Why do equations with two distinct variables with 2 distinct linear equations work?

navneet9431

New member
Joined
Oct 30, 2017
Messages
18
My question is about basic algebra. I am thinking about the "why" here and I'm looking for an intuitive answer.


If you have the following equations:
=>S+U=90
and
=>40S+25U=2625
you can then rewrite S=90−U and then substitute.

Now you have a single equation with one variable:
=>40(90−U)+25U=2625
=>3600−40U+25U=2625
=>−15U=−975
Hence, U=65


What's going on here? Ultimately, why does this always solve out? I realize single equations with one variable solve (there's gotta be some number that satisfies this equation), but why? What's going on? I guess by solving the equation, we're bypassing this repetitive process of trial and error of plugging in numbers and seeing if it equals 2625? Is that what "solving the equation" really means?

Note: I am a High School student and English is my second language.Thanks!
 
Last edited:
My question is about basic algebra. I am thinking about the "why" here and I'm looking for an intuitive answer.

If you have the following equations:
=>S+U=90
and
=>40S+25U=2625
you can then rewrite S=90−U and then substitute.

Now you have a single equation with one variable:
=>40(90−U)+25U=2625
=>3600−40U+25U=2625
=>−15U=−975
Hence, U=65

I made a little correction here so that we can see that your work isn't wrong. Of course, you have to finish and check:

S+65=90
S = 25

Check:

S+U=90 ==> 25 + 65 = 90, works
40S+25U=2625 ==> 40(25) + 25(65) = 1000 + 1625 = 2625, works

What's going on here? Ultimately, why does this always solve out? I realize single equations with one variable solve (there's gotta be some number that satisfies this equation), but why? What's going on? I guess by solving the equation, we're bypassing this repetitive process of trial and error of plugging in numbers and seeing if it equals 2625? Is that what "solving the equation" really means?

(Actually, it is not true that there has to be a solution; some equations have none.)

You could say that all of algebra is an easier way of doing something you could do by trial and error; but that doesn't explain anything. Yes, solving means finding a pair that satisfies both equations, which can sometimes be done by trial and error; but what are you really doing? Here is what is happening when you use the substitution method:

We want to find a pair of numbers such that both equations are true. That is, we suppose that (S, U) is a solution, meaning that both equations are true, and we work with these two true equations.

This assumption implies that S is equal to 90-U -- that is, these are two representations of the same number. So wherever S appears, we can replace it with (90-U). Right?

So in the second equation, 40S+25U=2625, we can replace S with (90-U), and this new equation will still be true.

Consequently, under the assumption that (S, U) is a solution of the system, we conclude that it must be (25, 65). And due to other details of our work, we know that this actually is a solution (but we checked it just in case we had made a mistake).

If we had found a contradiction, we would have known that in fact there is no solution. That can happen. And if we used trial and error, we would never realize that.

Do you see something interesting here? It turns out that algebra is not merely a way to find a solution (as trial and error is); it is a way to find out that this is the only solution (which trial and error can never do). So algebra is very powerful!
 
If you have the following equations:

. . .S+U=90

. . .40S+25U=2625

you can then rewrite S=90−U and then substitute....

Ultimately, why does this always solve out?
I'm not sure what you mean by "always solve out"...? If you mean "always arrive at a single x,y-point which solves the system", then these pairs of equations do not actually "always solve out". Sometimes there is no solution. (And sometimes there are infinitely-many solutions!)

A linear equation in two variables graphs as a straight line. Two straight lines are going to cross in one place (usually), no places (if parallel), or all places on the line (if the two equations represent the same line). (You can read a discussion of this here.) In short, "solutions" are "intersections". ;)
 
How does algebra guarantees that this is the only solution?

Thank you so much for such a wonderful answer!
Everything was easy for me to understand except the last few lines.
Do you see something interesting here? It turns out that algebra is not merely a way to find a solution (as trial and error is); it is a way to find out that this is the only solution (which trial and error can never do). So algebra is very powerful!
You said that algebra is a way to find out that this is the only solution.But, how does algebra tells us that (25, 65) are the only solutions?I think it never guarantees that (25, 65) are the only existing solutions for
=>S+U=90
and
=>40S+25U=2625.
Algebra just gives us a POSSIBLE solution of the equations we solved.
so,please tell me how does it guarantees that (25, 65) are the only solutions for equations =>40S+25U=2625 and =>S+U=90 ?

I will be thankful for help!
 
You said that algebra is a way to find out that this is the only solution. But, how does algebra tells us that (25, 65) are the only solutions?
What other solutions would be possible, aside of the one that makes both equations true at the same time? What other intersection point(s) do you expect there to be for these two straight lines?

