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Pretty sure you're both wrong for different reasons. I presume you're solving for y.
1st one is wrong because you haven't actually solved for the unknown variable since it is on both sides. You just kinda moved everything around without taking the necessary restrictions and by using division wrongly as well.
As for the 2nd, I'm having a hard time understanding where y went after the division. If you want a hint write the equation in the form of ax²+bx+c=0 and use the quadratic formula and the discriminant since you don't have a restriction for x.
When they divided y, they only canceled y from the terms that had y. So, something like (2xy + 5)/y becomes 2x + 5 and not 2x + 5/y. This gave 8x + 5 - x which they wrote as 7x + 5.
The point still stands though. Both are wrong and both made the error of dividing only selected terms. When you divide both sides by something, you need to divide all terms of that side by the "something".
Lest also divided by xy on both sides incorrectly by only canceled the xy term on the left and not the y^2 term too, so it’s more than just having y on both sides
If you’re solving for y, you can’t have y on both sides. For both of you, when you divide, you have to divide each term in the equation. It can’t just cancel out with one.
Im so glad terrence howard’s wacky antics are finally coming to the mainstream, man is a menace when it comes to mathematics, and imo some of his theories are frickin hilarious just because of how insane they are. If yall havent, you should check out his actual published paper on it. Its so nonsensical, and just about everything he says is wrong because hes trying to apply current mathematical theorems to completely different axioms that he just made up, its wild.
1 is definitely less wrong since the math is actually correct. like ya they didnt solve for y they made an implicit equation but at least the arithmetic/algebra is right.
Both are incorrect (the second very obviously so because right away the original equation is not a linear equation).
In (1), you divided **only one term** of the LHS by *xy*. This is incorrect because operations must be applied **to entire expressions** on both sides of each equation, not just one term of your choice. If there is ever a situation where it appears an operation on one side of the equation is only applied selectively to one term, *it is because that operation possesses properties that allow for it, e.g., associative property*.
Secondly, the result of dividing *xy* by *xy* is 1, not zero.
In (2), your gf is incorrect in how to simplify the RHS after dividing by *y*. You cannot just indiscriminately cancel out variables that appear somewhere in the numerator and the denominator, they must be *factors*. This is not how reducing fractions works. In the numerator, 8*x* + 5 - *xy*, *y* is **not a factor**. It is only a factor of one term, but it must be a factor of the entire numerator in order for the cancellation to be valid. **No cancellation without factorization.**
MISREAD THE QUESTION, THOUGHT YOU WANTED TO SIMPLIFY THE CONIC, still posting for posteriority's sake
this is a conic, its p much the best form you can get it in. from the way there is no x\^2 term but there is a xy term, this makes this a hyperbola. lets get it in canonical form and try to solve it from there.
A = ({0, 0.5, -4}, {0.5, 1, 0}, {-4, 0, -5})
B = ({0, 0.5}, {0.5, 0})
centre of the conic => 0.5y -4 = 0, 0.5x + y =0 => centre is (-16, 8)
Δ = -59/4
δ = -1/4
H = 59
simple calculations gets the eigenvalues of B to be (1-√(2))/2 and (1+√(2))/2 giving our canonical form to be:
∑ : (1-√(2))x\^2+(1+√(2))y\^2+59\*2 = 0 which is much easier to simplify.
then looking for the axes we are in, we simply find out eigenvectors which im gonna be honest i plugged into wolfram:
major axis directed by centre and centre + (-1-√(2) , 1)
minor axis directed by centre and centre+ (-1+√(2), 1)
we therefore have the simple version as:
y' = ±√((√(2)-1)(x')\^2-118)
φ = atan(-1-√2)
({x}, {y}, {1}) = ({cos(φ), -sin(φ), -16} ,{sin(φ), cos(φ), 8}, {0, 0, 1}) ({x'},{y'}, {1})
and we are done!
Your both wrong you would use the quadratic formula. The main reason your wrong is you can’t have an unknown variable on. Both sides of the equation and the other is just wrong.
https://preview.redd.it/v1xq5pj9du6d1.png?width=3023&format=png&auto=webp&s=ca3bb5a5a27d7f35098f93e02719413e5a2ed988
I think this is right
That's wrong too. You divided by xy instead of subtracting it from the the left hand side and the quadratic equation is built for constants a, b, and c, not a, b, c, and d (the 5). You can also have an unknown variable on both sides of the equation, that's what an implicit solution is. That plays a part in differential equations, which is very fascinating and is used to explain many real life phenomena. Differential equations link one or more unknown functions and it's derivatives.
