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x = 1 is fine AS WRITTEN.x/x-1 = 1/x-1 It would appear that x=1 but that would give us an undefined quantity. Explain please.

If you mean this: \(\displaystyle \dfrac{x}{x-1} = \dfrac{1}{x-1}\) then x = 1 is NOT fine.

Use parentheses to clarify intent.

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Presumably you solved by multiplying both sides by (x-1), or something equivalent. Then you get x = 1. But in doing this, you are assuming that you are not multiplying by zero, which would produce a non-equivalent equation; and in fact, when x = 1, x-1 = 0.x/(x-1) = 1/(x-1) It would appear that x=1 but that would give us an undefined quantity. Explain please.

So x = 1 is an

And the final answer is that there is no solution, since the only possible solution is not a solution.

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You can just use the quadratic formula to solve one equation for one variable, and substitute that into the other.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 = ?

I don't see a really nice way to directly isolate xy, but an alternative method I used is to add the two equations, and also subtract them as you did, and then use some factoring to combine them so that it eventually becomes possible to isolate either variable. In doing that, I was just trying things out without a specific goal, just to see what might happen.

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xy = 30 - y^2 = 20 - x^2Thank 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= ?

What can you tell us from there?

x(x+y) = 30

y(x+y) = 20

x+y = 30/x = 20/y

Anything strike you, yet?

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= ?

\(\displaystyle \text {Given: } x^2 + xy = 20 \text { and } y^2 + xy = 30.\)

\(\displaystyle x = 0 \implies x^2 + xy = 0 + 0 = 0 \ne 20 \implies \text { a contradiction.}\)

\(\displaystyle \therefore x \ne 0.\text { And so, } x^2 + xy = 20 \implies y = \dfrac{20 - x^2}{x}.\)

\(\displaystyle \therefore 30 = y^2 + xy = \left ( \dfrac{20 - x^2}{x} \right )^2 + x * \dfrac{20 - x^2}{x} \implies\)

\(\displaystyle 30 = \dfrac{400 - 40x^2 + x^4}{x^2} + 20 - x^2 = \dfrac{400 - 20x^2}{x^2} \implies\)

\(\displaystyle 30x^2 = 400 - 20x^2 \implies 50x^2 = 400 \implies x^2 = 8 = 4 * 2 \implies x = 2\sqrt{2} \text { or } -\ 2\sqrt{2}.\)

\(\displaystyle \therefore y = \dfrac{20 - 8}{2\sqrt{2}} = \dfrac{6}{\sqrt{2}} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2} \text { or}\)

\(\displaystyle y = -\ 3\sqrt{2}.\)

\(\displaystyle (2 \sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2}) = 4 * 2 + 6 * 2 = 8 + 12 = 20.\)

\(\displaystyle (3\sqrt{2})^2 + (2 \sqrt{2}) (3 \sqrt{2} = 9 * 2 + 6 * 2 = 18 + 12 = 30.\)

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.

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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.

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(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+ y = \(\displaystyle \pm\)√50 = \(\displaystyle \pm\)5√2 ................. (3)

Divide (1) by (2).............. y = 1.5 * x...........(4)

using (4) in (3)

2.5*x = \(\displaystyle \pm\)5√2

x = \(\displaystyle \pm\)2√2

y = (3/2) * (\(\displaystyle \pm\)2√2) = \(\displaystyle \pm\)(3)*√2 .................... [edited]

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Unfortunately, equations (1) and (2) are wrong!(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

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 ...

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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

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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.

Sitting in the corner for 12 hrs |

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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 ...

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

Would you please be so kind as to explain what is incorrect or incomplete in my prior post? Thank you.

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Maybe I'm not seeing something, butDr. 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 calculatedtwicemy 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.

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.

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, butthe 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.)

And it is interesting that the solution for xy is unique whereas the solutions for x and y are not.

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 theyAlso, 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.

Perhaps one reason for this approach is that I start off in word problems by identifying what is not known. This introduces systems of equations as soon as word problems are introduced. The substitutions arising from word problems are usually very simple.

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.

Thanks to all for your kind help. You have shown that there is more than one way to skin a cat.

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You can just use the quadratic formula to solve one equation for one variable, and substitute that into the other. ...

I'm not sure there's any difference of opinion. After all, I didNow 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."

What I like to encourage is,