I think it never guarantees that (25, 65) are the only existing solutions for
=>S+U=90
and
=>40S+25U=2625.
Algebra just gives us a POSSIBLE solution of the equations we solved.
What is your logical basis for this statement? By what logic would, for instance, "x = 3" have more than that one solution? Why would the system "x = 3, y = 2" have solutions other than the point (3, 2)?

I'm sure we'd love to help clear up the confusion, but I'm afraid I'm not clear on what is the source of that confusion...? Kindly please clarify. Thank you! ;)
 
Thank you so much for such a wonderful answer!
Everything was easy for me to understand except the last few lines.

You said that algebra is a way to find out that this is the only solution.But, how does algebra tells us that (25, 65) are the only solutions?I think it never guarantees that (25, 65) are the only existing solutions for
=>S+U=90
and
=>40S+25U=2625.
Algebra just gives us a POSSIBLE solution of the equations we solved.
so,please tell me how does it guarantees that (25, 65) are the only solutions for equations =>40S+25U=2625 and =>S+U=90 ?

The work I did showed that IF a given pair (S, U) is a solution of the system, THEN it must be (25, 65). If you don't see that, then you didn't really understand what I said.

This means that there can be no other solution! Do you see that logic? If not, by what logic do you conclude that there can be another solution?

Also, you should think about the graphical meaning of the system. A solution is an intersection of two lines. There can be no solution (if the lines are parallel), or one solution; if there are more than one solution (that is, more than one intersection of a pair of lines), then the lines must be the same line, and there are infinitely many solutions to the system. That is all that can happen.
 
we never proved that S=25 and U=65 is the ONLY possible solution

The work I did showed that IF a given pair (S, U) is a solution of the system, THEN it must be (25, 65). If you don't see that, then you didn't really understand what I said.

This means that there can be no other solution! Do you see that logic? If not, by what logic do you conclude that there can be another solution?

Also, you should think about the graphical meaning of the system. A solution is an intersection of two lines. There can be no solution (if the lines are parallel), or one solution; if there is more than one solution (that is, more than one intersection of a pair of lines), then the lines must be the same line, and there are infinitely many solutions to the system. That is all that can happen.
So, Ok!This is what I understood from your explanation.
First, we assume that equations =>S+U=90 and =>40S+25U=2625 are true, where S and U are its solutions.From this assumption, we get =>S=90-U
so,S and (90-U) are both representing the same number.Hence, we can replace "S" with "
(90-U)".
Thus, in the second equation, we can replace "S" with "(90-U)" and then by using simple logic we can reach the value U=65.
=>40S+25U=2625 we can
=>40(90−U)+25U=2625
=>3600−40U+25U=2625
=>−15U=−975
Hence, U=65
Again we can replace "U" with "65".So, replacing "U" with 65 in the first equation we get,
=> S=90-U=90-65=25.
And, we have arrived at *A* solution S=25 and U=65.
So, How does finding **a possible solution** guarantees that it is the **ONLY** possible solution?
As far as I see we have nowhere proved or used any logic to determine that S=25 and U=65 are the only possible solutions.(what we have done is that we have just found A POSSIBLE SOLUTION).
I am not saying that there must exist some other solutions also.But, I am just pointing out that we have never proved that S=25 and U=65 are the only possible solutions.

Consequently, under the assumption that (S, U) is a solution of the system, we conclude that it must be (25, 65). And due to other details of our work, we know that this actually is a solution (but we checked it just in case we had made a mistake).
What were "the other details of our work", about which you were talking?

I will be thankful for any HELP!
 
...
And, we have arrived at *A* solution S=25 and U=65.
So, How does finding **a possible solution** guarantees that it is the **ONLY** possible solution?
As far as I see we have nowhere proved or used any logic to determine that S=25 and U=65 are the only possible solutions.(what we have done is that we have just found A POSSIBLE SOLUTION).
I am not saying that there must exist some other solutions also. But, I am just pointing out that we have never proved that S=25 and U=65 are the only possible solutions.

We did prove that! Did you miss the "if ... then" in my answer?

The work we do in solving shows that IF (S,U) is a solution, THEN it must be (25, 65). Do you see this?

Consequently (this is the contrapositive, if you know that term) IF S and U are NOT 25 and 65, THEN they are NOT a solution. Do you see this?

This implies that there can be no solution other than the one we found. Any other pair of numbers would not be a soluNothing in what I said called it a "possible" solution; it is not just "a" solution. It is THE solution you arrive at, because you necessarily arrive at that particular one.

I wonder if you are just not accustomed to how "if" and "then" work in logic. Some of the terminology used, especially concepts like "only", can seem backward if you do not read them carefully; here is a discussion of a related idea.

What were "the other details of our work", about which you were talking?

You're the one who is saying that the work shows that it is a "possible solution"; that is correct - it is actually a solution (which is a very strong way to say it is "possible"). You tell me why that is so! (How does the work you do in solving any equation find a "possible" solution? Hint: think about equivalent equations.)
 
Top