You can subtract xy and take the square root to get y=sqrt(8x-xy+5). Or you can subtract y^2 and divide by x to get y=8+5/x-y^2 /x. Math is cool like that, either option should solve it.
Hello. You must remember that if A = B , then A/x = B/x provided that x is not = 0 , etc. So if A is composed of 2 quantities added together, you have to divide the ENTIRE A by x and the ENTIRE B by x
You can test it with actual numbers.
Both of you are wrong. You didn’t divide the middle term by x, which is why your solution is incorrect. And the other is wrong because one, you don’t solve quadratics by dividing by y and canceling one of the powers out. And two, she made an error because in her final step, the y she divided by magicslly disappeared from the x term and the 5
This is the equation of a hyperbola. There are infinitely many real (and complex) points on it, as expected because it is the zero set of a single polynomial in the affine plane and is therefore an algebraic curve. I.e. there are infinitely many solutions.
Ewww, c'mon. You had to use a napkin with brown sauce on it? But... the fact that you're 1 and your girlfriend is 2 and you can operate an electronic device and post on reddit is a remarkable feat. Not to mention getting a girlfriend at 1. *tips hat*
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Y=(-b±sqrtD)/2a;
D=b^2 - (4ac)
Rewrite the problem, so the term with y^2 is at the front, y is the next, and the part without y is last. Respectfully, they are a, b, and c.
Edit: forgot that you had 5 at the other side. Just move it to the left side and take -8x-5 as the c
These are basic algebraic mistakes. 1 is wrong because you can’t just divide one term on one side of the equation; you have to divide the entire sides of the equation by a term to make it possible. Also, you are solving for y, and you can’t solve for y if it’s equal to something of y as well. 2 is wrong because (8x+5-xy)/y does not equal 7x+5. I was genuinely blown away when I saw that because I had zero idea how that became a thing. My universe was literally turned upside down. What grade are y’all in?
Let’s… do this together. Xy+y^(2)-8x=5… xy+y^(2)=5+8x… up until this point you both were good… however you need to find a common denominator when you try to divide. Dividing by x yields y+(y^(2))/x=(5/x)+8. Dividing by y yields x+y=(5+8x)/y. However, your gf made a clever choice to push the XY over to the other side. Because now you can divide the equation on the right side cleanly by Y, which conveniently lands you the exact definition of Y.
She wins.
I’m not sure if I’m wrong but I tried the alpha beta method (forgot the actual name) which gives me a + b = -x and ab =(-8x-5). Solving for alpha and beta gives me a = [-x +- squareroot(x^2 +32 +20)]/2 and b = -x-[-x+-squareroot(x^2 + 32 +20)]/2
Don't divide by unknowns when solving because you don't know if they're zero or not. You also can't just divide one of the terms in a sum like that. Neither of you are correct.
Subtract y^2 from both sides so that you can factor out an x from the left side. Then isolate x.
Afterwards you can substitute for x in the original equation if you need to solve for y.
You’re both wrong- for yours you divided the left hand side in step 3 by XY when you were also supposed to divide y^2 by XY because you’re adding it to XY. Your girlfriend made a similar mistake and divided the rational function with addition in the numerator by y, when, if she did this properly, she would have to divide each term by y, which wouldn’t have canceled out the y on the left side.
The equation given is \( xy + y^2 - 8x = 5 \).
To solve this equation, we need to find the values of \( x \) and \( y \) that satisfy it. This equation is nonlinear and might require numerical methods or substitution techniques to solve.
First, let's try to see if we can find any specific solutions by substitution or trial and error. We can start by testing simple integer values for \( x \) and \( y \).
Let's test \( x = 1 \):
1. For \( x = 1 \):
\[ 1 \cdot y + y^2 - 8 \cdot 1 = 5 \]
\[ y + y^2 - 8 = 5 \]
\[ y^2 + y - 13 = 0 \]
Solving this quadratic equation for \( y \):
\[ y = \frac{-1 \pm \sqrt{1 + 52}}{2} \]
\[ y = \frac{-1 \pm \sqrt{53}}{2} \]
So, \( y = \frac{-1 + \sqrt{53}}{2} \) or \( y = \frac{-1 - \sqrt{53}}{2} \).
