x/x-1 = 1/x-1 It would appear that x=1 but that would give us an undefined quantity. Explain please.
x/(x-1) = 1/(x-1) It would appear that x=1 but that would give us an undefined quantity. Explain please.
Thank you for your help with this. Also may I ask about this one: if x^2 + xy = 20 and y^2 + xy = 30 what would xy = ?
Thank you for your help with this. Also may I ask about this one: if x squared +xy=20 and y squared +xy=30 what would xy= ?
If there is a solution to this system of equations, it can be found by substitution.Thank you for your help with this. Also may I ask about this one: if x squared +xy=20 and y squared +xy=30 what would xy= ?
An algebraic expression does not have an equal sign, that is by design it does not have an answer like x= 1 or x = 3. You can simplify an expression by combining like terms and/or factoring. What you gave us is an equation, since it has an equal sign, and one may or may not be able to solve it. In your case, as many have pointed out, this equation has no solution.x/x-1 = 1/x-1 It would appear that x=1 but that would give us an undefined quantity. Explain please.
(x^2 + xy) + (y^2 +xy) = 50xy = 30 - y^2 = 20 - x^2
What can you tell us from there?
x(x+y) = 20 ....................(1)
y(x+y) = 30 ....................(2)
x+y = 30/x = 20/y
Anything strike you, yet?
(x^2 + xy) + (y^2 +xy) = 50
x+ y = √50 = \(\displaystyle \pm\)5√2 ................. (3)
Divide (1) by (2).............. x = 1.5 * y...........(4)
using (4) in (3)
2.5*x = \(\displaystyle \pm\)5√2
x = \(\displaystyle \pm\)2√2
y = 2/3 * \(\displaystyle \pm\)2√2 = \(\displaystyle \pm\)(4/3)*√2
You made a silly error. I hope the (khan) academy doesn't hear about it.(x^2 + xy) + (y^2 +xy) = 50
x+ y = √50 = \(\displaystyle \pm\)5√2 ................. (3)
Divide (1) by (2).............. x = 1.5 * y...........(4)
using (4) in (3)
2.5*y = \(\displaystyle \pm\)5√2
y = \(\displaystyle \pm\)2√2
x = 2/3 * \(\displaystyle \pm\)2√2 = \(\displaystyle \pm\)(4/3)*√2
Yes ... I did - instead of starting from my own work I tried to steal somebody else's work ...You made a silly error. I hope the (khan) academy doesn't hear about it.
Shame, shame, double shame.........I tried to steal somebody else's work ...
Dr. Peterson, I am perplexed by your answer.Unfortunately, equations (1) and (2) are wrong!
But this is essentially one of the ways I was suggesting; I can see many approaches that will work.
But what is xy? No one has finished yet ...
Dr. Peterson, I am perplexed by your answer.
It is true that there are multiple methods that will work, but, as I explained before, substitution is a very general method that easily works for this problem. Moreover, it is generally considered adequate to find x and y rather than their product. Nevertheless, xy = 12 is explcitly calculated twice my prior post in this thread.
Would you please be so kind as to explain what is incorrect or incomplete in my prior post? Thank you.
So it does. I am so used to solving problems for the unknowns that habit took over. Thank you.Maybe I'm not seeing something, but the problem asks "what would xy = ?", and I don't think anyone ever concluded their answer by saying the answer to that question. If I were grading your answer on a test, I would take off something for not having explicitly answered the question asked. (I've been assuming that everyone, like me, was just leaving the last step for the OP to do, but I thought it was worth pointing this out.)
Now this represents a pedagogical difference of opinion. I like to start students off on systems of simultaneous equations with two general ideas, namely that there may not be any solution and that substitution is a very general method of solution. I agree that substitution is not always a quick method, but it stops students from saying that they "have no idea where to start."Also, I never said that anything you did was wrong; only that, when the question is about xy rather than x or y, I tend to wonder if there is a direct way to get that without first solving for x and y separately -- some contest-type problems have such shortcuts. And I haven't found a quick, elegant way to that answer. Often, substitution is the long way to a solution, so looking for other ways is a good habit, anyway.
If there is a solution to this system of equations, it can be found by substitution.
\(\displaystyle \text {Given: } x^2 + xy = 20 \text { and } y^2 + xy = 30.\)
\(\displaystyle x = 0 \implies x^2 + xy = 0 + 0 = 0 \ne 20 \implies \text { a contradiction.}\)
\(\displaystyle \therefore x \ne 0.\text { And so, } x^2 + xy = 20 \implies y = \dfrac{20 - x^2}{x}.\)
\(\displaystyle \therefore 30 = y^2 + xy = \left ( \dfrac{20 - x^2}{x} \right )^2 + x * \dfrac{20 - x^2}{x} \implies\)
\(\displaystyle 30 = \dfrac{400 - 40x^2 + x^4}{x^2} + 20 - x^2 = \dfrac{400 - 20x^2}{x^2} \implies\)
\(\displaystyle 30x^2 = 400 - 20x^2 \implies 50x^2 = 400 \implies x^2 = 8 = 4 * 2 \implies x = 2\sqrt{2} \text { or } -\ 2\sqrt{2}.\)
\(\displaystyle \therefore y = \dfrac{20 - 8}{2\sqrt{2}} = \dfrac{6}{\sqrt{2}} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2} \text { or}\)
\(\displaystyle y = -\ 3\sqrt{2}.\)
NOW CHECK.
\(\displaystyle (2 \sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2}) = 4 * 2 + 6 * 2 = 8 + 12 = 20.\)
\(\displaystyle (3\sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2} = 9 * 2 + 6 * 2 = 18 + 12 = 30.\)
So \(\displaystyle x = 2\sqrt{2} \text { and } y = 3\sqrt{2}\) is a valid answer.
And \(\displaystyle (-\ x)^2 + (-\ x)(-\ y) = x^2 + xy \text { and }(-\ y)^2 + (-\ x)(-\ y) = y^2 + xy.\)
So \(\displaystyle x = -\ 2\sqrt{2} \text { and } y = -\ 3\sqrt{2}\) is also a valid answer.
You can just use the quadratic formula to solve one equation for one variable, and substitute that into the other. ...
Now this represents a pedagogical difference of opinion. I like to start students off on systems of simultaneous equations with two general ideas, namely that there may not be any solution and that substitution is a very general method of solution. I agree that substitution is not always a quick method, but it stops students from saying that they"have no idea where to start."