are told that in the statement of the problem.

\displaystyle a + c = 61. You are told that in the statement of the problem.

\displaystyle \dfrac{b}{c} = \dfrac{5}{3}. You are told that in the statement of the problem.

Just translate the English into math.

As Subhotosh explained, we SYSTEMATICALLY eliminate variables when we have multiple variables and multiple equations. The third equation already does not have a so that seems like the easiest variable to eliminate. The easiest way to do so in this case (not the only way) is

\displaystyle (a + b) - (a + c) = 75 - 61 \implies b - c = 14 \implies b = c + 14.

We subtracted to get rid of a. Any questions about that?

Now we have two equations in two unknowns, namely

\displaystyle b - c = 14 \text { and } \dfrac{b}{c} = \dfrac{5}{3}.

That last equation is not linear, and I suspect that is what got you lost. But we can fix that issue by multiplying by c.

\displaystyle c * \dfrac{b}{c} = c * \dfrac{5}{3} \implies b = c * \dfrac{5}{3} \implies 3b = 5c.

\displaystyle b = c + 14 \text { and } 3b = 5c \implies 3(c + 14) = 5c.

thanks!!!

 

[lev888]

it shoud have been
i will isolate b for
75-b+c=61 so, (after subtracting from both sides/ simplifying and dividing both sides by -1)
b=c+14

Looks correct. You need to be more careful when substituting expressions: c+14 in c+14/c=5/3 should be (c+14). Somehow the parentheses appeared on the next line, but next time you may forget them.

 

[eddy2017]

Let us look at a solution. We are given that: \(A+B=75~\&~A+C=61\)
Subtracting those two we get \(B-C=14\) Now using the ratio we have:
\(\dfrac{B}{C}=\dfrac{5}{3}\text{ or }3B=5C\text{ or }C=\dfrac{3}{5}B\)
Substituting for \(C\) in the line two above we have \(B-\dfrac{3}{5}B=14\)
Multiplying through by \(5\) we get \(5B-3B=70\text{ or } 2B=70\text{ or } B=35\).
From that we get \(A=40~\&~C=21\).

this one was very straightfoward. i love it!. tx

 

[Deleted member 4993]

this is it. what you wrote. i need it clearly cos i am gonna file it.
Eddy

I may be confusing the issue, and if so I apologize.

\displaystyle a + b = 75. You are told that in the statement of the problem.

\displaystyle a + c = 61. You are told that in the statement of the problem.

\displaystyle \dfrac{b}{c} = \dfrac{5}{3}. You are told that in the statement of the problem.

Just translate the English into math.

As Subhotosh explained, we SYSTEMATICALLY eliminate variables when we have multiple variables and multiple equations. The third equation already does not have a so that seems like the easiest variable to eliminate. The easiest way to do so in this case (not the only way) is

\displaystyle (a + b) - (a + c) = 75 - 61 \implies b - c = 14 \implies b = c + 14.

We subtracted to get rid of a. Any questions about that?

Now we have two equations in two unknowns, namely

\displaystyle b - c = 14 \text { and } \dfrac{b}{c} = \dfrac{5}{3}.

That last equation is not linear, and I suspect that is what got you lost. But we can fix that issue by multiplying by c.

\displaystyle c * \dfrac{b}{c} = c * \dfrac{5}{3} \implies b = c * \dfrac{5}{3} \implies 3b = 5c.

\displaystyle b = c + 14 \text { and } 3b = 5c \implies 3(c + 14) = 5c.

thanks!!!

I strongly suggest you write down these "by hand" so that you would not have "\displaystyle" or "\frac" etc. In addition, writing by hand will reinforce will reinforce your learning.

 

[eddy2017]

I strongly suggest you write down these "by hand" so that you would not have "\displaystyle" or "\frac" etc. In addition, writing by hand will reinforce will reinforce your learning.

yes, i do, Doc. everything i write by hand. while i am working on the answers i write everything. i work problems on paper and then i type them to catalogue them for my future reference, but i love to write them. but i could not make up this piece: ???
Now we have two equations in two unknowns, namely

\displaystyle b - c = 14 \text { and } \dfrac{b}{c} = \dfrac{5}{3}. the { lost me.

That last equation is not linear, and I suspect that is what got you lost. But we can fix that issue by multiplying by c.

\displaystyle c * \dfrac{b}{c} = c * \dfrac{5}{3} \implies b = c * \dfrac{5}{3} \implies 3b = 5c.

by the way, i started watching a tutorial in latex on youtube. i registered for overleaf.com. i want to learn how to write that way.