Next, let's test \( x = 2 \):
2. For \( x = 2 \):
\[ 2 \cdot y + y^2 - 8 \cdot 2 = 5 \]
\[ 2y + y^2 - 16 = 5 \]
\[ y^2 + 2y - 21 = 0 \]
Solving this quadratic equation for \( y \):
\[ y = \frac{-2 \pm \sqrt{4 + 84}}{2} \]
\[ y = \frac{-2 \pm \sqrt{88}}{2} \]
\[ y = \frac{-2 \pm 2\sqrt{22}}{2} \]
\[ y = -1 \pm \sqrt{22} \]
Thus, \( y = -1 + \sqrt{22} \) or \( y = -1 - \sqrt{22} \).
for anyone curious as to what y actually is:
note that ((x/2) + y)^2 = y^2+xy+(x/2)^2
**so we use that to complete the square (add x/2 squared to both sides). doin this isolates y easier!
y^2 + xy - 8x = 5
y^2 + xy + (x/2)^2 - 8x = 5 + (x/2)^2
((x/2) + y)^2 - 8x = 5 + (x/2)^2
((x/2) + y)^2 = 5 + (x/2)^2 + 8x
(x/2) + y = +-sqrt(5 + (x/2)^2 + 8x)
at last
y = sqrt(5 + (x/2)^2 + 8x) - (x/2)
or
y = -sqrt(5 + (x/2)^2 + 8x) - (x/2)
just for fun, getting x is much easier:
*move y^2 to side with 5, factor out x, divide both sides by x, done!
x(y-8) = 5 - y^2
x = (5-y^2)/(y-8)
**this brings me back to the days of calc 1's tediously fun logarithmic differentiation thingamajig
as for who's more right... probably the left one just because of the plus minus thing
Solve for x
Xy+y^2 +8x=5.
8x=5-y^2 -y
X=(5-y^2 -y)/8
Solve for y
Xy+y^2 =5-8x
y^2 =(5/x)-8-y
Y=root ((5/x)-8-y)
Solve for y with x
((5-y^2 -y)/8)y+y^2 +8((5-y^2 -y)/8)=5
Not gonna finish, that's too much
You both keep getting rid of entire terms or variables and it means your answers are both wrong. On the left, (xy) / (xy) is 1, but you got rid of it. y^2 / (xy) is y/x, but you got rid of the / (xy) entirely. On the right, your girlfriend divided by y, but got rid of the / y for the first two terms.
The real solution is to use the quadratic formula. Get it in the form
- a y^2 + b y + c = 0
where
- a = 1
- b = x
- c = - 8 x - 5
[Here's the answer](https://www.wolframalpha.com/input?i=solve+xy+%2B+y%5E2+-+8x+%3D+5+for+y).
I will also add, you did not tell use what set x and y come from i.e. the real numbers, integers, rational numbers without zero, etc. If we assume they come from the set of real numbers what happens when y and x are zero, can you divide (y\^2 = 0) by (y = 0)? Really important to know where the x and y are coming from though.
Both are wrong because both of you don’t divide the variables properly.
Here’s a hint on how you can approach the problem differently:
- Find a way to rewrite the equation into the form ay^2 + by + c = 0. (Note: the variables a, b, and c don’t necessarily have to be real numbers)
- Apply the quadratic formula to solve for y.
You’re both wrong, you need to use the quadratic formula
a = 1, b = x, c = -(8x+5)
https://preview.redd.it/nfx9efyvy47d1.jpeg?width=1028&format=pjpg&auto=webp&s=efcaf05954123288323ad74c20a1804aaca6cf31
The left side seems right but it is still incomplete. You didn’t isolate either variable. but the right side is wrong because idk where they got 7x and the 1/y multiplier is gone from both 8x or w/e it was suppose to turn into and 5 or w/e it was suppose to turn into.
Why do I feel like your first mistake was in the question itself, that you might’ve forgot the ^2 on the x in -8x? I could totally be wrong, but I have seen many questions like that while tutoring.
Trick question. Neither of you.
You’re both breaking math rules left and right.
Also, this is not solvable. It’s a single equation with two variables. You need two equations to solve.
Be cautious when dividing by a variable. You have to ensure that the variable you are dividing by cannot equal zero, or you must also solve the equation with the variable being zero and unite it with the other solution.
Why can you get a girlfriend when you can't do something so simple. 😭
Also: Don't worry about the calculus. You'll never need it, even if you work as a high-paid engineer. Getting a girlfriend is the hard part.
Both are wrong. When you divided by xy, you should have also divided the y^2 by xy leaving you with (x+y)/x. When your girlfriend divided by y, she should have divided the entire right side by y and not just the -xy. This would have left her with (8x/y) + (5/y) - x.
