*I'll help you in Psych, English, and Philosophy in return Use the substitution method to solve the following system of equations.x + y = 32x + 2y = 10 (1, 2) (2, 3) No Solutions Infinitely Many Solutions
Incorrect
Something funny is going on with these questions. Are you sure you're putting down the equations correctly?
You're basically saying there is no solution, which I can see
Yes, I'm 100% sure. I don't see how a professor would put out questions that aren't set up correctly, but you know...
x + y = 32x + 2y = 10
"substitution method"
(x,y) = (-1/3,31/3)
If that is the case, then the system looks like this: \[x+y=10\] \[32x+2y=10\] By substitution, y=(10-x) and 32x+2(10-x)=10. Solving for x gives x= -1/3.
Now, this is absolutely right...
Both of you are right, but it all comes down to two digits, or no solution....
I see, From the first equa. x = ....? Then sub. the result into the second equa.
Or an INFINITE amount
I'm calling shenanigans on this situation.
Tell that to my professor, @CliffSedge
Give me his/her contact information and I will.
Change "no solution" to "none of the above"
Now it's your turn to help me in Philosophy: Do you think that Nihilism is a complete deconstruction of Platonic Idealism?
@InsanelyChaotic What the purpose of your showing in the OS site???
What do you think....?
The complete destruction of all meaning and all values would lead to an existence of apathy.....
Have you ever attempted to work on any math?
Of course I have. I've attempted all of these. What's your problem?
Tell your professor not give such confusing tasks..Tell him how he is wrong
*not to give
I'll be sure to
@AbhimanyuPudi and I have shown that this system does have a single, unique solution, so that eliminates the possibility of "no solution" or "infinite solutions." Testing the other two options, {(1,2), (2,3)}: {1,2}+{2,3}=32{1,2}+2{2,3} {3=36=10 False, 5=70=10 False} So none of the answer choices are valid.
Maybe you should email your professor the link to this thread and the other one. Let him/her see our critique.
@CliffSedge well you are doing a great job. you must be very patient i guess
again no option!!
@CliffSedge hahaha..He is gonna piss off
Eh, I think this stuff is fun, so it's worth sticking around to play with it.
" and I have shown that this system does have a single, unique solution" You said it yourself, there is no solution which is provided. @CliffSedge
You are patient, I could never be this patient with math
The answer is (E) "None of the Above."
I agree
If only!
I am not as patient as @CliffSedge is...i'll solve for solution but i don't have patience to get the right equations from the answer choices!!!
Do you both want an example?
Well, take this as a lesson in the importance of having a well-posed problem in mathematics. I think, really, the best thing you can do in this situation is to ignore the answer choices and provide the correct solution.
Ya..whats that example?
I believe that as well, but if that answer is not provided, which makes the student think that their answer is not valid
One sec
re: "I believe that as well, but if that answer is not provided, which makes the student think that their answer is not valid" And that is impetus to go back and check your work, but if you are confident after reworking it that your solution is correct, then you have to either ignore the choices given or assume that the question was misstated.
I agree once again @CliffSedge
5x + y=9 4x + 3y = 16 x= 1 y=4 (1,4) Pick either equation and solve for one of the variables. When you used the elimination method to solve systems of equations, you were able to choose the variable you wanted to eliminate. Here, you will choose the equation and the variable for which you’d like to solve. Since the y variable in the first equation does not have a coefficient, you might find it easiest to isolate. Although you’re welcome to try any of the others! This first equation is five x plus y equals nine. To isolate y, subtract five x from both sides of the equals sign. Now the variable y is written in terms of x. Substitute the expression for the isolated variable into the other equation. Since you know that y has the same value as negative 5 x plus 9, you can replace y in the second equation with that expression, to obtain a new equation. Now solve the new equation. Distribute the 3. Combine like terms. Isolate the variable by subtracting the constant from both sides of the equal sign. Simplify. Divide both sides by negative 11. x equals 1. Half way there! Substitute the value of x into one of the original equations and solve. You may choose either equation, but the second equation is shown here. Substitute one for the variable x. Simplify. Isolate the variable by subtracting the constant from both sides of the equal sign. Simplify. Divide both sides of the equation by three. You found the value for the second variable, y equals 4. The solution to this system of equations is the point 1 comma 4. Remember, this means when you substitute x equals 1 and y equals 4 into both equations you will get true equations. When this system is graphed, two lines cross at the point 1 comma 4.