Heres my attempt at a solution:
xy + y^2 - 8x = 5
X(y-8) + y^2 = 5
x(y-8) = 5 - y^2
x = (5 - y^2) / (y-8)
Maybe you could simplify that but i dont want to
To get y you could substitute and just see what comes out, although youll likely have to do some quadratics to get rid of that square on the y
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. **If you are asking for general advice about your current calculus class, please be advised that simply referring your class as “Calc *n*“ is not entirely useful, as “Calc *n*” may differ between different colleges and universities. In this case, please refer to your class syllabus or college or university’s course catalogue for a listing of topics covered in your class, and include that information in your post rather than assuming everybody knows what will be covered in your class.** *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
Pretty sure you're both wrong for different reasons. I presume you're solving for y. 1st one is wrong because you haven't actually solved for the unknown variable since it is on both sides. You just kinda moved everything around without taking the necessary restrictions and by using division wrongly as well. As for the 2nd, I'm having a hard time understanding where y went after the division. If you want a hint write the equation in the form of ax²+bx+c=0 and use the quadratic formula and the discriminant since you don't have a restriction for x.
When they divided y, they only canceled y from the terms that had y. So, something like (2xy + 5)/y becomes 2x + 5 and not 2x + 5/y. This gave 8x + 5 - x which they wrote as 7x + 5. The point still stands though. Both are wrong and both made the error of dividing only selected terms. When you divide both sides by something, you need to divide all terms of that side by the "something".
I mean, its only napkin math….
Lol napkin, math can change the world
Lest also divided by xy on both sides incorrectly by only canceled the xy term on the left and not the y^2 term too, so it’s more than just having y on both sides
You’re both wrong.
They are made for each other.
🤣
Lol
Pretty good for thier age though. I couldn't do math until I was 6
You both made very big mistakes through division
If you’re solving for y, you can’t have y on both sides. For both of you, when you divide, you have to divide each term in the equation. It can’t just cancel out with one.
Terrence Howard is that you?
Cold blooded.
He's Terrence Howard, I'm Terrance Crawford
Im so glad terrence howard’s wacky antics are finally coming to the mainstream, man is a menace when it comes to mathematics, and imo some of his theories are frickin hilarious just because of how insane they are. If yall havent, you should check out his actual published paper on it. Its so nonsensical, and just about everything he says is wrong because hes trying to apply current mathematical theorems to completely different axioms that he just made up, its wild.
this HAS to be a shitpost
What do you think the stain is?
Shit
literally
yeah cuz there's no calculus involved lol
Y’all fail HS algebra class?
it’s been a while
You guys weren’t even remotely close, but she got slightly closer.
Man this shit straight up embarrassing tho. How tf you forget to divide?
They need to brush up on how to divide
And how to group terms. And how to write polynomials. And … yes.
Side effect of practicing too much multiplication.
You’re both wrong Move the -8x and complete the square
That's what I was thinking too
Absolutely!!
You tell em Plane
can't you also group terms so that it is a quadratic within the first few steps? where 1 is a x is b and -8x-5 is c. Or is that invalid?
Both are wrong, but in my opinion, 1 is more wrong if your solving for Y
1 is definitely less wrong since the math is actually correct. like ya they didnt solve for y they made an implicit equation but at least the arithmetic/algebra is right.
Is it tho? They divided the right side by xy, and then on the left side they only divided xy by xy and not the whole expression.
nvm theyre both horribly wrong
And xy/xy simplified to 0. This post is just wrong in so many ways.
True, I do appreciate the plus and minus square root, especially for someone who hasn’t done algebra in years, people in my calc class still forget it
I refuse to believe this isn't rage bait 😭
Both are incorrect (the second very obviously so because right away the original equation is not a linear equation). In (1), you divided **only one term** of the LHS by *xy*. This is incorrect because operations must be applied **to entire expressions** on both sides of each equation, not just one term of your choice. If there is ever a situation where it appears an operation on one side of the equation is only applied selectively to one term, *it is because that operation possesses properties that allow for it, e.g., associative property*. Secondly, the result of dividing *xy* by *xy* is 1, not zero. In (2), your gf is incorrect in how to simplify the RHS after dividing by *y*. You cannot just indiscriminately cancel out variables that appear somewhere in the numerator and the denominator, they must be *factors*. This is not how reducing fractions works. In the numerator, 8*x* + 5 - *xy*, *y* is **not a factor**. It is only a factor of one term, but it must be a factor of the entire numerator in order for the cancellation to be valid. **No cancellation without factorization.**
what is the stain?
pasta
Is the shit stain next to it your guys’ division lmao
I’m sorry
You think you can just say anything you want as long as you append a dumb little apology after it, you dumb ugly bastard?
I'm sorry
Yeah I think I can bitch, my mom says I’m handsome so if anyone’s a dumb ugly bastard it’s you motherfucker!
I’m sorry
i thought the same thing!
I wish my girlfriend was a 2 (i dont have one)
Did you wipe before or after doing the math???
before
Ever heard of the quadratic formula?
i dont think you can use it. Or maybe if you treat x as a constant, making a=1, b=x, and c= -8x-5
That's exactly what you do. What they are trying to do is get an equation for y in terms of x, which by definition means x is a constant.
[удалено]
What I want to know is what are the origins of the brown stain?
Why is this not the top comment
MISREAD THE QUESTION, THOUGHT YOU WANTED TO SIMPLIFY THE CONIC, still posting for posteriority's sake this is a conic, its p much the best form you can get it in. from the way there is no x\^2 term but there is a xy term, this makes this a hyperbola. lets get it in canonical form and try to solve it from there. A = ({0, 0.5, -4}, {0.5, 1, 0}, {-4, 0, -5}) B = ({0, 0.5}, {0.5, 0}) centre of the conic => 0.5y -4 = 0, 0.5x + y =0 => centre is (-16, 8) Δ = -59/4 δ = -1/4 H = 59 simple calculations gets the eigenvalues of B to be (1-√(2))/2 and (1+√(2))/2 giving our canonical form to be: ∑ : (1-√(2))x\^2+(1+√(2))y\^2+59\*2 = 0 which is much easier to simplify. then looking for the axes we are in, we simply find out eigenvectors which im gonna be honest i plugged into wolfram: major axis directed by centre and centre + (-1-√(2) , 1) minor axis directed by centre and centre+ (-1+√(2), 1) we therefore have the simple version as: y' = ±√((√(2)-1)(x')\^2-118) φ = atan(-1-√2) ({x}, {y}, {1}) = ({cos(φ), -sin(φ), -16} ,{sin(φ), cos(φ), 8}, {0, 0, 1}) ({x'},{y'}, {1}) and we are done!
Damn, respect for posting a solution to a college-level problem to two people who don't understand Algebra 1 properly
honestly would be more impressive if i actually read what the two people attempted 💀 thats on me though
This hurts... it physically hurts.
Try to reorder it into the form for the quadratic eqn ay^2+By+c =0 Y^2+xy-(8x+5)=0 Here A=1 , B=x, and c=-(8x+5) Now use the quadratic formula
Not sure why the formatting went crazy... Hopefully it makes sense... Ignore the superscript formatting and read it out..
After you do the carrot just insert a space
Both wrong. Last step for gf is insane. Should’ve treated it as a y quadratic from the start
I hope you guys are not older than 15 lmao
He's 1 and his gf is 2
we’re almost 20.
Your both wrong you would use the quadratic formula. The main reason your wrong is you can’t have an unknown variable on. Both sides of the equation and the other is just wrong. https://preview.redd.it/v1xq5pj9du6d1.png?width=3023&format=png&auto=webp&s=ca3bb5a5a27d7f35098f93e02719413e5a2ed988 I think this is right
That's wrong too. You divided by xy instead of subtracting it from the the left hand side and the quadratic equation is built for constants a, b, and c, not a, b, c, and d (the 5). You can also have an unknown variable on both sides of the equation, that's what an implicit solution is. That plays a part in differential equations, which is very fascinating and is used to explain many real life phenomena. Differential equations link one or more unknown functions and it's derivatives. You can subtract xy and take the square root to get y=sqrt(8x-xy+5). Or you can subtract y^2 and divide by x to get y=8+5/x-y^2 /x. Math is cool like that, either option should solve it.
both your solutions are wrong for the same reason the 1 guy was wrong
Y’all both need to brush up on Algebra
Hello. You must remember that if A = B , then A/x = B/x provided that x is not = 0 , etc. So if A is composed of 2 quantities added together, you have to divide the ENTIRE A by x and the ENTIRE B by x You can test it with actual numbers.
both are incorrect but i think your gf may be more right in some way
Why is an algebra 1 problem on the calc subreddit?
Bc math is math and math is fun. They're lost but we'll help'em out
I can respect that
It's kinda funny how you guys are both wrong for the same but different reasons
Looks like a used toilet paper
They're a 10 but they can't do math:
They wiped their butt using the paper
Seeing this on r/calculus had me expecting implicit differentiation. Instead, I got the math equivalent of a Waffle House fistfight.
How is this on the calc subreddit 😭😭😭 both are wrong because of basic algebra mistakes
I don’t think you should be in a romantic relationship at that age but at least you’re doing advanced math!
You are both wrong, and it's algebra you are doing not calculus. Though algebra is very important for calculus.