I find it very odd that you had two problems in a row with that same flaw.
And that is impetus to go back and check your work, but if you are confident after reworking it that your solution is correct, then you have to either ignore the choices given or assume that the question was misstated. You are correct, but I've always personally believed in the professor being correct
Quotation marks above*
Never assume the professor is correct. Trust the math, not the person.
Hahaha..believe me, I never trusted what professors say unless i learn or do it. And i strongly belive in whatever i get as answer.
Both of you are very confident in math
*believe
Thank you for showing that, I understand :)
Mathematics is the only area of human thought where things can be proven to be true with certainty. (There's a little philosophy back atcha)
Where are you guys from?
But back to the example: 5x + y = 9 4x + 3y = 16 subtract -5x on both sides y= -5x +9
Krypton.
its not a little philosophy...lol
@CliffSedge Touche
That was a touch of phil!
(Epistemology is another hobby of mine. I highly recommend the book, "The Beginning of Infinity" by David Deutch) But anyway, back to your example...
"5x + y = 9 4x + 3y = 16 subtract -5x on both sides y= -5x +9" This is all correct so far, and the next step is to solve 4x+3y=16 for x, then sub in the y from eq.1 to eq.2
I'll definitely check it out, thank you. I've been looking for good books to read. Most of my friends suggest Twilight and the Greys book
y= -5 x +9 4x + 3y = 16
y= -5x +9 4x + 3 (-5x+9) = 16
4x = 16 4x - 15x +27 = 16 -11 x +27 = 16 (subtract "-27" on both sides)
-11x = -11 (divide "-11" on both sides)
x= 1
Right-o...
x= 1 4x +3y = 16 4 (1) + 3y = 16 4 +3y= 16 (subtract -4 on both sides)
Equals: 3y=12
3y/3 = 12/3 y= 4
*solution: (1, 4)
Are there any errors....?
No, that's the usual way to do it.
Nope..
Then why is the other question difficult?
The other questions were very different for an important reason.
To get me to "think"?
If you separate them into two equations, the other question would be as easy as the one you solved now
No, I think they were just wrong.
You both have opposing views
I'm fairly sure i didn't separate them correctly.
the questions were not wrong, the answer choices were wrong
The other questions were of the form ax + by = cx + dy = e, which is called a compound equality and is generally bad form; it's a like run-on sentence in English grammar.
But it is a part of math; whether it is good or bad
Sure, but we see the trouble it is leading to.
I'm not a fan of run-on sentences
You get the same form of equations when you go higher in your education...while solving 3D-geometry
Trouble is a common thing for learners...Overcoming them is very important!!!
You both should get an A+
Thank you but why? I am a high-school graduate. So, its quite natural that I know better than school students. You will be better than me if you work hard..just as you solved the above problem
I'm still going with the assumption that the system of equations was not stated correctly. We saw that x+y=32x+2y=10 yielded, via separation, x+y=10 32x+2y=10 a solution that was not an option. I find it more likely that the system is actually supposed to be something else. Let's assume for the moment, that x+y=32x+2y=10 is actually a case of not separating the two equations properly. What if it is actually x+y=3 2x+2y=10 ? Then you will find that there is no solution. Shall I test this hypothesis on the other question?
Hypothesis was tested, and I believe it to be correct. The system was not clearly delineated. They are not compound equalities. They were simply typographical errors; probably from not properly copying the text.
NO SOLUTION!!!!! :)
Join our real-time social learning platform and learn together with your friends!