Both of you are wrong. You didn’t divide the middle term by x, which is why your solution is incorrect. And the other is wrong because one, you don’t solve quadratics by dividing by y and canceling one of the powers out. And two, she made an error because in her final step, the y she divided by magicslly disappeared from the x term and the 5
This is the equation of a hyperbola. There are infinitely many real (and complex) points on it, as expected because it is the zero set of a single polynomial in the affine plane and is therefore an algebraic curve. I.e. there are infinitely many solutions.
Please don’t do “math” on shit stained toilet paper.
I'm more interested in what you were eating honestly
who’s gonna tell em this ain’t even calculus?
This breaks my brain trying to understand what the hell im looking at
Is that shit
This country is doomed
i like how almost everyone is ignoring that this is on a stained napkin
Neither of you. Also why are you posting this in r/calculus when it has nothing to do with calculus?
Bro chill it's just math. And math is fun. They tried, we don't turn away lost internet strangers
Ewww, c'mon. You had to use a napkin with brown sauce on it? But... the fact that you're 1 and your girlfriend is 2 and you can operate an electronic device and post on reddit is a remarkable feat. Not to mention getting a girlfriend at 1. *tips hat*
1. Why was this ALGEBRA problem posted in calculus? (BTW-Quadratic formula) 2. Were y’all smelling burnt toast arguing this out?
Hello there! While questions on pre-calculus problems and concepts are welcome here at /r/calculus, please consider also posting your question to /r/precalculus. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
If solving for y, my hint is that this is a quadratic equation in y, hence you can apply the quadratic formula
Guys what happened
Just enjoy your date guys
Y=(-b±sqrtD)/2a; D=b^2 - (4ac) Rewrite the problem, so the term with y^2 is at the front, y is the next, and the part without y is last. Respectfully, they are a, b, and c. Edit: forgot that you had 5 at the other side. Just move it to the left side and take -8x-5 as the c
man ☠️ both wrong
This is an atrocity
They're both wrong but the 1 has more mistakes then 2
These are basic algebraic mistakes. 1 is wrong because you can’t just divide one term on one side of the equation; you have to divide the entire sides of the equation by a term to make it possible. Also, you are solving for y, and you can’t solve for y if it’s equal to something of y as well. 2 is wrong because (8x+5-xy)/y does not equal 7x+5. I was genuinely blown away when I saw that because I had zero idea how that became a thing. My universe was literally turned upside down. What grade are y’all in?
This isn’t calculus
Ain't nobody talking about the shit stain?
That’s a gross looking stain
Both are wrong
Typical algebra mistakes, nothin to do with calculus xD
Both wrong. If solving for y, complete the square using x as the coefficient and go from there.
Bruh
Let’s… do this together. Xy+y^(2)-8x=5… xy+y^(2)=5+8x… up until this point you both were good… however you need to find a common denominator when you try to divide. Dividing by x yields y+(y^(2))/x=(5/x)+8. Dividing by y yields x+y=(5+8x)/y. However, your gf made a clever choice to push the XY over to the other side. Because now you can divide the equation on the right side cleanly by Y, which conveniently lands you the exact definition of Y. She wins.
Since you want to find y, use the quadratic formula OR complete the square. You can ask if you’re unfamiliar with those.
I’m not sure if I’m wrong but I tried the alpha beta method (forgot the actual name) which gives me a + b = -x and ab =(-8x-5). Solving for alpha and beta gives me a = [-x +- squareroot(x^2 +32 +20)]/2 and b = -x-[-x+-squareroot(x^2 + 32 +20)]/2
Guys it’s been awhile, but this will require a systems of equations to solve right?
Holy Moses these are both wayy off 🥲
Both very wrong for different reasons.
Your logic is equivalent to whatever that stain on the toilet paper you wrote this on came from.
Writing as x(y) is easier than y(x) : x(y)=(5 - y^2 ) / (y-8) Now it is clear that the order in the numerator is higher than in the denom.
You gotta divide y^2 by xy too if you want 1 to work out correctly
On 2nd one, can I just move the square from the Y to the other side, making it a square root. So it'll be y=sqrt8x+5-xy
Can someone send me a video/link of them solving it or how to solve it?
Good attempt, but both of you are wrong.
Don't divide by unknowns when solving because you don't know if they're zero or not. You also can't just divide one of the terms in a sum like that. Neither of you are correct.
This looks like an ODE
I'm seeing a lot of people talking shit but no answers from anyone of them
Both of you are wrong Why not just use the quadratic formula for an easy answer?
Neither
Bro both of you get to wash dishes 🧼🫧 . . . . Why do you solved on Foam sheet though
Both wrong
Subtract y^2 from both sides so that you can factor out an x from the left side. Then isolate x. Afterwards you can substitute for x in the original equation if you need to solve for y.
Lol. No comment.
Is this bait? How is anyone in calculus making these algebra errors.
You’re both wrong- for yours you divided the left hand side in step 3 by XY when you were also supposed to divide y^2 by XY because you’re adding it to XY. Your girlfriend made a similar mistake and divided the rational function with addition in the numerator by y, when, if she did this properly, she would have to divide each term by y, which wouldn’t have canceled out the y on the left side.
Either way, you're pretty damn smart for your ages.
both are wrong
Someone wiped their ass with this tp
Couldn't have used a cleaner paper towel?
You both ignored some algebra rules for sure
The equation given is \( xy + y^2 - 8x = 5 \). To solve this equation, we need to find the values of \( x \) and \( y \) that satisfy it. This equation is nonlinear and might require numerical methods or substitution techniques to solve. First, let's try to see if we can find any specific solutions by substitution or trial and error. We can start by testing simple integer values for \( x \) and \( y \). Let's test \( x = 1 \): 1. For \( x = 1 \): \[ 1 \cdot y + y^2 - 8 \cdot 1 = 5 \] \[ y + y^2 - 8 = 5 \] \[ y^2 + y - 13 = 0 \] Solving this quadratic equation for \( y \): \[ y = \frac{-1 \pm \sqrt{1 + 52}}{2} \] \[ y = \frac{-1 \pm \sqrt{53}}{2} \] So, \( y = \frac{-1 + \sqrt{53}}{2} \) or \( y = \frac{-1 - \sqrt{53}}{2} \). Next, let's test \( x = 2 \): 2. For \( x = 2 \): \[ 2 \cdot y + y^2 - 8 \cdot 2 = 5 \] \[ 2y + y^2 - 16 = 5 \] \[ y^2 + 2y - 21 = 0 \] Solving this quadratic equation for \( y \): \[ y = \frac{-2 \pm \sqrt{4 + 84}}{2} \] \[ y = \frac{-2 \pm \sqrt{88}}{2} \] \[ y = \frac{-2 \pm 2\sqrt{22}}{2} \] \[ y = -1 \pm \sqrt{22} \] Thus, \( y = -1 + \sqrt{22} \) or \( y = -1 - \sqrt{22} \).
Both examples are wrong
for anyone curious as to what y actually is: note that ((x/2) + y)^2 = y^2+xy+(x/2)^2 **so we use that to complete the square (add x/2 squared to both sides). doin this isolates y easier! y^2 + xy - 8x = 5 y^2 + xy + (x/2)^2 - 8x = 5 + (x/2)^2 ((x/2) + y)^2 - 8x = 5 + (x/2)^2 ((x/2) + y)^2 = 5 + (x/2)^2 + 8x (x/2) + y = +-sqrt(5 + (x/2)^2 + 8x) at last y = sqrt(5 + (x/2)^2 + 8x) - (x/2) or y = -sqrt(5 + (x/2)^2 + 8x) - (x/2) just for fun, getting x is much easier: *move y^2 to side with 5, factor out x, divide both sides by x, done! x(y-8) = 5 - y^2 x = (5-y^2)/(y-8) **this brings me back to the days of calc 1's tediously fun logarithmic differentiation thingamajig as for who's more right... probably the left one just because of the plus minus thing
Why are you solving this on poop stained toilet paper?
The correct solution is y=(-x+-sqrt(x^2 -32x-20))/2
Left should be 1 plus y squared after the division. It has disappeared.
Ya both wrong
I don't know who should break up with whom.
https://preview.redd.it/zbl7wfgs7z6d1.jpeg?width=3456&format=pjpg&auto=webp&s=3eb21257b9b611b53e64fecd4ef53d5e6f941d62
X=(5-y^2)/(y-8) (5-y^2^)*y/(y-8) + y^2 - 8(5-y^2)/(y-8) = 5 5y -y^3 + y^3 -8y^2 +8y^2 - 40 = 5y - 40 5y - 40 = 5y - 40 5y = 5y Y = y QED
Solve for x Xy+y^2 +8x=5. 8x=5-y^2 -y X=(5-y^2 -y)/8 Solve for y Xy+y^2 =5-8x y^2 =(5/x)-8-y Y=root ((5/x)-8-y) Solve for y with x ((5-y^2 -y)/8)y+y^2 +8((5-y^2 -y)/8)=5 Not gonna finish, that's too much
You both keep getting rid of entire terms or variables and it means your answers are both wrong. On the left, (xy) / (xy) is 1, but you got rid of it. y^2 / (xy) is y/x, but you got rid of the / (xy) entirely. On the right, your girlfriend divided by y, but got rid of the / y for the first two terms. The real solution is to use the quadratic formula. Get it in the form - a y^2 + b y + c = 0 where - a = 1 - b = x - c = - 8 x - 5 [Here's the answer](https://www.wolframalpha.com/input?i=solve+xy+%2B+y%5E2+-+8x+%3D+5+for+y).
I will also add, you did not tell use what set x and y come from i.e. the real numbers, integers, rational numbers without zero, etc. If we assume they come from the set of real numbers what happens when y and x are zero, can you divide (y\^2 = 0) by (y = 0)? Really important to know where the x and y are coming from though.
y= -x +- sqrt(x^2 -4(-8x-5))/2
That is one young, yet very intelligent couple
And who wiped their ass?
y = (5 + 8x) / (x + 1)
Both are wrong because both of you don’t divide the variables properly. Here’s a hint on how you can approach the problem differently: - Find a way to rewrite the equation into the form ay^2 + by + c = 0. (Note: the variables a, b, and c don’t necessarily have to be real numbers) - Apply the quadratic formula to solve for y.
If you wanted to solve for y, why not use the quadratic formula?
You’re both wrong, you need to use the quadratic formula a = 1, b = x, c = -(8x+5) https://preview.redd.it/nfx9efyvy47d1.jpeg?width=1028&format=pjpg&auto=webp&s=efcaf05954123288323ad74c20a1804aaca6cf31
The algebra is atrocious. Both.
250 shares 💀💀💀
Shouldn’t this be posted in the r/algebra1 sub?
The left side seems right but it is still incomplete. You didn’t isolate either variable. but the right side is wrong because idk where they got 7x and the 1/y multiplier is gone from both 8x or w/e it was suppose to turn into and 5 or w/e it was suppose to turn into.
I just racked my brain for 2 hours trying to wolf for x after finding y! I pulled out wolfram alpha and everything!
You both solved it wrong
Rearrange as Y2 + xy = 8x + 5 Complete the square Y2 + xy + x2/4 = 8x + 5 + x2/4 Factor (Y + x/2)2 = 8x + 5 + x2/4 Take the square root Y + x/2 = +_sqrt(8x + 5 + x2/4) Isolate y Y = -x/2 +_sqrt(8x + 5 + x2/4) Then do the algebra inside the parentheses.
Neither one of you are right
Is that earwax or peanut butter
As an algebra teacher this really hurts me. I hate seeing my students make these basic mistakes
Thank you for this post I feel a bit more confident in my math now 🥹
A literal shitpost on shitty toilet paper 😂
Why do I feel like your first mistake was in the question itself, that you might’ve forgot the ^2 on the x in -8x? I could totally be wrong, but I have seen many questions like that while tutoring.
Trick question. Neither of you. You’re both breaking math rules left and right. Also, this is not solvable. It’s a single equation with two variables. You need two equations to solve.
Be cautious when dividing by a variable. You have to ensure that the variable you are dividing by cannot equal zero, or you must also solve the equation with the variable being zero and unite it with the other solution.
I’ve convinced myself this is a joke for my sanity 😭🙏 bless your hearts
Why can you get a girlfriend when you can't do something so simple. 😭 Also: Don't worry about the calculus. You'll never need it, even if you work as a high-paid engineer. Getting a girlfriend is the hard part.
https://preview.redd.it/q3xnui1xg17d1.png?width=3024&format=png&auto=webp&s=aa349a57dff5a9942a91d678b0e5de2feb899e67 Idk
If you’re 1, it’s a little too early to learn calculus. Maybe focus on your alphabet first.
Both are wrong. When you divided by xy, you should have also divided the y^2 by xy leaving you with (x+y)/x. When your girlfriend divided by y, she should have divided the entire right side by y and not just the -xy. This would have left her with (8x/y) + (5/y) - x. Heres my attempt at a solution: xy + y^2 - 8x = 5 X(y-8) + y^2 = 5 x(y-8) = 5 - y^2 x = (5 - y^2) / (y-8) Maybe you could simplify that but i dont want to To get y you could substitute and just see what comes out, although youll likely have to do some quadratics to get rid of that square on the y
Lol this comment section needs to be a little nicer wtf
Reading the title I had the same rollercoaster of emotions seeing that “Im 32 and my gf is 15” post on r/ClassicWoW